Generation of the 2-Torsion Part of the Brauer Group of a Local Quintic by Quaternion Algebras

Genady Margolin  Ulf Rehmann  Vyacheslav Yanchevskii

1 Introduction

Let X be a smooth projective connected variety over field k of characteristic zero, and K = k(X) be its function field. It is known (see e.g. [6]) that there is a natural inclusion of Br X in Br(K) and this inclusion identifies Br X with unramified Brauer group Br_{nr}(K|k). Below we shall write Br X instead of Br_{nr}(K|k) keeping in mind this identification.

The group Br X is of great importance for many problems. It is enough to mention the problems of rationality of varieties, the problem of Brauer-Manin obstructions to various Hasse principles and, finally, the classical problem of the description of finite-dimensional division algebras over function fields. (see [6],[10], [11].)

In this paper we are interested in the problem of description of Br X for a smooth geometrically connected projective curves over local fields, more precisely, for finite extensions k of Q_{p}. The classical class field theory gives us a complete description of Brauer group Br(k) of k (and also of finite dimensional central division algebras over k). But if k(X) is a function field of smooth geometrically connected projective variety over k, the situation is much more complicated. In spite of many results obtained in connection with the developing of multidimensional class field theory ([8],[9]), unfortunately, we do not have a complete description as in the classical situation even in case of curves X (a ”complete” description means here a presentation of each element of Br(k(X)) by the corresponding central division algebra over k(X), in other words, it means a description of all central division algebras over k(X)). Traditions of local-to-global principle prescribe us before the investigation of a global situation to describe all local ones. In case of curves X over k this requires to investigate the groups Br(k_{v}(X_{v})) for all v, where k_{v} is a completion of k at absolute value v on k and X_{v} = X ×_{k} k_{v}. By Tsen’s theorem Br(k_{v}(X_{v})) is trivial in case of complex v and its structure is well-known for real v according to results in [1], [2], [3] and [7].

Thus, in this situation the case of non-Archimedean v is the main unknown one. In the following, by a local field we shall understand a finite extension of the field of p-adic numbers Q_{p}. Let now X be a smooth geometrically connected projective curve over a local field k. Then it follows from [4] that there exists the exact sequence

0 ⟶ Br X ⟶ Br(k(X)) ⟶^{ϕ} ∐_{P ∈X} χ(G_{P}) ⟶^{ψ} χ(G) ⟶ 0.

Here G denotes the Galois group of the separable closure of k, G_{P} = Gal((k_{P})_{s}|k_{P}), k_{P} is the residue field at P ∈ X and χ(G_{P}), χ(G) are the groups of continuous characters of G_{P} and G respectively (for definitions of ϕ and ψ see [4]). In view of this sequence the group Im ϕ is known, so the group Br X is of main interest for the description of Br(k(X)). The structure of Br X is well-known in case of curves of genus zero, but even in case of elliptic curves the complete structure of Br X is unknown. More generally, one can consider a smooth connected projective model X of an affine curve defined by the equation y^{2} = f(x), where f(x) is a polynomial with coefficients in k without multiple roots and one can ask for the structure of Br X. One of the aims of our paper is to answer some questions related to this. Since Br X is an abelian periodic group, the main problem of its description is to find the m-torsion part of it as an abstract abelian group for any m. Among the groups _{m}Br X the group _{2}Br X is of particular importance, because its elements are represented by central division algebras with involutions, and this group is, according to the result of Merkurjev [12], related to the theory of quadratic forms by the isomorphism _{2}Br(k(x)) ≅ I^{2}/I^{3}, where I is the fundamental ideal of Witt ring of k(X). Another important problem is to find an explicit presentation of all elements of _{m}Br X (or at least to find such presentation for generators of _{m}Br X) by central division algebras.

Of course, solutions of these two problems strongly depend on the curve X. Both problems have trivial solutions in case of curves of genus zero. For genus one curves one has more complicated situation. If X is an elliptic curves both these problems were solved completely in [13] for non-dyadic fields k. There are also a few preliminary results in case of dyadic fields and in case of principal homogeneous spaces which will be published elsewhere. For hyperelliptic curves only the case of non-dyadic local curves with good reduction is considered (see [14]).

Our main object of consideration is the group _{2}Br X for X being a smooth projective model of the affine curve defined by the equation y^{2} = f(x), where f(x) is a polynomial of the fifth degree with coefficients in k without multiple roots. The presentation of generators of _{2}Br X by quaternion algebras depends on the irreducible factors of f(x) over k. We will consider all possible decomposition cases of f(x) step by step. The main result is the complete list of quaternion algebras representing generators of the group _{2}Br X.

2 Splitting type (1,1,1,1,1).

Preliminary results. Let k be a local field of characteristic 0 with residue field k̄. Let π ∈ k denote a prime element, and let O_{k}, O_{k}^{∗}, πO_{k} be its valuation ring, the group of units and the maximal ideal of the valuation ring. For u ∈ O_{k}^{∗} let ū denote its class modulo πO_{k}. We will consider also a hyperelliptic curve C defined over k by the equation y^{2} = f(x), where f(x) ∈ O_{k}[x] is a monic polynomial without multiple roots and deg f = 5. Let us assume also that f(x) = ∏_{i=1}^{m} f_{i}(x), f_{i}(x) ∈ O_{k}[x] are monic irreducible polynomials. In what follows let α be a fixed unit of k which is not a square in k, ḡ(x) will denote the reduction of a polynomial g(x) ∈ O_{k}[x]. For a, b ∈ k^{∗} we write a ∼ b if a and b belong to the same class modulo (k^{∗})^{2}.

In our further considerations we will need the following lemmas.

Lemma 1

| _{2}Br C| = 2^{m},

particularly, in this section

| _{2}Br C| = 32.

For the proof see theorem 1 and §4 in [14].

Lemma 2 Let g(x) ∈ k[x] be a monic divisor of f(x), g(x) ∉ k and let either n be odd or deg g(x) be even. Then for any a ∈ k^{∗} the quaternion algebra

(\frac{a , g(x)}{k(C)})

is unramified and not isomorphic to the scalar algebra

(\frac{π , α}{k(C)}) .

The proof is similar to the proof of lemma 7 in [14].

Lemma 3 Let g(x) ∈ O_{k}[x] be a monic divisor of f(x). If ḡ ∉ k̄[x]^{2} and \overline{f(x)/g(x)} ∉ k̄[x]^{2}, then the quaternion algebra

(\frac{π , g(x)}{k(C)})

is non-trivial.

For the proof see [14], proposition 4.

Lemma 4 Let g(x) ∈ O_{k}[x] be a monic divisor of f(x). If the quaternion algebra

Ā = (\frac{ᾱ , ḡ(x)}{k̄(x, \sqrt{f̄(x)})})

is non-trivial one, then the algebra

A = (\frac{α , g(x)}{k(C)})

is also non-trivial.

Proof. If A ∼ 1, then k(x)(\sqrt{f(x)}) is a splitting field for the algebra

A_{0} = (\frac{α , g(x)}{k(x)}) ,

therefore this field is isomorphic to a maximal subfield of A_{0}. Then

f = ε_{1}^{2}α + ε_{2}^{2}g − ε_{3}^{2}αg, ε_{i} ∈ k(x).

If ε_{i} = p_{i}/q_{i}, p_{i} ∈ k[x], q_{i} ∈ k[x] let s_{i} = cont_{π}p_{i} − cont_{π}q_{i}. We have

f = π^{2s_{1}} \begin{matrix}p′_{1} & 2 \\ q′_{1} & 2 \end{matrix} α + π^{2s_{2}} \begin{matrix}p′_{2} & 2 \\ q′_{2} & 2 \end{matrix} g − π^{2s_{3}} \begin{matrix}p′_{3} & 2 \\ q′_{3} & 2 \end{matrix} αg,

where p′_{i} and q′_{i} are primitive polynomials. If h is the greatest common divisor of q_{1}, q_{2}, q_{3} and h is primitive polynomial, then

h^{2}f = π^{2s_{1}}μ_{1}^{2}α + π^{2s_{2}}μ_{2}^{2}g − π^{2s_{3}}μ_{3}^{2}αg (∗)

and all the polynomials μ_{i}, h are primitive. In particular, μ_{i}, h ∈ O_{k}[x]. By comparing the π-contents of both sides of the previous equality we obtain that all s_{i} cannot be positive. If all of them are equal to 0, then we have

f̄ = ε̄_{1}^{2}ᾱ + ε̄_{2}^{2}ḡ − ε̄_{3}^{2}ᾱḡ

and Ā ∼ 1. If s_{i_{0}} = min_{1≤i≤3}{s_{i}} < 0, then after comparing the π-contents of both sides of (*) we obtain that there exists j_{0} ≠ i_{0} such that s_{i_{0}} = s_{j_{0}}. If s_{1} = s_{2} < s_{3}, then after taking the reduction μ̄_{1}^{2}ᾱ + μ̄_{2}^{2}ḡ = 0. If s_{1} = s_{3} < s_{2}, then μ̄_{1}^{2}ᾱ − μ̄_{3}^{2}ᾱḡ = 0. If s_{2} = s_{3} < s_{1}, then μ̄_{2}^{2}ḡ − μ̄_{3}^{2}ᾱḡ = 0. Finally, let s_{1} = s_{2} = s_{3}, so μ̄_{1}^{2}ᾱ + μ̄_{2}^{2}ḡ − μ̄_{3}^{2}ᾱḡ = 0. In all these cases we have that there exists θ ∈ Ā such that Nrd(θ) = 0 and this is a contradiction with Ā ≁ 1.

Lemma 5 Let V be a variety defined over a finite field F_{q}. Then for any integral positive number N there exists an integral positive n_{0} such that for any n > n_{0} the subset V (F_{q^{n}}) of all F_{q^{n}}-rational points of V consists of more then N elements.

Proof is straightforward in view of the Lang-Weil theorem.

Lemma 6 Let K be an algebraically closed field and f(x), g(x) ∈ K[x]. Then the system

\{ \begin{matrix}y^{2} = f(x) \\ z^{2} = g(x) \end{matrix}

defines a variety V in P^{3}(K) if and only if f(x) ∉ K[x]^{2}, g(x) ∉ K[x]^{2} and f(x)g(x) ∉ K[x]^{2}.

Proof. It is sufficient to prove that the K-algebra K[V] = K[x, y, z]/(y^{2} − f(x), z^{2} − g(x)) has no zero divisors. Let us consider the 4-dimensional commutative algebra A over the field K(x) with the basis 1, i, j, k and with the multiplication rule i^{2} = f, j^{2} = g, ij = ji = k. The associativity of the multiplication of the basis elements is evident. Let us consider the map

ϕ : K[V] ⟶ A, ϕ([p(x, y, z)]) = p(x, i, j), p ∈ K[x, y, z].

Because of ϕ((y^{2} − f(x), z^{2} − g(x))) = 0, ϕ is correctly defined. Furthermore, ϕ is a ring homomorphism. Let us show that ϕ is an embedding. Indeed, let p ∈ K[x, y, z]. We have p(x, y, z) = p_{0}(x) + p_{1}(x)y + p_{2}(x)z + p_{3}(x)yz + q(x, y, z)(y^{2} − f(x)) + r(x, y, z)(z^{2} − g(x)). Then ϕ([p]) = p_{0} + p_{1}i + p_{2}j + p_{3}k. If ϕ([p]) = 0, then p_{0} = p_{1} = p_{2} = p_{3} = 0 and p ∈ (y^{2} − f(x), z^{2} − g(x)), so [p] = 0. Since K[V] ∈ A it is enough to prove that A has no zero divisors. Let I = {a ∈ A, ab = 0 for some b ∈ A, b ≠ 0} and a ∈ I, a ≠ 0. If a = a_{0} + a_{1}i + a_{2}j + a_{3}k, then

a(α + βk) = (αa_{0} + βa_{3}fg) + (αa_{1} + βa_{2}g)i + (αa_{2} + βa_{1}f)j + (αa_{3} + βa_{0})k ∈ I.
ai = a_{1}f + a_{0}i + a_{3}fj + a_{2}k ∈ I.
aj = a_{2}g + a_{3}gi + a_{0}j + a_{1}k ∈ I.
ak = a_{3}fg + a_{2}gi + a_{1}fj + a_{0}k ∈ I.

Since a ≠ 0, then we have that among of a_{l} there exists a non-zero coefficient. Then after replacement of a by ai, aj, or ak if it is necessary we can assume that a_{0} ≠ 0. If a_{3} ≠ 0 let us replace a by a′ = a(α + βk), where β = −αa_{3}/a_{0}. After the replacement we have a′_{3} = 0. If a′_{0} = 0, then a_{0}^{2} = a_{3}^{2}fg. Hence fg ∈ K[x]^{2} and we have a contradiction. Thus a′_{0} ≠ 0 and one can assume from the very beginning that a = 1 + a_{1}i + a_{2}j. If ab = 0, b ∈ I, then in view of the previous arguments we can suppose that b = 1 + x_{1}i + x_{2}j. Therefore,

(1 + a_{1}i + a_{2}j)(1 + x_{1}i + x_{2}j) = (1 + a_{1}x_{1}f + a_{2}x_{2}g) + (a_{1} + x_{1})i + (a_{2} + x_{2})j + (a_{2}x_{1} + a_{1}x_{2})k = 0,

so

\{ \begin{matrix}x_{1} = −a_{1}, \\ x_{2} = −a_{2}, \\ a_{1}a_{2} = 0, \\ a_{1}^{2}f + a_{2}^{2}g = 1. \end{matrix}

If a_{1} = 0, then g ∈ K[x]^{2} and if a_{2} = 0, then f ∈ K[x]^{2}. The lemma is proved.

The reduction theorem. From now on we will assume that f(x) ∈ k[x] completely splits over k. Let k(C) be the function field of C. If f is of odd degree, we may assume without loss of generality that f(x) ∈ O_{k}[x] and f(x) is monic. We also may assume that f(x) = x(x − π^{ε}u)g(x), where 0 ≤ ε ≤ 1, u ∈ O_{k}^{∗}, and g(x) ∈ O_{k}[x].

Proof: By a linear transformation we obtain that f(x) is divisible by x, hence

f(x) = x(x − π^{l}u) ∏_{i=1}^{n−2}(x − π^{k_{i}}u_{i}),

where 0 ≤ l ≤ k_{1} ≤ k_{n} and u, u_{1}, …, u_{n} ∈ O_{k}^{∗}. Let l = 2m + ε with ε ∈ {0, 1} and x = π^{2m}x′. Then

y^{2} = π^{10m}x′(x′ − π^{ε}u) ∏_{i=1}^{n−2}(x′ − π^{k_{i}−2m}u_{i}),

and the transformation y = y′π^{5m} will lead us to the equation of the desired type.

Let now f be of degree 5. The aim of this section is to prove that by means of some special transformation of variables we may restrict our consideration to some special forms of f(x). More precisely, we prove the

Theorem 1 Let f(x) be as above, such that, in addition f̄(x) has multiple roots over k̄ (bad reduction). Then f(x) can be transformed by an appropriate transformation x ↦ x + m, where m ∈ πO_{k} into one of the following forms.

I. f(x) = x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}),

where u, u_{i} ∈ O_{k}^{∗}, 1 ≤ k_{1} ≤ k_{2} ≤ k_{3}. If k_{i} = 1, then ū_{i} ≠ ū.

If k_{i} = k_{j}, i ≠ j, then ū_{i} ≠ ū_{j}.

I′_{∗}. f(x) = x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{1}}(u_{1} + π^{s}v))(x − π^{k_{3}}u_{3}),

where 1 < k_{1} < k_{3}, u, v, u_{1}, u_{3} ∈ O_{k}^{∗} and s > 0.

I′_{∗∗}. f(x) = x(x − πu)(x − π(u + π^{s}v))(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}),

where 1 ≤ k_{2} ≤ k_{3}, 1 < k_{3}, u, v, u_{2}, u_{3} ∈ O_{k}^{∗}.

If k_{2} = 1, then ū ≠ ū_{2}, if k_{2} = k_{3}, then ū_{2} ≠ ū_{3}, and s > 0.

II. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3})(x + e),

where 1 ≤ k_{1} ≤ k_{2} ≤ k_{3}, u_{i}, e ∈ O_{k}^{∗}, and if k_{i} = k_{j}, i ≠ j, then ū_{i} ≠ ū_{j}.

II′. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{1}}(u_{1} + π^{s}v))(x − π^{k_{3}}u_{3})(x + e),

where 1 ≤ k_{1} < k_{3}, u_{i}, v ∈ O_{k}^{∗}, s > 0.

III. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e)(x + e − π^{l}v),

where 1 ≤ k_{1} ≤ k_{2}, u_{i}, e ∈ O_{k}^{∗}, l > 0, and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}.

IV. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e_{1})(x + e_{2}),

where 1 ≤ k_{1} ≤ k_{2}, u_{i}, e_{i} ∈ O_{k}^{∗}, ē_{1} ≠ ē_{2}, and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}.

V. f(x) = x(x − π^{k_{1}}v_{1})(x + u)(x + u − π^{k_{2}}v_{2})(x + e),

where k_{1}, k_{2} ≥ 1, u, e, v_{i} ∈ O_{k}^{∗}, ū ≠ ē.

V I. f(x) = x(x − π^{k}u)(x + e_{1})(x + e_{2})(x + e_{3}),

where k ≥ 1, u, e_{i} ∈ O_{k}^{∗}, ē_{i} ≠ ē_{j} if i ≠ j.

The proof of the theorem is based on the following lemmas.

Lemma 7 Let f̄(x) = x^{5}. Then by an appropriate transformation x ↦ x + n, n ∈ πO_{k} the polynomial f(x) can be transformed to one of the forms I, I′_{∗}, I′_{∗∗}.

Proof. Assume that f(x) cannot be transformed to the form I. Then by an appropriate transformation it can be transformed to a polynomial of the form

x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}) (∗∗)

where u, u_{i} ∈ O_{k}^{∗}, 0 < k_{1} ≤ k_{2} ≤ k_{3} > 1, all elements ū_{i} such that k_{i} = k_{3} are pairwise unequal and in case k_{1} = k_{2} = 1 ū_{1} ≠ ū_{2}. Indeed, by above considerations

f(x) = x(x − πv)(x − π^{k_{1}}v_{1})(x − π^{k_{2}}v_{2})(x − π^{k_{3}}v_{3}),

where v, v_{i} ∈ O_{k}^{∗} and 0 < k_{1} ≤ k_{2} ≤ k_{3}. If k_{3} = 1, then the change of variables x ↦ x + πv_{3} transforms f(x) to the polynomial

x(x − π(v − v_{3}))(x − π(v_{1} − v_{3}))(x − π(v_{2} − v_{3}))(x − π(−v_{3})).

By assumption this polynomial is not of the form I, so at least one of the elements \overline{v − v_{3}}, \overline{v_{1} − v_{3}}, \overline{v_{2} − v_{3}} is zero and this means that for the transformed polynomial after renumeration the exponent k_{3} is greater than 1. Thus we may assume from the beginning that k_{3} > 1.

Let now k_{2} = k_{3}. Then the transformation x ↦ x + π^{k_{3}}v_{3} leads us to the polynomial

x(x − π(v − π^{k_{3}−1}v_{3}))(x − π^{k_{1}}(v_{1} − π^{k_{3}−k_{1}}v_{3}))(x − π^{k_{3}}(−v_{3}))(x − π^{k_{3}}(v_{2} − v_{3})).

If k_{1} ≠ k_{3}, then this polynomial is of the form (∗∗). If k_{1} = k_{3}, then we have the polynomial

x(x − π(v − π^{k_{3}−1}v_{3}))(x − π^{k_{3}}(v_{1} − v_{3}))(x − π^{k_{3}}(−v_{3}))(x − π^{k_{3}}(v_{2} − v_{3})).

If v̄_{1} − v̄_{3} ≠ v̄_{2} − v̄_{3}, then this polynomial is also of the form (∗∗). If v̄_{1} − v̄_{3} = v̄_{2} − v̄_{3} = 0 then we have either the above case k_{1} ≠ k_{2} = k_{3} or case k_{1} < k_{2} < k_{3} and in the last case our polynomial is again of the form (∗∗). Thus v̄_{1} − v̄_{3} = v̄_{2} − v̄_{3} ≠ 0. In this case it is enough to use the transformation x ↦ x + π^{k_{3}}(v_{1} − v_{3}). Hence in all possible cases f(x) can be transformed to a polynomial of the form (∗∗). Thus we have: in case k_{1} = k_{2} = 1 f(x) can be transformed to

x(x − πu)(x − π(u + π^{s}v))(x − πu_{2})(x − π^{k_{3}}u_{3}),  (1)

s > 0, k_{3} > 1, u, u_{i}, v ∈ O_{k}^{∗}, ū ≠ ū_{2}.

In case k_{2} > 1, k_{1} = 1 f(x) can be transformed to

x(x − πu)(x − π(u + π^{s}v))(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}),  (2)

(if k_{2} = k_{3}, ū_{2} ≠ ū_{3}, s > 0).

The remaining case is

x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{1}}(u_{1} + π^{s}v))(x − π^{k_{3}}u_{3}),  (3)

where 1 < k_{1} < k_{3}, s > 0. Combining together cases 1 and 2 we will obtain case of the form I′_{∗∗} and case 3 is the case of the form I′_{∗}.

Lemma 8 f̄(x) = x^{4}(x + ē), where ē ≠ 0. Then by an appropriate transformation x ↦ x + m, m ∈ πO_{k} f(x) can be transformed to one of forms II, II′.

Proof. Let

f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3})(x + e),

where u_{i}, e ∈ O_{k}^{∗} and 0 < k_{1} ≤ k_{2} ≤ k_{3}.

Firstly, let k_{3} = 1 and all ū_{i} be pairwise unequal. Then f(x) is of the form II. Otherwise there exist i ≠ j such that u_{i} ≠ u_{j} but ū_{i} = ū_{j}. Then the transformation x ↦ x + πu_{i} leads us to the case where k_{3} > 1. Let now k_{3} > 1. Then there are the following possibilities for k_{1}, k_{2}, k_{3}.

(i) k_{1} = k_{2} = 1,

(ii) 1 = k_{1} < k_{2},

(iii) 1 < k_{1} = k_{2} = k_{3},

(iv) 1 < k_{1} < k_{2},

(v) 1 < k_{1} = k_{2} < k_{3}.

Let us consider all possibilities step by step.

In case i) if ū_{1} ≠ ū_{2}, then f(x) is of the form II. Otherwise after the transformation x ↦ x + πu_{1} we will have a polynomial of the form II′.

In case ii) if k_{2} = k_{3} and ū_{2} = ū_{3}, then the transformation x ↦ x + π^{k_{3}}u_{3} leads us to a polynomial of the form II. In case ii) if k_{2} ≠ k_{3} or k_{2} = k_{3} but ū_{2} ≠ ū_{3}, then f(x) is of the form II.

In case iii) if u_{i} (i = 1, 2, 3) are pairwise unequal, then f(x) is of the form II. Otherwise, without loss of generality let us suppose ū_{1} = ū_{3}. Then the transformation x ↦ x + π^{k_{3}}u_{3} leads us to the polynomial

x(x − π^{k_{3}}(−u_{3}))(x − π^{k_{3}}(u_{1} − u_{3}))(x − π^{k_{3}}(u_{2} − u_{3}))(x + e + π^{k_{3}}u_{3}).

This polynomial either is of the form II or can be transformed to one of cases iv), v).

In case iv) if either k_{2} ≠ k_{3} or k_{2} = k_{3} but ū_{2} ≠ ū_{3} f(x) is of the form II. Otherwise, the transformation x ↦ x + π^{k_{3}}u_{3} leads us to case iv) where k_{2} ≠ k_{3}.

In case v) it is clear that f(x) is either of the form II or II′.

Lemma 9 By means of an appropriate transformation x ↦ x + m, m ∈ πO_{k} f(x) can be transformed to a polynomial of the form:

III. x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e)(x + e − π^{l}v),

where u_{i}, e ∈ O_{k}^{∗}, l > 0, 1 ≤ k_{1} ≤ k_{2}

and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}

(case f̄(x) = x^{3}(x + ē)^{2}, ē ≠ 0).

IV. x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e_{1})(x + e_{2}),

where u_{i}, e_{i} ∈ O_{k}^{∗}, ē_{1} ≠ ē_{2}, 1 ≤ k_{1} ≤ k_{2} and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}

(case f̄(x) = x^{3}(x + ē_{1})(x + ē_{2}), ē_{i} ≠ 0, ē_{1} ≠ ē_{2}).

V. x(x − π^{k_{1}}v_{1})(x + u)(x + u − π^{k_{2}}v_{2})(x + e),

where u, e, v_{i} ∈ O_{k}^{∗}, k_{1}, k_{2} ≥ 1, ū ≠ ē

(case f̄(x) = x^{2}(x + ū)^{2}(x + ē), ēū ≠ 0, ū ≠ ē).

V I. x(x − π^{k}u)(x + e_{1})(x + e_{2})(x + e_{3}),

where u, e_{i} ∈ O_{k}^{∗}, ē_{i} ≠ ē_{j} if i ≠ j

(case f̄(x) = x^{2}(x + ē_{1})(x + ē_{2})(x + ē_{3}), ē_{1}ē_{2}ē_{3} ≠ 0, ē_{i} ≠ ē_{j} if i ≠ j).

Proof. Let f̄(x) = x^{3}(x + ē)^{2}. If either k_{1} ≠ k_{2} or k_{1} = k_{2} but ū_{1} ≠ ū_{2}, then f(x) is of the form III. Otherwise the transformation x ↦ x + π^{k_{2}}u_{2} leads us to case k_{1} ≠ k_{2}, so we have transformed f(x) to the form III.

Let f̄(x) = x^{3}(x + ē_{1})(x + ē_{2}). Then similarly to the previous case either f(x) is of the form III or one may use the transformation x ↦ x + π^{k_{2}}u_{2}.

The two last splitting cases of f̄(x) are obvious. Now our theorem follows from lemmas 7–9 because all possible cases of splitting of f̄(x) are considered. It follows from the theorem that in the case of bad reduction of C given by equation y^{2} = f(x) to

describe quaternion generation of _{2}Br C we may restrict our attention to f(x) given by forms I–V I, I′_{∗}, I′_{∗∗}, II′. For the further considerations we will need the following list of subcases.

The cases of consideration.

I. f(x) = x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}),

where u, u_{i} ∈ O_{k}^{∗}, 1 ≤ k_{1} ≤ k_{2} ≤ k_{3}. If k_{i} = 1, then ū_{i} ≠ ū.

If k_{i} = k_{j}, i ≠ j, then ū_{i} ≠ ū_{j}.

I1. k_{1} ≡ k_{2}(2).

I2. I2′. k_{1} ≢ k_{2}(2), k_{1} ≡ k_{3}(2).

I2′′. k_{1} ≢ k_{2}(2), k_{1} ≢ k_{3}(2), k_{2} ≠ k_{3}, −uu_{1}u_{2} ∼ 1.

I3. k_{1} ≢ k_{2}(2), k_{1} ≢ k_{3}(2) and either k_{2} = k_{3} or −uu_{1}u_{2} ≁ 1.

I′_{∗}. f(x) = x(x − πu)(x − π^{k_{1}}u_{1})(x − π^{k_{1}}(u_{1} + π^{s}v))(x − π^{k_{3}}u_{3}),

where u, v, u_{i} ∈ O_{k}^{∗}, 1 < k_{1} < k_{3} and s > 0.

I′_{∗∗}. f(x) = x(x − πu)(x − π(u + π^{s}v))(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}),

where u, v, u_{i} ∈ O_{k}^{∗}, 1 ≤ k_{2} ≤ k_{3}, 1 < k_{3} and s > 0.

If k_{2} = 1, then ū ≠ ū_{2}, and if k_{2} = k_{3}, then ū_{2} ≠ ū_{3}.

I′_{∗∗}1. k_{2} ≡ 1(2).

I′_{∗∗}2. k_{2} ≡ 0(2) and either k_{3} ≡ 1(2) or (k_{3} ≡ 0(2) but k_{2} ≠ k_{3} and −u_{2} ∼ 1).

I′_{∗∗}3. k_{2} ≡ k_{3} ≡ 0(2) and either k_{2} = k_{3} or −u_{2} ≁ 1.

II. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3})(x + e),

where u_{i}, e ∈ O_{k}^{∗}, 1 ≤ k_{1} ≤ k_{2} ≤ k_{3}, if k_{i} = k_{j}, i ≠ j, then ū_{i} ≠ ū_{j}.

II1. e ∼ 1.

II1, 1. k_{1} ≢ k_{2}(2).

II1, 2. Either k_{1} ≡ k_{2}(2), k_{1} ≢ k_{3}(2) or k_{1} ≡ k_{2} ≡ k_{3}(2) but k_{2} ≠ k_{3} and u_{1}u_{2} ∼ 1.

II1, 3. k_{1} ≡ k_{2} ≡ k_{3}(2) and either k_{2} = k_{3} or u_{1}u_{2} ≁ 1.

II2. e ≁ 1.

II2, 1. k_{1} ≢ k_{2}(2).

II2, 2. k_{1} ≡ k_{2}(2), k_{1} ≢ k_{3}(2) or k_{1} ≡ k_{2} ≡ k_{3}(2) but k_{2} ≠ k_{3} and u_{1}u_{2} ≁ 1.

II2, 3. k_{1} ≡ k_{2} ≡ k_{3}(2) and either k_{2} = k_{3} or u_{1}u_{2} ∼ 1.

II′. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{1}}(u_{1} + π^{s}v))(x − π^{k_{3}}u_{3})(x + e), where u_{i}, e, v ∈ O_{k}^{∗}, 1 ≤ k_{1} < k_{3}, s > 0. II′1. e ∼ 1. II′2, 1. e ≁ 1, k_{1} ≢ k_{3}(2). II′2, 2. e ≁ 1, k_{1} ≡ k_{3}(2).

III. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e)(x + e − π^{l}v), where u_{i}, e, v ∈ O_{k}^{∗}, 1 ≤ k_{1} ≤ k_{2}, l > 0 and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}. III1. −e ≁ 1, k_{1} ≢ 0(2). III2. −e ∼ 1, k_{1} ≢ 0(2). III3. −e ≁ 1 and either (k_{1} ≡ 0(2), k_{2} ≢ 0(2)) or

(k_{1} ≡ k_{2} ≡ 0(2) and k_{1} < k_{2}, −u_{1} ∼ 1). III4. −e ∼ 1 and either (k_{1} ≡ 0(2), k_{2} ≢ 0(2)) or

(k_{1} ≡ k_{2} ≡ 0(2) and k_{1} < k_{2}, −u_{1} ∼ 1). III5. −e ≁ 1, k_{1} ≡ k_{2} ≡ 0(2) and either k_{1} = k_{2} or −u_{1} ≁ 1. III6. −e ∼ 1, k_{1} ≡ k_{2} ≡ 0(2) and either k_{1} = k_{2} or −u_{1} ≁ 1.

IV. f(x) = x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})(x + e_{1})(x + e_{2}), where u_{i}, e_{i} ∈ O_{k}^{∗}, ē_{1} ≠ ē_{2}, 1 ≤ k_{1} ≤ k_{2}, and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}. IV 1. k_{1} ≡ k_{2} ≡ 0(2) and either k_{1} = k_{2} or −e_{1}e_{2}u_{1} ≁ 1. IV 2. The other cases.

V. f(x) = x(x − π^{k_{1}}u_{1})(x + u)(x + u − π^{k_{2}}u_{2})(x + e), where u, e, v_{i} ∈ O_{k}^{∗}, ū ≠ ē, k_{1}, k_{2} > 0. V 1. e ≁ 1, e − u ≁ 1. V 2. e ≁ 1, e − u ∼ 1. V 3. e ∼ 1, e − u ≁ 1. V 4. e ∼ 1, e − u ∼ 1.

V I. f(x) = x(x − π^{k}u)(x + e_{1})(x + e_{2})(x + e_{3}), where u, e_{i} ∈ O_{k}^{∗}, ē_{i} ≠ ē_{j} if i ≠ j. V I1. e_{1}e_{2}e_{3} ≁ 1. V I2. e_{1}e_{2}e_{3} ∼ 1.

Presentation by quaternion algebras.

Theorem 2 Let C be given by the equation y^{2} = f(x), where f(x) is of form I. Then the class of (π, α) and the classes of the following quaternion algebras generate _{2}Br C.

I1. {(α, l_{i})}, where l_{i} is a monic linear divisor of f(x).

I2. (α, x), (α, x − πu), (α, x − π^{k_{1}}u_{1}), (π, x).

I3. (α, x), (α, x − πu), (π, x − π^{k_{3}}u_{3}), (π, x).

Proof. By lemma 1 the order of _{2}Br C is 32, so to prove the theorem it is enough to prove that in each case I1, I2, I3 the classes of the corresponding algebras generate the subgroup H of _{2}Br C of order 16 such that [(π, α)] ∉ H.

Observe firstly that in order to prove that some algebra is non-trivial it is enough to prove that some of its completions is non-trivial. We have

(α, x)_{x−πu} ∼ (α, πu) ≁ 1,
(α, x − π^{k_{i}}u_{i})_{x−πu} ∼ (α, πu − π^{k_{i}}u_{i}) ∼ (α, π) ≁ 1 (i = 1, 2, 3),
(α, x − πu)_{x} ∼ (α, πu) ≁ 1.

Let now {i, j, t, m, n} = {1, 2, 3, 4, 5}. Then (α, l_{i}l_{j}l_{t}l_{m}l_{n}) ∼ 1, so that (α, l_{i}l_{j}l_{t}l_{m}) ∼ (α, l_{n}) ≁ 1 and (α, l_{i}l_{j}l_{t}) ∼ (α, l_{m}l_{n}). It follows from the above consideration that to prove that the order of the group H is 16 it is enough to prove that all the algebras (α, l_{i}l_{j}) are non-trivial. If l_{i} = x − πu, then

(α, (x − πu)l_{j})_{x−πu} ∼ (α, l_{t}l_{m}l_{n})_{x−πu} ∼ (α, π^{3}) ≁ 1.

Let us consider the remaining cases, where l_{i}, l_{j} ≠ x − πu.

(α, x(x − π^{k_{1}}u_{1}))_{x−π^{k_{1}} u_{1}} ∼ (α, (x − πu)(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x−π^{k_{1}} u_{1}} ∼ (α, π^{2k_{1}+1}) ≁ 1.

(α, x(x − π^{k_{3}}u_{3}))_{x−π^{k_{3}} u_{3}} ∼ (α, π^{k_{1}+k_{2}+1}) ∼ (α, π) ≁ 1.

(α, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2}))_{x−π^{k_{1}} u_{1}} ∼ (α, π^{k_{1}}u_{1}(π^{k_{1}}u_{1} − πu)(π^{k_{1}}u_{1} − π^{k_{3}}u_{3})) ∼ (α, π) ≁ 1.
(α, (x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3}))_{x−π^{k_{1}} u_{1}} ∼ (α, π^{k_{1}}u_{1}(π^{k_{1}}u_{1} − πu)(π^{k_{1}}u_{1} − π^{k_{2}}u_{2})) ∼ (α, π) ≁ 1.

(α, x(x − π^{k_{2}}u_{2}))_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{1}+k_{2}+1}) ∼ (α, π) ≁ 1.

(α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}}u_{2}(π^{k_{2}}u_{2} − πu)(π^{k_{2}}u_{2} − π^{k_{1}}u_{1})) ∼ (α, π) ≁ 1.

To finish the consideration of case I1 one needs to observe that by the lemma 2 all the algebras from the list in case I1 are unramified and non-isomorphic to (π, α).

In case I2 let H be the subgroup of _{2}Br C generated by the classes of algebras (α, x), (α, x − πu), (α, x − π^{k_{1}}u_{1}), (π, x). All these algebras are unramified over k(C). Let us prove that H is of order 16. Just by the same way as in the previous case one can prove that the algebras (α, x − πu), (α, x(x − π^{k_{1}}u_{1})), (α, x), (α, x−π^{k_{1}}u_{1}), (α, x(x−π^{k_{1}}u_{1})), (α, x(x−πu)), (α, (x−πu)(x−π^{k_{1}}u_{1})) are non-trivial. To prove that [(α, x)], [(α, x − πu)], [(α, x − π^{k_{1}}u_{1})] generate a group of order 8 one needs only to prove that (α, x(x − πu)(x − π^{k_{1}}u_{1})) ∼ (α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3})) ≁ 1.

In case I2′ we have (α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x−π^{k_{3}} u_{3}} ∼ (α, π^{k_{1}+k_{3}+1}) ≁ 1.

In case I2′′ k_{2}+1 < k_{3} and (α, (x−π^{k_{2}}u_{2})(x−π^{k_{3}}u_{3}))_{x−π^{k_{2}+1}} ∼ (α, (π^{k_{2}+1}−π^{k_{2}}u_{2})(π^{k_{2}+1}−π^{k_{3}}u_{3})) ∼ (α, π) ≁ 1 since f(π^{k_{2}+1}) ∼ −π^{k_{1}+k_{2}+1}uu_{1}u_{2} ∼ −uu_{1}u_{2} ∼ 1.

To prove that (π, x) ≁ 1 let us consider a curve (−1)^{k_{2}} ᾱūū_{1}x^{2} − ūū_{1}ū_{2} = y^{2} defined over the residue field k̄ of k. This curve has a smooth point (ã, b̃) such that ã, b̃ ∈ k̄(ã), where k̄(ã) is an extension of k̄ of some prime odd degree. Then for the unramified extension N over k with residue field k̄(ã) there is a lift (a, b) ∈ N^{2} of (ã, b̃) to a point of a curve (−1)^{k_{2}}αuu_{1}x^{2} − uu_{1}u_{2} = y^{2}. Set w = (−1)^{k_{2}}αa^{2} and θ = π^{k_{2}}w. It is easy to see that f(θ) ∼ (−1)^{k_{2}}αuu_{1}a^{2} − uu_{1}u_{2} ∼ 1. Then it follows that

(\frac{π , x}{k(x, \sqrt{f(x)})})_{x−θ} ∼ (\frac{π , θ}{N(\sqrt{f(θ)})⟨x − θ⟩}) ∼ (\frac{π , (−1)^{k_{2}}w}{N⟨x − θ⟩} ) ∼ (\frac{π , α}{N⟨x − θ⟩}) ≁ 1.

Hence (π, x) ≁ 1. Since the algebra

(\frac{π , x}{k(\sqrt{α})(x, \sqrt{f(x)})})

belongs again to case I2 (with constant field k(\sqrt{α}), then k(\sqrt{α})(\sqrt{f(x)}) does not split the algebra (π, x) but it splits all the algebras (α, x), (α, x−πu), (α, x−π^{k_{1}}u_{1}), so it follows that elements [(α, x)], [(α, x− πu)], [(α, x − π^{k_{1}}u_{1})] and [(π, x)] generate a group H of order 16. In view of lemma 2 [(π, α)] ∉ H. This finishes the consideration of case I2.

In case I3 one can prove similar to case I2 that [(α, x)], [(α, x − πu)] generate a group of order 4. As to the non-triviality of (π, x) one can prove this just in the same way as in the previous case if k_{2} < k_{3}. If k_{2} = k_{3} consider the curve over k̄ defined by the equation

(−1)^{k_{2}} ᾱūū_{1} (x^{2} − \frac{(−1)^{k_{2}} ū_{2}}{ᾱ} ) (x^{2} − \frac{(−1)^{k_{2}} ū_{3}}{ᾱ} ) = y^{2}.

Then there exists an extension k̄(ã)|k̄ of odd prime degree with a smooth point (ã, b̃) such that b̃ ∈ k̄(ã). Let (a, b) ∈ N^{2} be a lifted point on the curve defined by the equation

(−1)^{k_{2}}αuu_{1} (x^{2} − \frac{(−1)^{k_{2}}u_{2}}{α} ) (x^{2} − \frac{(−1)^{k_{2}}u_{3}}{α} ) = y^{2},

where N|k is the unramified extension with residue field k̄(ã), w = (−1)^{k_{2}}αa^{2}. Set θ = π^{k_{2}}w. Then f(θ) ∼ w(w − u_{2})(w − u_{3})uu_{1} ∼ 1 and consequently

(\frac{π , x}{k(x, \sqrt{f(x)})})_{x−θ} ∼ (\frac{π , θ}{N(\sqrt{f(θ)})⟨x − θ⟩}) ∼ (\frac{π , α}{N⟨x − θ⟩}) ≁ 1.

Thus (π, x) ≁ 1.

Let N be an extension of k as above for a curve defined by the equation uu_{1}((−1)^{k_{2}}αx^{2} − u_{2}) = y^{2}. Then such a curve has a point (a, b) over N. In case k_{2} < k_{3} for θ = π^{k_{2}}(−1)^{k_{2}}αa^{2} we have f(θ) ∼ 1 and

(\frac{π , x − π^{k_{3}}u_{3}}{k(x, \sqrt{f(x)})} )_{x−θ} ∼ (\frac{π , θ}{N(\sqrt{f(θ)})⟨x − θ⟩}) ∼ (\frac{π , α}{N⟨x − θ⟩}) ≁ 1.

In case k_{2} = k_{3} one need to consider instead of the latter curve, a curve defined by the equation

y^{2} = (−1)^{k_{2}}αuu_{1}((−1)^{k_{2}}αx^{2} − u_{2})((−1)^{k_{2}}αx^{2} − u_{3})

and proceed by analogy as in case k_{2} < k_{3}.

One can prove in a similar way that (π, x(x − π^{k_{3}}u_{3})) ≁ 1. If k_{2} < k_{3} consider a curve y^{2} = α(x^{2} + u_{2}/(uu_{1}))(x^{2} + (u_{2} − u_{3})/(uu_{1})) and in case k_{2} = k_{3} a curve y^{2} = αx^{2} − u_{3} and proceed by analogy as in the previous cases.

To finish the consideration of case I3 let us prove that groups ⟨[(α, x)], [(α, x − πu)]⟩ and ⟨[(π, x)], [(π, x − π^{k_{3}}u_{3})]⟩ have trivial intersection.

Since the extension of k by \sqrt{α} leaves us either in case I2 or in case I3, the algebra (π, x) does not split by k(\sqrt{α})(x, \sqrt{f(x)}) and consequently [(π, x)] ∉ ⟨[(α, x)], [(α, x − πu)]⟩. The transformation x ↦ x + π^{k_{3}}u_{3} leaves us again either in case I2 or in case I3, so that [(π, x − π^{k_{3}}u_{3})] ∉ ⟨[(α, x)], [(α, x − πu)]⟩. Let us consider the algebra (π, x(x − π^{k_{3}}u_{3})). If k_{2} = k_{3}, then the extension of constants by \sqrt{α} shows us that [(π, x(x − π^{k_{3}}u_{3}))] ∉ ⟨[(α, x)], [(α, x − πu)]⟩. Let now k_{2} ≠ k_{3}. Comparing completions of (π, x(x − π^{k_{3}}u_{3})) and (α, x − πu):

(π, x(x − π^{k_{3}}u_{3}))_{x−π^{k_{2}} u_{2}} ∼ (π, π^{k_{2}}u_{2}(π^{k_{2}}u_{2} − π^{k_{3}}u_{3})) ∼ 1,
(α, x − πu)_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}}u_{2} − πu) ∼ (α, π) ≁ 1,

so one can see that (π, x(x − π^{k_{3}}u_{3})) and (α, x − πu) are non-isomorphic. Comparing completions of (π, x(x−π^{k_{3}}u_{3})) and (α, x) in x−πu one can see that these algebras are also non-isomorphic. And finally we have for i ∈ {1, 2}, k_{i} ≡ 0(2) that

(α, x(x − πu))_{x−π^{k_{i}} u_{i}} ∼ (α, π^{k_{i}+1}) ≁ 1 and (π, x(x − π^{k_{3}}u_{3}))_{x−π^{k_{i}} u_{i}} ∼ (π, (π^{k_{i}}u_{i})^{2}) ∼ 1.

Hence (α, x(x − πu)) ≁ (π, x(x − π^{k_{3}}u_{3})). This completes the proof.

In cases I′_{∗} and I′_{∗∗} we have

Theorem 3 Let C be given by the equation y^{2} = f(x), where f(x) is of form I′. Then the class of (π, α) and the classes of the following quaternion algebras generate _{2}Br C.

I′_{∗}. {(α, l_{i})}, where l_{i} is a monic linear divisor of f.

I′_{∗∗}1. {(α, l_{i})}, where l_{i} is a monic linear divisor of f.

I′_{∗∗}2. (α, x), (α, x − πu), (α, x − πu′), (π, x), u′ = u + π^{s}v.

I′_{∗∗}3. (α, x), (α, x − πu), (π, x − π^{k_{3}}u_{3}), (π, x).

Proof. Similarly to case I one can prove that (α, l_{i}) ≁ 1 for a monic linear divisor l_{i} of f(x). In case I′_{∗} if l_{i} = x − πu and i ≠ j we have similarly to case I (α, l_{i}l_{j}) ≁ 1. Furthermore,

(α, x(x − π^{k_{1}}u_{1}))_{x} ∼ (α, π^{k_{1}+k_{3}+1}), (α, x(x − π^{k_{1}}u_{1}))_{x−π^{k_{3}} u_{3}} ∼ (α, π^{k_{1}+k_{3}}).

One of the algebras (α, π^{k_{1}+k_{3}}) and (α, π^{k_{1}+k_{3}+1}) is non-trivial, therefore (α, x(x−π^{k_{1}}u_{1})) ≁ 1. Similarly one can prove that (α, x(x − π^{k_{1}}(u_{1} + π^{s}v))) ≁ 1. We have also

(α, x(x − π^{k_{3}}u_{3}))_{x} ∼ (α, π^{2k_{1}+1}) ≁ 1

and

(α, (x − π^{k_{1}}u_{1})(x − π^{k_{1}}u′_{1}))_{x−π^{k_{1}} u_{1}} ∼ (α, x(x − πu)(x − π^{k_{3}}u_{3}))_{x−π^{k_{1}} u_{1}} ≁ 1.

For the algebra (α, (x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3})) we have

(α, (x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3}))_{x−π^{k_{1}} u_{1}} ∼ (α, x(x − π^{k_{1}}u′_{1})(x − πu))_{x−π^{k_{1}} u_{1}} ∼ (α, π^{s+1}),
(α, (x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3}))_{x−π^{k_{1}} u′_{1}} ∼ (α, π^{2k_{1}+s}) ∼ (α, π^{s}).

One of the algebras (α, π^{s}), (α, π^{s+1}) is non-trivial, so that (α, (x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3})) ≁ 1.

In a similar way one proves that (α, (x − π^{k_{1}}u′_{1})(x − π^{k_{3}}u_{3})) ≁ 1. Now by analogy to case I we complete the proof.

Let us consider case I′_{∗∗}. We have

(α, (x − πu)(x − πu′))_{x−πu} ∼ (α, x(x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x−πu} ∼ (α, π^{3}) ≁ 1.

We have also (α, x(x − πu)) ≁ 1, since

(α, x(x − πu))_{x−πu} ∼ (α, π^{s+1}) and (α, x(x − πu))_{x−πu′} ∼ (α, π^{s}).

Similarly (α, x(x − πu′)) ≁ 1.

Consider now the algebra (α, (x − πu)(x − π^{k_{2}}u_{2})). We have (α, (x − πu)(x − π^{k_{2}}u_{2})) ≁ 1, since

(α, (x − πu)(x − π^{k_{2}}u_{2}))_{x−πu} ∼ (α, π^{s+1}) and (α, (x − πu)(x − π^{k_{2}}u_{2}))_{x−πu′} ∼ (α, π^{s}).

Similarly (α, (x − πu′)(x − π^{k_{2}}u_{2})) ≁ 1.

Furthermore, (α, (x − πu)(x − π^{k_{3}}u_{3})) ≁ 1, since

(α, (x − πu)(x − π^{k_{3}}u_{3}))_{x−πu} ∼ (α, π^{s+1}) and (α, (x − πu)(x − π^{k_{3}}u_{3}))_{x−πu′} ∼ (α, π^{s}).

Similarly (α, (x − πu′)(x − π^{k_{3}}u_{3})) ≁ 1.

We have also (α, x(x − π^{k_{2}}u_{2})) ≁ 1, since

(α, x(x − π^{k_{2}}u_{2}))_{x−π^{k_{2}} u_{2}} ∼ (α, (x − πu)(x − πu′)(x − π^{k_{3}}u_{3}))_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}})

and k_{2} is odd.

Then

(α, x(x − π^{k_{3}}u_{3}))_{x} ∼ (α, (x − πu)(x − πu′)(x − π^{k_{2}}u_{2}))_{x} ∼ (α, π^{k_{2}})

and k_{2} is odd.

Finally, (α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3})) ≁ 1, since

(α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x−π^{k_{2}} u_{2}} ∼ (α, x(x − πu)(x − πu′))_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}}) ≁ 1.

It is clear from the above considerations that similarly to case I one can complete the proof in case I′_{∗∗}.

Let us consider the next case. We can conclude that the group ⟨[(α, x)], [(α, x − πu)], [(α, x − πu′)]⟩ is of order 8 in case I′_{∗∗}2 and ⟨[(α, x)], [(α, x − πu)]⟩ is of order 4 in case I′_{∗∗}3. The non-triviality of the algebra (π, x) in both cases can be established by considerations similar to the ones in the proof of the theorem 2. As to the algebra (π, x − π^{k_{3}}u_{3}) in case I′_{∗∗}3 one can use the replacement x′ = x − π^{k_{3}}u_{3} and reduces this case to the case of the algebra (π, x). The non-triviality of the algebra (π, x(x − π^{k_{3}}u_{3})) in case I′_{∗∗}3 can be obtained by using arguments similar to the ones in case I. To complete the proof in case I′_{∗∗}2 let us observe that the extension of constants by \sqrt{α} does not split (π, x) but splits any algebra of the form (α, g(x)).

To complete the proof of the theorem in case I′_{∗∗}3 we need to prove that the groups ⟨[(α, x)], [(α, x − πu)]⟩ and ⟨[(π, x)], [(π, x − π^{k_{3}}u_{3})]⟩ have a trivial intersection. But in case I′_{∗∗}3, k_{2} = k_{3} the extension of constants by \sqrt{α} leaves us in case I′_{∗∗}3 and this completes the proof in case k_{2} = k_{3}. If we are in case I′_{∗∗}3, k_{2} ≠ k_{3} then the similar extension of constants leads us to case I′_{∗∗}2 so that [(π, x)] ∉ ⟨[(α, x)], [(α, x − πu)]⟩. After the replacement x′ = x − π^{k_{3}}u_{3} we are left again in the case I′_{∗∗}3, then [(π, x−π^{k_{3}}u_{3})] ∉ ⟨[(α, x)], [(α, x−πu)]⟩. The proof of the fact that [(π, x(x−π^{k_{3}}u_{3}))] ∉ ⟨[(α, x)], [(α, x− πu)]⟩ is just the same as the corresponding part of the proof of the theorem 2. This completes the proof of the theorem.

In case II we have

Theorem 4 Let C be given by the equation y^{2} = f(x), where f(x) is of form II. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

II1, 1. (α, x), (α, x − π^{k_{1}}u_{1}), (α, x − π^{k_{2}}u_{2}), (π, x).

II1, 2. (α, x), (α, x − π^{k_{1}}u_{1}), (π, x − π^{k_{1}}u_{1}), (π, x).

II1, 3. (α, x), (π, x − π^{k_{1}}u_{1}), (π, x − π^{k_{2}}u_{2}), (π, x).

II2, 1. (α, x), (α, x − π^{k_{3}}u_{3}), (π, x + e), (π, x).

II2, 2. (α, x), (π, x − π^{k_{1}}u_{1}), (π, x + e), (π, x).

II2, 3. (π, x − π^{k_{1}}u_{1}), (π, x − π^{k_{2}}u_{2}), (π, x − π^{k_{3}}u_{3}), (π, x).

Proof. Observe first of all that in case II (α, x + e) ∼ 1 for any e ∈ O_{k}^{∗} and (π, x + e) ≁ 1 if and only if e ∈ O_{k}^{∗} \ (O_{k}^{∗})^{2}. Indeed, one needs only to prove that for any irreducible polynomial p(x) ∈ k[x] we have (α, x + e)_{p(x)} ∼ 1, and for any irreducible p(x) we have (π, x + e)_{p(x)} ∼ 1 if e ∈ (O_{k}^{∗})^{2} but there exists p(x) such that (π, x + e)_{p(x)} ≁ 1 if e ∉ (O_{k}^{∗})^{2}.

Assume firstly that θ is a root of p(x). Then

(α, x + e)_{p(x)} ∼ (α, θ + e) ∼ (α, θ(θ − π^{k_{1}}u_{1})(θ − π^{k_{2}}u_{2})(θ − π^{k_{3}}u_{3}))

and if θ ∉ πO_{k}, then (α, x + e)_{p(x)} ∼ (α, θ^{4}) ∼ 1. Let θ ∈ πO_{k}, then

(α, x + e)_{p(x)} ∼ (α, θ + e) ∼ (α, e) ∼ 1.

Assume that e ∈ (O_{k}^{∗})^{2}, then similar arguments show that (π, x + e) ∼ 1. In case e ∉ (O_{k}^{∗})^{2} we have (π, x + e)_{x} ∼ (π, e) ≁ 1.

Consider case II1. Let l_{0} = x, l_{j} = x − π^{k_{j}}u_{j}, j = 1, 2, 3. Since f̄(x)/l̄_{j}(x), l̄_{j}(x) ∉ k̄[x]^{2}, j = 0, 1, 2, 3, then by lemma 3 (π, l_{j}) ≁ 1. Furthermore,

(\frac{ᾱ , l̄_{i}(x)}{k̄(x)(\sqrt{f̄(x)})}) = (\frac{ᾱ , x}{k̄(x)(\sqrt{x + ē})}) .

Since in case II1 e ∼ 1 we have

(\frac{ᾱ , x}{k̄(x)(\sqrt{x + ē})})_{x} ∼ (\frac{ᾱ , x}{k̄⟨x⟩}) ≁ 1.

Hence

(\frac{ᾱ , x}{k̄(x)(\sqrt{x + ē})}) ≁ 1

and in view of lemma 4 (α, l_{i}) ≁ 1, (i = 0, 1, 2, 3). Thus all the algebras listed in cases II1, 1, II1, 2, II1, 3 are non-trivial. Since (α, x + e) ∼ 1, then (α, l_{0}l_{1}l_{2}) ∼ (α, l_{3}) ≁ 1 and hence in order to prove that ⟨[(α, l_{0})], [(α, l_{1})], [(α, l_{2})]⟩ is a group of order 8 it is enough to prove that (α, l_{0}l_{1}) ≁ 1, (α, l_{0}l_{2}) ≁ 1, (α, l_{1}l_{2}) ≁ 1. We have

(α, l_{0}l_{1})_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}}π^{k_{1}}) ∼ (α, π) ≁ 1,
(α, l_{0}l_{2})_{x−π^{k_{2}} u_{2}} ∼ (α, l_{1}l_{3})_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{1}}π^{k_{2}}) ≁ 1, (α, l_{1}l_{2})_{x} ∼ (α, π^{k_{1}}π^{k_{2}}) ≁ 1.

It is clear that the extension of constants by \sqrt{α} leaves us in case II1 and this implies that [(π, x)] ∉ ⟨[(α, l_{0})], [(α, l_{1})], [(α, l_{2})]⟩, so that case II1, 1 is considered.

In case II1, 2 let us show firstly that (α, x(x − π^{k_{1}}u_{1})) ≁ 1 and (π, x(x − π^{k_{1}}u_{1})) ≁ 1. If k_{2} ≢ k_{3}(2) we have

(α, x(x − π^{k_{1}}u_{1}))_{x} ∼ (α, (x − π^{k_{2}}u_{2})(x − π^{k_{3}}u_{3}))_{x} ∼ (α, π^{k_{2}+k_{3}}) ≁ 1.

If k_{2} ≡ k_{3}(2), then

f(π^{k_{2}+1}) ∼ π^{k_{2}+1}(−π^{k_{1}}u_{1})(−π^{k_{2}}u_{2})π^{k_{2}+1} ∼ u_{1}u_{2} ∼ 1

and this implies

(α, x(x − π^{k_{1}}u_{1}))_{x−π^{k_{2}+1}} ∼ (α, π^{k_{1}+k_{2}+1}) ≁ 1,

so that (α, x(x − π^{k_{1}}u_{1})) ≁ 1.

To prove (π, x(x − π^{k_{1}}u_{1})) ≁ 1 let us find a polynomial p(x) such that (π, x(x − π^{k_{1}}u_{1}))_{p} ≁ 1. Let p(x) be a minimal polynomial of element θ = π^{k_{3}}w, where w is a unit from the algebraic closure of k. Then

f(θ) ∼ π^{k_{3}}w(−π^{k_{1}}u_{1})(−π^{k_{2}}u_{2})π^{k_{3}}(w − u_{3}) ∼ u_{1}u_{2}w(w − u_{3}).

Assume now that k(w)|k is an unramified prime odd degree extension (or its degree is 1). Then

(π, x(x − π^{k_{1}}u_{1}))_{p(x)} ∼ (\frac{π , (−1)^{k_{1}+k_{3}+1}u_{1}w}{k(w)⟨x − θ⟩\sqrt{u_{1}u_{2}w(w − u_{3})}}) .

And if u_{1}u_{2}w(w − u_{3}) ∼ 1 and (−1)^{k_{1}+k_{3}+1}u_{1}w ∼ α then (π, x(x − π^{k_{1}}u_{1}))_{p(x)} ≁ 1. To find such w let us consider a curve over k̄ given by the equation y^{2} = ū_{1}ū_{2}x^{2} + (−1)^{k_{1}+k_{3}} ū_{2}ū_{3}ᾱ.

There exists a finite odd degree extension of k̄, where our curve has a smooth point (ã, b̃) which can be lifted to a point (a, b) on the curve y^{2} = u_{1}u_{2}x^{2} + (−1)^{k_{1}+k_{3}}u_{2}u_{3}α. And we will be done if put w = (−1)^{k_{1}+k_{3}+1}u_{1}αa^{2}. It follows immediately from the above considerations that ⟨[(α, x)], [(α, x − π^{k_{1}}u_{1})]⟩ and ⟨[(π, x)], [(π, x − π^{k_{1}}u_{1})]⟩ are groups of order 4. Since the extension of constants by \sqrt{α} leaves us in case II1, 2 these two groups have the trivial intersection and this completes the consideration of case II1, 2.

In case II1, 3 in order to prove that ⟨[(π, x)], [(π, x − π^{k_{1}}u_{1})] [(π, x − π^{k_{2}}u_{2})]⟩ is a group of order 8 in view of (π, x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})) ∼ (π, x − π^{k_{3}}u_{3}) ≁ 1 it is enough to establish that the algebras (π, x(x − π^{k_{1}}u_{1})), (π, x(x − π^{k_{2}}u_{2})), (π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})) are non-trivial. If k_{2} < k_{3}, then the non-triviality of the first algebra can be checked similar to the previous case. Let now k_{2} = k_{3} and assume that w is a unit from the algebraic closure of k with the above mentioned properties, θ = π^{k_{3}}w, and p(x) is the minimal polynomial of θ. Then we have f(θ) ∼ w(π^{k_{3}−k_{1}}w − u_{1})(w − u_{2})(w − u_{3}) and

(π, x(x − π^{k_{1}}u_{1}))_{p(x)} ∼ (\frac{π , w(wπ^{k_{3}−k_{1}} − u_{1})}{k(w)⟨x − θ⟩(\sqrt{w(π^{k_{3}−k_{1}}w − u_{1})(w − u_{2})(w − u_{3})})})

and if

\{ \begin{matrix}w(π^{k_{3}−k_{1}}w − u_{1}) ∼ α \\ (w − u_{2})(w − u_{3}) ∼ α, \end{matrix}

then (π, x(x − π^{k_{1}}u_{1}))_{p(x)} ≁ 1.

Assume that k_{1} < k_{3}, then one can take w = −αu_{1}a^{2}, where (a, b) is a point on a curve y^{2} = α(x^{2} + u_{2}/(αu_{1}))(x^{2} + u_{3}/(αu_{1})).

If k_{1} = k_{2} = k_{3}, then we have a system

\{ \begin{matrix}w(w − u_{1}) ∼ α \\ (w − u_{2})(w − u_{3}) ∼ α \end{matrix}

The existence of w in this case can be established by inspection of points on the curve defined by the following system in view of lemmas 5 and 6.

\{ \begin{matrix}y^{2} = ᾱx(x − ū_{1}) \\ z^{2} = ᾱ(x − ū_{2})(w − ū_{3}). \end{matrix}

If k_{2} ≠ k_{3}, then the non-triviality of the algebra (π, x(x − π^{k_{2}}u_{2})) can be checked by analogy as in the previous cases.

Let k_{2} = k_{3} and k_{1} < k_{2}. If as above θ = wπ^{k_{3}}, then we have f(θ) ∼ π^{k_{3}}w(−π^{k_{1}}u_{1})π^{k_{3}}(w−u_{2})π^{k_{3}}(w− u_{3}) ∼ −wu_{1}(w − u_{2})(w − u_{3}) and for minimal polynomial p(x) of θ over k

(π, x(x − π^{k_{2}}u_{2}))_{p(x)} ∼ (\frac{π , w(w − u_{2})}{k(w)⟨p⟩(\sqrt{f(θ)})}) .

The non-triviality of the algebra (π, x(x − π^{k_{2}}u_{2}))_{p(x)} will be follow from the existence of solution of the system

\{ \begin{matrix}w(w − u_{2}) ∼ α \\ −u_{1}(w − u_{3}) ∼ α \end{matrix}

and k(w)|k is of odd prime degree or its degree is 1. The existence of such w as usual can be deduced by the inspection of points of the reduced curve.

Let us show that (π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})) ≁ 1. If k_{2} ≠ k_{3}, then

(π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2}))_{x} ∼ (π, π^{k_{1}+k_{2}}u_{1}u_{2}) ∼ (π, u_{1}u_{2}) ≁ 1.

Let k_{2} = k_{3} and k_{1} < k_{3} (if k_{1} = k_{2}, then one can prove everything as in one of the previous cases since (π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})) ∼ (π, x(x − π^{k_{3}}u_{3}))).

By analogy with the above considerations one can find a unit w such that k(w)|k is of odd prime degree and such that

\{ \begin{matrix}w(w − u_{3}) ∼ α \\ −u_{1}(w − u_{2}) ∼ α. \end{matrix}

Then for the minimal polynomial p(x) of θ = π^{k_{3}}w we have (π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2}))_{p(x)} ≁ 1, so that (π, (x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2})) is also non-trivial.

Thus ⟨[(π, x)], [(π, x−π^{k_{1}}u_{1})], [(π, x−π^{k_{2}}u_{2})]⟩ is a group of order 8. Since the extension of constants by \sqrt{π} leaves us in case II1, 3, we conclude that [(α, x)] does not belong to this group and the proof of case II1 is finished.

In case II2 the algebras (π, x), (π, x − π^{k_{i}}u_{i}), (π, x(x + e)) and (π, (x + e)(x − π^{k_{i}}u_{i})) are non-trivial by the lemma 3.

Let us consider case II2, 1. There exists i ∈ {1, 2} such that k_{i} is odd. Then we have

(α, x)_{x−π^{k_{i}} u_{i}} ≁ 1, (α, x − π^{k_{3}}u_{3})_{x−π^{k_{i}} u_{i}} ≁ 1,
(α, x(x − π^{k_{3}}u_{3}))_{x} ∼ (α, (x + e)(x − π^{k_{1}}u_{1})(x − π^{k_{3}}u_{3}))_{x} ∼ (π, π^{k_{1}+k_{2}}) ≁ 1,

so that ⟨[(α, x)], [(α, x − π^{k_{3}}u_{3})]⟩ is of order 4. From the above considerations it follows that ⟨[(π, x)], [(π, x + e)]⟩ is also of order 4. After the extension of constants to k(\sqrt{α}) we have case II1, 1, so that

(π, x) ⊗ k(\sqrt{α})(x, \sqrt{f(x)}) ≁ 1 and
(π, x(x + e)) ⊗ k(\sqrt{α})(x, \sqrt{f(x)}) ∼ (π, x) ⊗ k(\sqrt{α})(x, \sqrt{f(x)}) ≁ 1

and therefore [(π, x)], [(π, x(x + e))] ∉ ⟨[(α, x)], [(α, x − π^{k_{3}}u_{3})]⟩. To complete the inspection of case II2, 1 we need only to prove that [(π, x + e)] ∉ ⟨[(α, x)], [(α, x − π^{k_{3}}u_{3})]⟩.

Let k_{j} is even, j ∈ {1, 2}. Then

(π, x + e)_{x−π^{k_{j}} u_{j}} ∼ (π, e) ≁ 1, (α, x)_{x−π^{k_{j}} u_{j}} ∼ (α, π^{k_{j}}u_{j}) ∼ 1,
(α, x − π^{k_{3}}u_{3})_{x−π^{k_{j}} u_{j}} ∼ 1, (α, x(x − π^{k_{3}}u_{3}))_{x−π^{k_{j}} u_{j}} ∼ 1,

and case II2, 1 is considered.

In case II2, 1 in order to see that the group ⟨[(π, x − π^{k_{1}}u_{1})], [(π, x + e)], [(π, x)]⟩ has order 8 it is enough to prove that (π, x(x − π^{k_{1}}u_{1})(x + e)) ≁ 1 and (π, x(x − π^{k_{1}}u_{1})) ≁ 1. This follows from the above considerations. But the extension of constants by \sqrt{α} leads us to case II1, 2 and we saw that in this case (π, x + e) ∼ 1, (π, x(x − π^{k_{1}}u_{1})) ≁ 1 and therefore (π, x(x − π^{k_{1}}u_{1})(x + e)) ≁ 1 over k(C).

Let us prove now that (α, x) ≁ 1. If k_{2} ≢ k_{3}(2) and k_{i} (i ∈ {2, 3}) is odd, then (α, x)_{x−π^{k_{i}} u_{i}} ∼ (α, π^{k_{i}}) ≁ 1 therefore without loss of generality we can assume k_{1}, k_{2}, k_{3} to be even (k_{2} < k_{3}). Then

f(π^{k_{2}+1}) ∼ π^{k_{2}+1}(−π^{k_{1}}u_{1})(−π^{k_{2}}u_{2})π^{k_{2}+1}e ∼ u_{1}u_{2}e ∼ 1 and

(α, x)_{x−π^{k_{2}+1}} ∼ (\frac{α , π^{k_{2}+1}}{k⟨x − π^{k_{2}+1}⟩(\sqrt{f(π^{k_{2}+1})})}) ∼ (\frac{α , π}{k⟨x − π^{k_{2}+1}⟩}) ≁ 1.

To complete the consideration of case II2, 2 we need to prove that [(α, x)] ∉ ⟨[(π, x − π^{k_{1}}u_{1})], [(π, x + e)], [(π, x)]⟩.

First of all (α, x) ≁ (π, x) because the extension of constants by \sqrt{α} leads us to case II1, 2, where (π, x) ≁ 1. Furthermore, the same extension of constants shows that (α, x) ≁ (π, x(x + e)) since in case II1, 2 we have (π, x + e) ∼ 1. By the same way we have

(α, x) ≁ (π, x − π^{k_{1}}u_{1}), (α, x) ≁ (π, (x − π^{k_{1}}u_{1})(x + e)),
(α, x) ≁ (π, x(x − π^{k_{1}}u_{1})) and (α, x) ≁ (π, x(x + e)(x − π^{k_{1}}u_{1})).

Consider the algebra (π, x + e). If among of k_{1}, k_{2}, k_{3} there exists an even k_{i}, then (α, x)_{x−π^{k_{i}} u_{i}} ∼ 1, but (π, x + e)_{x−π^{k_{i}} u_{i}} ∼ (π, e) ≁ 1. Let now all k_{i} be odd. Then in view of k_{2} + 1 < k_{3} we have f(π^{k_{2}+1}) ∼ u_{1}u_{2}e ∼ 1 and (α, x)_{x−π^{k_{2}+1}} ∼ (α, π^{k_{2}+1}) ∼ 1, (π, x + e)_{x−π^{k_{2}+1}} ∼ (π, e) ≁ 1 and again (α, x) ≁ (π, x + e). This completes the consideration of case II2, 2.

Consider case II2, 3. Let l_{0} = x, l_{j} = x − π^{k_{j}}u_{j} (j = 1, 2, 3). Let us observe that all algebras listed in II2, 3 are non-trivial because of lemma 3, so to prove that the group generated by them is of order 16 it is enough to check that all of them are pairwise non-isomorphic.

First of all it is true for (π, l_{i}(x + e)) because of lemma 3. For the algebra (π, l_{0}l_{3}) if k_{2} ≠ k_{3} we have (π, l_{0}l_{3})_{x} ∼ (π, u_{1}u_{2}e) ≁ 1. If k_{2} = k_{3}, k_{1} < k_{2} one can find a unit w such that k(w)|k is of odd prime degree and

\{ \begin{matrix}w(w − u_{3}) ∼ α \\ −u_{1}e(w − u_{2}) ∼ α. \end{matrix}

Then for the minimal polynomial p(x) of π^{k_{3}}w over k one has (π, l_{0}l_{3})_{p(x)} ≁ 1. In case k_{1} = k_{2} = k_{3} one can use a unit w such that

\{ \begin{matrix}w(w − u_{3}) ∼ α \\ (w − u_{1})(w − u_{2}) ∼ α. \end{matrix}

Then again (π, l_{0}l_{3})_{p(x)} ≁ 1. Similar arguments prove that (π, l_{0}l_{1}) and (π, l_{0}l_{2}) are non-trivial in case k_{2} < k_{3}. Indeed, in first case one can use a unit w such that

\{ \begin{matrix}w ∼ −u_{1}α \\ w − u_{3} ∼ −u_{2} \end{matrix}

and in the second one such a unit that

\{ \begin{matrix}w ∼ −u_{2}α \\ w − u_{3} ∼ −u_{1}. \end{matrix}

Then in both cases for the minimal polynomial p(x) of π^{k_{3}}w over k one has (π, l_{0}l_{1})_{p(x)} ≁ 1 and (π, l_{0}l_{2})_{p(x)} ≁ 1. Let now k_{2} = k_{3}. Then in the case of the algebra (π, l_{0}l_{1}) we need to use a unit w,

\{ \begin{matrix}w ∼ −αu_{1} \\ (w − u_{2})(w − u_{3}) ∼ 1, \end{matrix}

if k_{1} < k_{2} and

\{ \begin{matrix}w(w − u_{1}) ∼ α \\ (w − u_{2})(w − u_{3}) ∼ 1, \end{matrix}

if k_{1} = k_{2} = k_{3}. In the case of algebra (π, l_{0}l_{2}) one can work similarly.

Consider algebras (π, l_{1}l_{3}), (π, l_{2}l_{3}), (π, l_{1}l_{2}). After replacing x′ = x − π^{k_{3}}u_{3} we are again in case II2, 3 and with new notations algebras (π, l_{1}l_{3}), (π, l_{2}l_{3}) look like algebras (π, l′_{0}l′_{1}), (π, l′_{0}l′_{2}), where l′_{0} = x′ and l′_{i} = l_{i} + π^{k_{3}}u_{3}. But the last algebras are non-trivial in view of the above arguments. In case k_{2} = k_{3} for the algebra (π, l_{1}l_{2}) one can renumber l_{1}, l_{2}, l_{3} as follows: l′_{2} = l_{3}, l′_{3} = l_{2}, l′_{1} = l_{1}, so that (π, l_{1}l_{2}) = (π, l′_{1}l′_{3}) and the last algebra is non-trivial. Finally, let k_{1} < k_{3}. Then

(π, l_{1}l_{2})_{x−π^{k_{1}} u_{1}} ∼ (π, l_{0}l_{3}(x + e))_{x−π^{k_{1}} u_{1}} ∼ (π, e) ≁ 1.

The theorem is proved.

If f(x) is of the form II′ we have the following statement.

Theorem 5 Let C be given by the equation y^{2} = f(x), where f(x) is of form II′. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

II′1. (α, x), (α, x − π^{k_{1}}u_{1}), (π, x − π^{k_{1}}u_{1}), (π, x).

II′2, 1. (α, x), (π, x − π^{k_{1}}u_{1}), (π, x), (π, x + e).

II′2, 2. (π, x − π^{k_{1}}u_{1}), (π, x − π^{k_{1}}u′_{1}), (π, x + e), (π, x), u′_{1} = u_{1} + π^{s}v.

Proof. As in the proof of theorem 4 one can easy prove that (α, x + e) ≁ 1 for any e ∈ O_{k}^{∗} and (π, x + e) ≁ 1 if and only if e ∈ O_{k}^{∗} \ (O_{k}^{∗})^{2}. Similarly to this proof one can check that the algebras (π, x), (π, x − π^{k_{1}}u_{1}), (π, x − π^{k_{1}}u′_{1}), (π, x − π^{k_{3}}u_{3}) are non-trivial.

If e ∼ 1, then one has as in the proof of theorem 4 that the algebras (α, x − π^{k_{1}}u_{1}), (α, x) are non-trivial. Hence all algebras (except (α, x) in case II′2, 1) listed in case II′ are non-trivial.

Consider case II′1. First of all observe that (α, l_{0}l_{1}) ≁ 1. Indeed, if k_{1} ≢ k_{3}(2), then

(α, l_{0}l_{1})_{x} ∼ (α, (x − π^{k_{1}}u′_{1})l_{3}(x + e))_{x} ∼ (α, π^{k_{1}+k_{3}}) ≁ 1.

If k_{1} ≡ k_{3}(2), then since f(π^{k_{1}+1}) ∼ e ∼ 1 we have

(α, l_{0}l_{1})_{x−π^{k_{1}+1}} ∼ (\frac{α , π}{k⟨x − π^{k_{1}+1}⟩}) ≁ 1.

As to the algebra (π, l_{0}l_{1}) one can find a unit w generating an odd degree extension over k such that

\{ \begin{matrix}w ∼ (−)^{s}αu_{1} \\ w − v ∼ (−1)^{s}αu_{1}. \end{matrix}

Let θ = π^{k_{1}}(u_{1} + π^{s}w), p = Irr_{k(w)|k}(θ). Then

f(θ) ∼ π^{k_{1}}u_{1}(π^{k_{1}}u_{1} + π^{k_{1}+s}w − π^{k_{1}}u_{1})(π^{k_{1}}u_{1} + π^{k_{1}+s}w − π^{k_{1}}u_{1} − π^{k_{1}+s}v) ∼ w(w − v) ∼ 1

and

(π, l_{0}l_{1})_{p} ∼ (π, π^{s}wu_{1}) ∼ (π, (−1)^{s}u_{1}w) ∼ (π, α) ≁ 1.

Thus groups ⟨[(α, l_{0})], [(α, l_{1})]⟩ and ⟨[(π, l_{0})], [(π, l_{1})]⟩ are of order 4. And since the extension of constants by \sqrt{α} leaves us in case II′1 we have that these two groups have trivial intersection, so case II′1 is considered.

Now we are in case II′2. The extension of k by \sqrt{α} leads us to case II′1 and splits the algebra (π, x + e), hence the group ⟨[(π, l_{0})], [(π, l_{1})], [(π, x + e)]⟩ is of order 8.

Now if i ∈ {1, 3} and k_{i} is odd, then (α, x)_{x−π^{k_{i}} u_{i}} ∼ (α, π^{k_{i}}) ≁ 1 and therefore (α, x) ≁ 1. To complete the consideration of case II′2, 1 we need only to prove that [(α, x)] ∉ ⟨[(π, l_{0})], [(π, l_{1})], [(π, x + e)]⟩. The extension of constants by \sqrt{α} shows us that [(α, x)] ∉ ⟨[(π, l_{0})], [(π, l_{1})]⟩ and (α, x) ≁ (π, l_{0}(x + e)), (α, x) ≁ (π, l_{1}(x + e)) and (α, x) ≁ (π, l_{0}l_{1}(x + e)). Furthermore, for even k_{i} ∈ {k_{2}, k_{3}} (α, x)_{l_{i}} ∼ 1 and (π, x + e)_{l_{i}} ≁ 1, so (α, x) ≁ (π, x + e). Case II′2, 1 is done.

Consider case II′2.2. In this case it is enough to prove that all listed algebras with the algebra (π, x − π^{k_{3}}u_{3}) are pairwise non-isomorphic. Furthermore, (π, l_{i}(x + e)) ≁ 1 and (π, l′_{1}(x + e)) ≁ 1 as was noted before. We also have

(π, l_{0}l_{3})_{x} ∼ (π, l_{1}l′_{1}(x + e))_{x} ∼ (π, e) ≁ 1,

so that (π, l_{0}l_{3}) ≁ 1. One can also prove similarly to the previous case that (π, l_{0}l_{1}) ≁ 1 and (π, l_{0}l′_{1}) ≁ 1. To prove that (π, l_{1}l_{3}) ≁ 1 (similarly (π, l′_{1}l_{3}) ≁ 1) it is enough to observe that after replacement x′ = x − π^{k_{3}}u_{3} our algebra looks like (π, l̃_{0}l̃_{1}), where l̃_{0} = x′ and l̃_{1} = l_{1} + π^{k_{3}}u_{3}. Finally,

(π, l_{1}l′_{1})_{l_{1}} ∼ (π, l_{0}l_{3}(x + e))_{l_{1}} ∼ (π, e) ≁ 1.

The theorem is proved.

In case where f(x) is of the form III we have

Theorem 6 Let C be given by the equation y^{2} = f(x), where f(x) is of form III. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

III1. (α, x), (α, x − π^{k_{2}}u_{2}), (π, x + e), (π, x + e − π^{l}v).

III2. (α, x), (α, x − π^{k_{2}}u_{2}), (π, x + e), (α, x + e).

III3. (α, x), (π, x), (π, x + e), (π, x + e − π^{l}v).

III4. (α, x), (π, x), (π, x + e), (α, x + e).

III5. (π, x), (π, x − π^{k_{2}}u_{2}), (π, x + e), (π, x + e − π^{l}v).

III6. (π, x), (π, x − π^{k_{2}}u_{2}), (π, x + e), (α, x + e).

Proof. Let G_{1} = ⟨[(π, x+e)], [(π, x+e′)]⟩, where e′ = e−π^{l}v. Let us prove that in cases III1, III3, III5 G_{1} is of order 4. Indeed, in these cases (π, x + e) ≁ 1 and (π, x + e′) ≁ 1 according to lemma 3. Furthermore,

(π, (x + e)(x + e′))_{x+e} ∼ (π, x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2}))_{x+e} ∼ (π, −e) ≁ 1.

Now let G_{2} = ⟨[(π, x + e)], [(α, x + e)]⟩, then in cases III2, III4, III6 G_{2} is also of order 4. Indeed, (α, x+e) ≁ 1 by lemma 4 and (π, x+e) ≁ 1 by lemma 3. And (α, x+e) ≁ (π, x+e) since k(\sqrt{α})(x, \sqrt{f(x)}) does not split (π, x + e).

In cases III1, III2 let G_{3} = ⟨[(α, x)], [(α, x − π^{k_{2}}u_{2})]⟩. Then G_{3} is of order 4. Indeed,

(α, x)_{x−π^{k_{1}} u_{1}} ∼ (α, π^{k_{1}}) ≁ 1, (α, x − π^{k_{2}}u_{2})_{x−π^{k_{1}} u_{1}} ∼ (α, π) ≁ 1,
(α, x(x − π^{k_{2}}u_{2}))_{x} ∼ (α, (x − π^{k_{1}}u_{1})(x + e)(x + e′))_{x} ∼ (α, −π^{k_{1}}u_{1}ee′) ≁ 1.

In cases III3, III4 let G_{4} = ⟨[(α, x)], [(π, x)]⟩ and let us show that G_{4} is of order 4. For the algebra (α, x) we have: if k_{2} is odd, then

(α, x)_{x−π^{k_{2}} u_{2}} ∼ (α, π^{k_{2}}u_{2}) ≁ 1.

If k_{2} is even, then

f(π^{k_{1}+1}) ∼ π^{k_{1}+1}(−π^{k_{1}}u_{1})π^{k_{1}+1} ∼ −u_{1} ∼ 1

and

(α, x)_{x−π^{k_{1}+1}} ∼ (\frac{α , π}{k⟨x − π^{k_{1}+1}⟩}) ≁ 1.

As for the algebra (π, x), one can find a unit w such that k(w)|k is an unramified odd prime degree extension and

\{ \begin{matrix}w ∼ α \\ w − u_{1} ∼ 1. \end{matrix}

In view of f(π^{k_{1}}w) ∼ w − u_{1} we have

(π, x)_{p(x)} ∼ (\frac{π , π^{k_{1}}w}{k(w)⟨p⟩} ) ≁ 1,

where p(x) is a minimal polynomial of π^{k_{1}}w over k. Now since k(\sqrt{α})(x, \sqrt{f(x)}) does not split (π, x), we have (π, x) ≁ (α, x).

Finally, let G_{5} = ⟨[(π, x)], [(π, x − π^{k_{2}}u_{2})]⟩ in cases III5 and III6. We prove that G_{5} is of order 4. To prove that (π, x) is non-trivial it is enough to repeat previous arguments with the last w. Similarly, with the same w in case k_{1} < k_{2} we have

(π, x − π^{k_{2}}u_{2})_{x−π^{k_{1}} w} ∼ (π, x)_{x−π^{k_{1}} w} ≁ 1.

If k_{1} = k_{2} let w satisfies the following property

\{ \begin{matrix}w − u_{2} ∼ α \\ w(w − u_{1}) ∼ α. \end{matrix}

Then (π, x − π^{k_{2}}u_{2})_{x−π^{k_{1}} w} ≁ 1. For the algebra (π, x(x − π^{k_{2}}u_{2})) in case k_{1} ≠ k_{2} we have

(π, x(x − π^{k_{2}}u_{2}))_{x} ∼ (π, (x − π^{k_{1}}u_{1})(x + e)(x + e))_{x} ∼ (π, −u_{1}) ≁ 1.

If k_{1} = k_{2} one can find a unit w with the property

\{ \begin{matrix}w(w − u_{2}) ∼ α \\ w − u_{1} ∼ α. \end{matrix}

Then (π, x(x − π^{k_{2}}u_{2}))_{x−π^{k_{1}} w} ≁ 1. Thus G_{5} is of order 4.

To complete the proof of the theorem it remains to show that all the following groups are trivial 1. G_{1} ∩ G_{3} in case III1,

2. G_{2} ∩ G_{3} in case III2,

3. G_{1} ∩ G_{4} in case III3,

4. G_{2} ∩ G_{4} in case III4,

5. G_{1} ∩ G_{5} in case III5,

6. G_{2} ∩ G_{5} in case III6.

In case III6 the extension of constants by \sqrt{π} leaves us in this case, so that [(α, x)] ∉ G_{5} and [(α, x) ⊗ (π, x + e)] ∉ G_{5}. Let g(x) ∈ {x, x − π^{k_{2}}u_{2}, x(x − π^{k_{2}}u_{2}). Then (π, g(x)(x + e)) ≁ 1 in view of ḡ(x)(x + ē), f̄(x)/(ḡ(x)(x + ē)) ∉ k̄[x]^{2} and by lemma 3. This implies G_{2} ∩ G_{5} is trivial.

In case III4 since the extension of constants by \sqrt{α} leaves us in case III4, then in order to prove that G_{2} ∩ G_{4} is trivial it is enough to check that (α, x(x + e)) ≁ 1 and (π, x(x + e)) ≁ 1. But (α, x(x + e)) ≁ 1, since (ᾱ, x(x + ē)) ≁ 1 and (π, x(x + e)) ≁ 1, since x(x + ē), f̄/(x(x + ē)) ∉ k̄[x]^{2}.

In case III2 since the extension of constants by \sqrt{α} leaves us in this case, we have [(π, x+e)], [(πα, x+ e)] ∉ G_{3}. And it remains to show that [(α, x + e)] ∉ G_{3}. But for any g ∈ {x, x − π^{k_{2}}u_{2}, x(x − π^{k_{2}}u_{2})} (α, g(x + e)) ≁ 1 since (ᾱ, ḡ(x + ē)) ≁ 1.

In case III1 the extension of constants by \sqrt{α} leads us to case III2, since [(π, x+e)] ∉ G_{3}. The same is true for [(π, x + e′)]. Now it remains to prove that [(π, (x + e)(x + e′))] ∉ G_{3}. We have

(α, x)_{x−π^{k_{1}} u_{1}} ∼ (α, x − π^{k_{2}}u_{2})_{x−π^{k_{1}} u_{1}} ≁ 1 but (π, (x + e)(x + e′))_{x−π^{k_{1}} u_{1}} ∼ (π, ee′) ∼ 1.

Furthermore,

(α, x(x − π^{k_{2}}u_{2}))_{x} ∼ (α, (x − π^{k_{1}}u_{1})(x + e)(x + e′))_{x} ∼ (α, −π^{k_{1}}u_{1}ee′) ≁ 1

and (π, (x + e)(x + e′))_{x} ∼ (π, ee′) ∼ 1. Thus [(π, (x + e)(x + e′))] ∉ G_{3}.

In case III3 the extension of constants by \sqrt{α} leads us to case III4. So this extension does not split the algebras (π, x + e), (π, x + e′). Let us show that (α, x) ≁ (π, (x + e)(x + e′)). In case k_{2} is odd we have

(π, (x + e)(x + e′))_{x−π^{k_{2}} u_{2}} ∼ 1 and (α, x)_{x−π^{k_{2}} u_{2}} ≁ 1.

In case k_{2} is even

(π, (x + e)(x + e′))_{x−π^{k_{1}+1}} ∼ 1 and (α, x)_{x−π^{k_{1}+1}} ≁ 1.

Thus [(α, x)] ∉ G_{1}. We have

(π, x) ≁ (π, x + e) and (π, x) ≁ (π, x + e′),

since x(x + ē), f̄(x)/(x(x + ē)) ∉ k̄[x]^{2}. Let w be a unit such that (π, x)_{x−π^{k_{1}} w} ≁ 1, then (π, (x + e)(x + e′))_{x−π^{k_{1}} w} ∼ 1. This implies [(π, x)] ∉ G_{1}. The remaining possibility is [(πα, x)] ∈ G_{1}. Observe firstly that (πα, x)_{x−π^{k_{1}} w} ≁ 1, hence (πα, x) ≁ (π, (x + e)(x + e′)). Furthermore, if (πα, x) ∼ (π, x + e), then (α, x) ∼ (π, x(x + e)). But k(\sqrt{α})(x, \sqrt{f(x)}) splits (α, x) and does not split (π, x(x + e)). Similarly (πα, x) ≁ (π, x + e′), so G_{1} ∩ G_{4} is trivial.

Now pass to remaining case III5. If g ∈ {x, x − π^{k_{2}}u_{2}, x(x − π^{k_{2}}u_{2})}, then (π, g(x + e) ≁ 1 by lemma 3. Thus [(π, x + e)] ∉ G_{5}. Similarly [(π, x + e′)] ∉ G_{5}. Let (π, g) ∼ (π, (x + e)(x + e′)). On the other hand, if w is a unit such that (π, x)_{x−π^{k_{1}} w} ≁ 1, then since (π, (x + e)(x + e′))_{x−π^{k_{1}} w} ∼ 1 we have g ≠ x. In the same way one can prove that g ≠ x − π^{k_{2}}u_{2}. Finally, if g = x(x − π^{k_{2}}u_{2}), then in case k_{1} = k_{2} for a unit w such that (π, x(x − π^{k_{2}}u_{2}))_{x−π^{k_{1}} w} ≁ 1 we have (π, (x + e)(x + e′))_{x−π^{k_{1}} w} ∼ 1 and in case k_{1} ≠ k_{2} (π, (x + e)(x + e′))_{x} ∼ 1, (π, x(x − π^{k_{2}}u_{2}))_{x} ≁ 1. Thus [(π, (x + e)(x + e′))] ∉ G_{4}. The theorem is proved.

In case where f(x) is of the form IV we have the following statement.

Theorem 7 Let C be given by the equation y^{2} = f(x), where f(x) is of form IV. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

IV 1. (π, x), (π, x − π^{k_{2}}u_{2}), (π, x + e_{1}), (π, x + e_{2}).

IV 2. (π, x), (π, x + e_{1}), (π, x + e_{2}), (α, x).

Proof. Observe that G = ⟨[(π, x)], [(π, x + e_{1})], [(π, x + e_{2})], ⟩ is of order 8. First of all x, x + ē_{1}, x + ē_{2}, x(x+ē_{1}), x(x+ē_{2}), (x+ē_{1})(x+ē_{2}), f̄(x)/x, f̄(x)/(x+ē_{1}), f̄(x)/(x+ē_{2}), f̄(x)/(x(x+ē_{1})), f̄(x)/(x(x+ ē_{2})), f̄(x)/((x + ē_{1})(x + ē_{2})) ∉ k̄[x]^{2} and by lemma 3 (π, x) ≁ 1, (π, x + e_{1}) ≁ 1, (π, x + e_{2}) ≁ 1, (π, x(x + e_{1})) ≁ 1, (π, x(x + e_{2})) ≁ 1, (π, (x + e_{1})(x + e_{2})) ≁ 1, so G is of order 8.

If one of the k_{i} is odd, we have (α, x)_{x−π^{k_{i}} u_{i}} ≁ 1. If k_{1} ≡ k_{2} ≡ 0(2), k_{1} < k_{2} and −u_{1}e_{1}e_{2} ∈ (k^{∗})^{2}, then k_{1} < k_{1} + 1 < k_{2} and (α, x)_{x−π^{k_{1}+1}} ≁ 1, since f(π^{k_{1}+1}) ∼ −u_{1}e_{1}e_{2}. Thus in case IV 2 (α, x) ≁ 1. Since the extension of constants by \sqrt{α} leaves us in case IV, we conclude that [(α, x)] ∉ G.

Consider case IV 1. If k_{1} ≠ k_{2}, we have

(π, x(x − π^{k_{2}}u_{2}))_{x} ∼ (π, (x − π^{k_{1}}u_{1})(x + e_{1})(x + e_{2}))_{x} ∼ (π, −u_{1}e_{1}e_{2}) ≁ 1.

If k_{1} = k_{2} let w be a unit such that k(w)|k is an unramified prime degree extension and

\{ \begin{matrix}w(w − u_{2}) ∼ α \\ w − u_{1} ∼ αe_{1}e_{2}. \end{matrix}

Then (π, x(x − π^{k_{2}}u_{2}))_{x−π^{k_{2}} w} ≁ 1. This implies (π, x) ≁ (π, x − π^{k_{2}}u_{2}). By lemma 3 the algebra (π, x − π^{k_{2}}u_{2}) is also non-isomorphic to (π, x + e_{1}), (π, x + e_{2}), (π, x(x + e_{2})), (π, x(x + e_{2})), (π, x(x + e_{1})(x + e_{2})). Finally, let k_{1} ≠ k_{2}. Then

(π, (x − π^{k_{2}}u_{2})(x + e_{1})(x + e_{2}))_{x−π^{k_{1}} w} ∼ (π, x(x − π^{k_{1}}u_{1}))_{x−π^{k_{1}} w} ≁ 1

if w is a unit with the above properties and

\{ \begin{matrix}w ∼ αe_{1}e_{2} \\ w − u_{1} ∼ e_{1}e_{2}. \end{matrix}

In case k_{1} = k_{2} we only need to change the latter conditions by

\{ \begin{matrix}w(w − u_{1}) ∼ α \\ w − u_{2} ∼ αe_{1}e_{2}. \end{matrix}

In any case (π, x − π^{k_{2}}u_{2}) ≁ (π, (x + e_{1})(x + e_{2})). This implies [(π, x − π^{k_{2}}u_{2})] ∉ G and completes the proof of the theorem.

If f(x) is of the form V we have

Theorem 8 Let C be given by the equation y^{2} = f(x), where f(x) is of form V. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

V 1. (π, x), (π, x − π^{k_{1}}v_{1}), (π, x + u), (π, x + u − π^{k_{2}}v_{2}).

V 2. (π, x), (π, x − π^{k_{1}}v_{1}), (π, x + u), (α, x + u).

V 3. (π, x), (α, x), (π, x + u), (π, x + u − π^{k_{2}}v_{2}).

V 4. (π, x), (α, x), (π, x + u), (α, x + u).

Proof. Let l_{1}, l_{2}, l_{3}, l_{4}, l_{5} be the different linear monic divisors of f(x). If l_{i} ≠ x+e we have by lemma 3 in view of l̄_{i}, f̄(x)/l̄_{i} ∉ k̄[x]^{2} that (π, l_{i}) ≁ 1. By the same reason (π, l_{i}l_{j}) ≁ 1 if l̄_{i} ≠ l̄_{j}. Let now e ≁ 1, then (π, x(x−π^{k_{1}}v_{1}))_{x+αv_{1}π^{k_{1}+2}} ≁ 1, since f(−αv_{1}π^{k_{1}+2}) ∼ αe ∼ 1. Thus if e ≁ 1, then (π, x(x−π^{k_{1}}v_{1})) ≁ 1. The change of variables x′ = x + u leads us to the equation

y^{2} = x′(x′ − π^{k_{2}}v_{2})(x′ − u)(x′ − u − π^{k_{1}}v_{1})(x′ + (e − u))

and we have (π, (x+u)(x+u−π^{k_{2}}v_{2})) = (π, x′(x′ −π^{k_{2}}v_{2})), so by the previous arguments (π, (x+u)(x+ u − π^{k_{2}}v_{2})) ≁ 1. Observe also that if e ∼ 1, then

(\frac{ᾱ , x}{k̄(x, \sqrt{f̄(x)})})_{x} ∼ (\frac{ᾱ , x}{k̄⟨x⟩}) ≁ 1.

It follows by lemma 4 that in case e ∼ 1 (α, x) ≁ 1. Similarly if e − u ∼ 1, then (α, x + u) ≁ 1.

Now, to complete the consideration of case V I we need only to prove that (π, l_{i}l_{j}l_{r}) ≁ 1, where l_{i}, l_{j}, l_{r} are not equal to each other and to x + e and (π, x(x − π^{k_{1}}v_{1})(x + u)(x + u − π^{k_{2}}v_{2})) ≁ 1 if e ≁ 1. But the last statement is equivalent to (π, x + e) ≁ 1 which is valid in view of (π, x + e)_{x} ≁ 1. As to the previous one, we have l̄_{i}l̄_{j}l̄_{r}, f̄(x)/(l̄_{i}l̄_{j}l̄_{r}) ∉ k̄[x]^{2}, so lemma 3 works.

In case V 2 it is enough to observe that by above arguments the group ⟨[(π, x)], [(π, x−π^{k_{1}}v_{1})], [(π, x+ u)]⟩ is of order 8 and the field k(\sqrt{π})(x, \sqrt{f(x)}) does not split the algebra (α, x + u).

Case V 3 can be considered by analogy with case V 2.

We saw that the group ⟨[(π, x)], [(π, x + u)]⟩ is of order 4. To prove that ⟨[(α, x)], [(α, x + u)]⟩ is also of order 4 we need to check that (α, x(x + u)) ≁ 1. But this is true in view of lemma 4 since

(\frac{ᾱ , x(x + ū)}{k̄(x, \sqrt{f̄(x)})})_{x} ∼ (\frac{ᾱ , x}{k̄⟨x⟩}) ≁ 1.

Finally, since the extension of constants by \sqrt{α} leaves us in case V 4 the groups ⟨[(π, x)], [(π, x + u)]⟩ and ⟨[(α, x)], [(α, x + u)]⟩ have the trivial intersection. The theorem is proved.

In case where f(x) is of the form V I we have

Theorem 9 Let C be given by the equation y^{2} = f(x), where f(x) is of form V I. Then the class of (π, α) and the classes of the following algebras generate _{2}Br C.

V I1. {(π, l_{i})}_{i}, l_{i} is a monic linear divisor of f.

V I2. {(π, l_{i})}_{i}, l_{i} is a monic linear divisor of f, l_{i} ≠ x − π^{k}u and (α, x).

Proof. In case V I1 it is enough to check that for any monic linear divisor of f(x) (π, l_{i}) ≁ 1 and (π, l_{i}l_{j}) ≁ 1 if i ≠ j. We have that (π, l_{i}) ≁ 1 by lemma 3. By the same reason (π, l_{i}l_{j}) ≁ 1 if l_{i}l_{j} ≠ x(x − π^{k}u). If l_{i}l_{j} = x(x − π^{k}u), then (π, x(x − π^{k}u))_{x} ∼ (π, e_{1}e_{2}e_{3}) ≁ 1.

In case V 2 arguments similar to the previous ones shows that the group H generated by [(π, l_{i})], where l_{i} runs through the set of all monic linear divisors not equal to x − π^{k}u is of order 8. To complete the consideration of the case let us show that (α, x) ≁ 1. If k is odd we have (α, x)_{x−π^{k}u} ≁ 1, if not, then (α, x)_{x−π} ≁ 1, since f(π) ∼ π(π − π^{k}u)e_{1}e_{2}e_{3} ∼ e_{1}e_{2}e_{3} ∼ 1. Finally, the extension of constants by \sqrt{α} leaves us in case V I2, so all non-trivial algebras from H are non-trivial after this extension but it is not the case for (α, x). The theorem is proved.

3 Splitting type (1,1,1,2).

This section is devoted to case deg f_{1} = deg f_{2} = deg f_{3} = 1, deg f_{4} = 2 and the reduction is bad.

Preliminary results. The evident list of all cases under consideration according to the reduction type of f(x) is as follows.

I. f(x) = (x^{2} − π^{k}u)(x − e_{1})(x − e_{2})(x − e_{3}),

π^{k}u ≁ 1, u, e_{i} ∈ O_{k}^{∗}, k > 0, ē_{i} ≠ ē_{j} if i ≠ j.

II. f(x) = (x^{2} − π^{k}u)(x − e)(x − e′)(x − e_{1}),

π^{k}u ≁ 1, u, e, e′, e_{1} ∈ O_{k}^{∗}, k > 0, ē = ē′ ≠ ē_{1}.

III. f(x) = (x^{2} − π^{k}u)(x − e_{1})(x − e_{2})(x − e_{3}),

u ∈ O_{k}^{∗}, k > 0, π^{k}u ≁ 1, ē_{1} = ē_{2} = ē_{3} ≠ 0.

IV. f(x) = (x^{2} + ax + b)x(x − π^{k}u)(x + e),

u, e ∈ O_{k}^{∗}, k > 0, x^{2} + āx + b̄ is irreducible over k̄. V. f(x) = (x^{2} + ax + b)x(x − π^{k_{1}}u_{1})(x − π^{k_{2}}u_{2}),

u_{i} ∈ O_{k}^{∗}, 0 < k_{1} ≤ k_{2}, and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}, x^{2} + āx + b̄ is irreducible over k̄.

V I. f(x) = (x^{2} − π^{k}u)(x − π^{m}v)(x + e_{1})(x + e_{2}),

π^{k}u ≁ 1, k, m > 0, u, v, e_{i} ∈ O_{k}^{∗}, ē_{1} ≠ ē_{2}. V II. f(x) = (x^{2} − π^{k}u)(x − π^{m}v)(x + e)(x + e′),

π^{k}u ≁ 1, k, m, l > 0, u, v, e ∈ O_{k}^{∗}, ē = ē′. V III. f(x) = (x^{2} − π^{k}u)(x − π^{m_{1}}v_{1})(x − π^{m_{2}}v_{2})(x + e),

π^{k}u ≁ 1, 0 < m_{1} ≤ m_{2}, k > 0, u, v_{i}, e ∈ O_{k}^{∗}. IX. f(x) = (x^{2} − π^{k}u)(x − π^{m_{1}}v_{1})(x − π^{m_{2}}v_{2})(x − π^{m_{3}}v_{3}),

π^{k}u ≁ 1, 0 < m_{1} ≤ m_{2} ≤ m_{3}, k > 0, u, v_{i} ∈ O_{k}^{∗}. Let us prove firstly some preliminary results.

Lemma 10

| _{2}Br C| = 16.

Lemma 11 let u ∈ O_{k}^{∗} is a unit such that u ≁ 1, then

i) for any w ∈ O_{k}^{∗} w ± \sqrt{u} ∼ 1 in k(\sqrt{u}) iff w^{2} − u ∼ 1 in k, ii) for any local field k there exists w ∈ O_{k}^{∗} such that w^{2} − u ≁ 1, iii) for any v ∈ O_{k}^{∗} there exists w ∈ O_{k}^{∗} such that v^{2} − w ∈ O_{k}^{∗} and v^{2} − w ≁ 1.

Proof. i) follows from the surjectivity on units of the norm homomorphism N_{k(\sqrt{u})|k} and for ii) and iii) see [13], lemma 1.

Lemma 12 Let g(x) ∈ k[x] be a monic polynomial, and let either n be odd or deg g(x) be even. Then for any b ∈ k(C)^{∗} the quaternion algebra

(\frac{b , g(x)}{k(C)})

is not isomorphic to the scalar algebra

(\frac{π , α}{k(C)}) .

To check this fact it is enough to observe that

(\frac{b , g(x)}{k(C)})_{∞} ∼ 1 and (\frac{π , α}{k(C)})_{∞} ≁ 1.

Lemma 13 Let K be an algebraically closed field and f(x), g(x), h(x) ∈ K[x]. Then the system

\{ \begin{matrix}y^{2} = f(x), \\ z^{2} = g(x), \\ t^{2} = h(x) \end{matrix}

defines a variety V in P^{4}(K) if and only if all polynomials f(x), g(x), h(x), f(x)g(x), f(x)h(x), g(x)h(x) and f(x)g(x)h(x) are not in K[x]^{2}.

Proof. Consider the homomorphism

ϕ : K[x, y, z, t] ⟶ K(x, \sqrt{f(x)}, \sqrt{g(x)}, \sqrt{h(x)}),

defined by the rule

x ↦ x, y ↦ \sqrt{f(x)}, z ↦ \sqrt{g(x)}, t ↦ \sqrt{h(x)}.

Then the ideal I = (y^{2} − f(x), z^{2} − g(x), t^{2} − h(x)) ⊂ Ker ϕ. Let F = F_{0}(x) + F_{1}(x)y + F_{2}(x)z + F_{3}(x)t + F_{4}(x)yz + F_{5}(x)yt + F_{6}(x)zt + G(x, y, z, t) ∈ Ker ϕ, where G(x, y, z, t) ∈ I. So ϕ(F) = F_{0}(x) + F_{1}(x)\sqrt{f} + F_{2}(x)\sqrt{g} + F_{3}(x)\sqrt{h} + F_{4}(x)\sqrt{fg} + F_{5}(x)\sqrt{fh} + F_{6}(x)\sqrt{gh} = 0. Let us check that the extension K(x)(\sqrt{f(x)}, \sqrt{g(x)}, \sqrt{h(x)})|K(x) is of degree 8. Indeed, K(x)(\sqrt{f(x)})|K(x) is of degree 2. If \sqrt{g(x)} ∈ K(x)(\sqrt{f(x)}), then g(x) = a(x)^{2} + b(x)^{2}f(x) + 2a(x)b(x)\sqrt{f(x)}, where a(x), b(x) ∈ K(x). So a(x)b(x) = 0 and either g(x) = a(x)^{2} or g(x) = b(x)^{2}f(x). And this is impossible in view of g(x) ∉ K[x]^{2} and f(x)g(x) ∉ K[x]^{2}. Hence K(x)(\sqrt{f(x)}, \sqrt{g(x)})|K(x) is of degree 4. Let now \sqrt{h(x)} ∈ K(x)(\sqrt{f(x)}, \sqrt{g(x)}), i.e. \sqrt{h(x)} = a_{0}(x) + a_{1}(x)\sqrt{f(x)} + a_{2}(x)\sqrt{g(x)} + a_{3}(x)\sqrt{f(x)g(x)}. We have h = (a_{0}^{2} + a_{1}^{2}f + a_{2}^{2}g + a_{3}^{2}fg) + 2(a_{0}a_{1} + a_{2}a_{3}g)\sqrt{f} + 2(a_{0}a_{2} + a_{1}a_{3}f)\sqrt{g} + 2(a_{0}a_{3} + a_{1}a_{2})\sqrt{fg}. Therefore

\{ \begin{matrix}a_{0}a_{1} + a_{2}a_{3}g = 0, \\ a_{0}a_{2} + a_{2}a_{3}f = 0, \\ a_{0}a_{3} + a_{1}a_{2} = 0. \end{matrix}

We have h(x) ∉ K(x)(\sqrt{f(x)g(x)}), so that one of a_{1}, a_{2}, say, a_{1} is not equal to 0. Then

\{ \begin{matrix}a_{2} = −a_{0}a_{3}/a_{1}, \\ a_{0}(a_{3}^{2}g − a_{1}^{2}) = 0, \\ a_{3}(a_{1}^{2}f − a_{0}^{2}) = 0, \end{matrix}

Furthermore, a_{3}^{2}g − a_{1}^{2} ≠ 0, since a_{1} ≠ 0 and g(x) ∉ K[x]^{2}. Then a_{0} = a_{2} = a_{3} = 0 and \sqrt{h(x)} ∈ K(x)(\sqrt{f(x)}). This contradiction proves that the extension

K(x)(\sqrt{f(x)}, \sqrt{g(x)}, \sqrt{h(x)})|K(x)

is of degree 8. The evident basis of this extension is 1, \sqrt{f(x)}, \sqrt{g(x)}, \sqrt{h(x)}, \sqrt{f(x)g(x)}, \sqrt{f(x)h(x)}, \sqrt{g(x)h(x)}, \sqrt{f(x)g(x)h(x)}, so F_{i}(x) = 0 for any i and Ker ϕ = I. Then I is prime and V is a variety.

Lemma 14 Let f(x) = (x^{2} −π^{k}u)g(x), g(x) ∈ O_{k}[x] without multiple roots, ḡ(x) ∉ k̄[x]^{2}, u ∈ O_{k}^{∗}, k > 0, π^{k}u ≁ 1 and E = g(0) ∈ O_{k}^{∗}. Then the following algebra is nontrivial, unramified and not isomorphic to the scalar algebra (π, α).

1. (−πEu, x) if k ≡ 1 (mod 2).

2. (α, x − π^{k/2}w) if k ≡ 0 (mod 2) and E ∼ 1, where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1.

3. (π, x^{2} − π^{k}u) if k ≡ 0 (mod 2) and E ≁ 1.

Proof.

(−πEu, x)_{x} ∼ (\frac{−πEu , x}{k⟨x⟩(\sqrt{−πEu})}) ∼ 1,

so this algebra is unramified. Let K be an unramified odd degree extension of k such that there exists a unit τ ∈ O_{K}^{∗} with the property τ ≁ 1, g(τ) ∼ 1 and p(x) = Irr_{K|k}(τ). Such K exists in view of lemma 6 from the previous section. Then

(−πEu, x)_{p} ∼ (\frac{−πEu , τ}{K⟨p⟩(\sqrt{g(τ)})}) ∼ (\frac{π , τ}{K⟨p⟩}) ≁ 1.

The unit w in 2) exists in view of lemma 11. We have

(α, x − π^{k/2}w)_{x−π^{k/2}w} ∼ (\frac{α , x − π^{k/2}w}{k⟨x − π^{k/2}w⟩(\sqrt{(π^{k}w^{2} − π^{k}u)E})}) ∼

∼ (\frac{α , x − π^{k/2}w}{k⟨x − π^{k/2}w⟩(\sqrt{α})}) ∼ 1,

so the algebra (α, x − π^{k/2}w) is unramified. Furthermore,

(α, x − π^{k/2}w)_{x^{3}−π} ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨x^{3} − π⟩(\sqrt{(π^{2/3} − π^{k}u)E})}) ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨x^{3} − π⟩}) ≁ 1.

Lemma 2 provides the algebra (π, x^{2} − π^{k}u) to be unramified. Moreover,

(π, x^{2} − π^{k}u)_{x−π^{k/2}w} ∼ (\frac{π , α}{k⟨x − π^{k/2}w⟩(\sqrt{αE})}) ∼ (\frac{π , α}{k⟨x − π^{k/2}w⟩}) ≁ 1.

To complete the proof it is enough to check that all algebras are not isomorphic to the scalar algebra, and this is true in view of lemma 12.

Presentation by algebras.

Theorem 10 Let f be as in cases I or II. Then the group _{2}Br C is generated by the class of the algebra (π, α) and classes of the following algebras.

• Case I.

1. (e_{1}e_{1}e_{3}uπ, x), (π, x − e_{1}), (π, x − e_{2}), if k ≡ 1 (mod 2).

2. (α, x − π^{k/2}w), (π, x − e_{1}), (π, x − e_{2}), if k ≡ 0 (mod 2), −e_{1}e_{2}e_{3} ∼ 1.

3. (π, x^{2} − π^{k}u), (π, x − e_{1}), (π, x − e_{2}), if k ≡ 0 (mod 2), −e_{1}e_{2}e_{3} ≁ 1.

• Case II.

1. (e_{1}uπ, x), (π, x − e), (π, x − e′), if k ≡ 1 (mod 2), e − e_{1} ≁ 1.

2. (e_{1}uπ, x), (π, x − e), (α, x − e), if k ≡ 1 (mod 2), e − e_{1} ∼ 1.

3. (α, x − π^{k/2}w), (π, x − e), (π, x − e′), if k ≡ 0 (mod 2), −e_{1} ∼ 1, e − e_{1} ≁ 1.

4. (α, x − π^{k/2}w), (π, x − e), (α, x − e), if k ≡ 0 (mod 2), −e_{1} ∼ 1, e − e_{1} ∼ 1.

5. (π, x^{2} − π^{k}u), (π, x − e), (π, x − e′), if k ≡ 0 (mod 2), −e_{1} ≁ 1, e − e_{1} ≁ 1.

6. (π, x^{2} − π^{k}u), (π, x − e), (α, x − e), if k ≡ 0 (mod 2), −e_{1} ≁ 1, e − e_{1} ∼ 1,

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1.

Proof. In case I all algebras (π, x − e_{i}) and (π, (x − e_{i})(x − e_{j})), i ≠ j are unramified and not isomorphic to (π, α) in view of lemma 2. They are nontrivial by lemma 3.

In case II let us prove that if e−e_{1} ≁ 1, then the algebras (π, x−e), (π, x−e′) and (π, (x−e)(x−e′)) are unramified, nontrivial and not isomorphic to (π, α). If e − e_{1} ∼ 1, then the same is true for the algebras (π, x − e), (α, x − e) and (πα, x − e). Indeed, if e − e_{1} ≁ 1, then (π, (x − e)(x − e′))_{x−e} ∼ (π, (x^{2} − π^{k}u)(x − e_{1}))_{x−e} ∼ (π, e − e_{1}) ≁ 1. Let now e − e_{1} ∼ 1, then

(ᾱ, x − ē)_{x−ē} ∼ (\frac{ᾱ , x − ē}{k̄⟨x − ē⟩(\sqrt{ē − ē_{1}})}) ≁ 1,

and in view of lemma 4 (α, x − e) ≁ 1. Finally, after the replacement ν = πα of prime element we have (πα, x − e) = (ν, x − e) ≁ 1.

Because of the polynomial f in cases I and II satisfies the conditions of lemma 12 it is enough to prove that algebra from this lemma is not isomorphic to the algebras already considered in the current proof.

Consider firstly case I. If k ≡ 1 (mod 2), then one can find such w_{1}, w_{2} and w_{3} from O_{K}^{∗} (K as above) that w_{i} ∼ 1, w_{i} − e_{i} ≁ 1, (w_{i} − e_{1})(w_{i} − e_{2})(w_{i} − e_{3}) ∼ 1. Indeed, for example, in case i = 1 it is enough to solve the following system

\{ \begin{matrix}w_{1} = x^{2}, \\ y^{2} = α(x^{2} − e_{1}) \\ z^{2} = α(x^{2} − e_{2})(x^{2} − e_{3}), \end{matrix}

and lemma 6 works. Let p_{i} = Irr_{K|k}(w_{i}). Then for any i we have (−πEu, x)_{p_{i}} ∼ 1, (π, x − e_{i})_{p_{i}} ≁ 1 and (π, (x − e_{1})(x − e_{2}))_{p_{3}} ≁ 1.

If E = −e_{1}e_{2}e_{3} ∼ 1, then the algebra (α, x − π^{k/2}w) splits by the extension of constants by \sqrt{α} but this is not true for the algebras (π, x − e_{i}) and (π, (x − e_{1})(x − e_{2})).

Let k ≡ 0 (mod 2), −e_{1}e_{2}e_{3} ≁ 1, then (π, x^{2} − π^{k}u) ≁ (π, x − e_{i}), (π, (x − e_{1})(x − e_{2})) in view of lemma 3.

Consider now case II. The extension by \sqrt{α} does not split (−πEu, x), therefore (−πEu, x) ≁ (α, x−e). Let w ∈ O_{K}^{∗} such that w ∼ 1, w − e ≁ 1 and (w − e)(w − e′)(w − e_{1}) ∼ w − e_{1} ∼ 1, p = Irr_{K|k}(w). Then (−πEu, x)_{p} ∼ 1 and (π, x − e)_{p} ≁ 1. Changing the prime element ν = πα one can check that (−πEu, x) ≁ (πα, x − e). In the same way we have (−πEu, x) ≁ (π, x − e′). Let us show finally that (−πEu, x) ≁ (π, (x − e)(x − e′)). If q = Irr_{K|k}(w), where w ≁ 1, w̄ ≠ ē and w − e_{1} ∼ 1, then (−πEu, x)_{q} ≁ 1, (π, (x − e)(x − e′))_{q} ∼ (π, (w − e)^{2}) ∼ 1.

Let now k ≡ 0 (mod 2), E ∼ −e_{1} ∼ 1. We have (α, x−π^{k/2}w) ≁ (π, x−e), (π, x−e′), (π, (x−e)(x−e′)), since the extension of constants by \sqrt{π} does not split the first algebra but splits the others. After this extension (πα, x − e) ∼ (α, x − e). It only remains to check that (α, x − π^{k/2}w) ≁ (α, x − e) if e − e_{1} ∼ 1. It is true since (α, x − π^{k/2}w)_{x^{3}−π} ≁ 1, but (α, (x − e)(x − e′))_{x^{3}−π} ∼ 1.

Finally, let k ≡ 0 (mod 2), −e_{1} ≁ 1. We have (π, (x^{2} − π^{k}u)(x − e)), (π, (x^{2} − π^{k}u)(x − e′)) ≁ 1. Moreover, (π, (x^{2} − π^{k}u)(x − e)(x − e′))_{x−e} ∼ (π, e − e_{1}) ≁ 1 if e − e_{1} ≁ 1. The extension of constants by \sqrt{π} does not split (α, x − e), so that (π, x^{2} − π^{k}u) ≁ (α, x − e). After the same extension (πα, x − e) ∼ (α, x − e) ≁ 1, so (πα, x − e) ≁ (π, x^{2} − π^{k}u). The theorem is proved.

Without loss of generality one can assume that in case III f(x) = (x^{2} − π^{k}u)(x − e_{1})(x − e_{2})(x − e_{3}), where e_{1} = e, e_{2} = e + π^{k_{1}}u_{1}, e_{3} = e + π^{k_{2}}u_{2}, e, u_{i} ∈ O_{k}^{∗}, 0 < k_{1} ≤ k_{2} and if k_{1} = k_{2}, then ū_{1} ≠ ū_{2}.

Theorem 11 Let f be as in case III. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

1. (euπ, x), (α, x − e_{1}), (α, x − e_{2}),

if k ≡ 1 (mod 2), k_{1} ≡ 1 (mod 2).

2. (euπ, x), (α, x − e_{1}), (π, x − e_{1}),

if k ≡ 1 (mod 2), k_{1} ≡ 0 (mod 2) and either k_{2} ≡ 1 (mod 2) or k_{1} < k_{2}, −u_{1} ∼ 1.

3. (euπ, x), (π, x − e_{1}), (π, x − e_{2}),

if k ≡ 1 (mod 2), k_{1} ≡ k_{2} ≡ 0 (mod 2) and either k_{1} = k_{2} or −u_{1} ≁ 1.

4. (α, x − π^{k/2}w), (α, x − e_{1}), (α, x − e_{2}),

if k ≡ 0 (mod 2), −e ∼ 1, k_{1} ≡ 1 (mod 2).

5. (α, x − π^{k/2}w), (α, x − e_{1}), (π, x − e_{1}),

if k ≡ 0 (mod 2), −e ∼ 1, k_{1} ≡ 0 (mod 2) and either k_{2} ≡ 1 (mod 2) or k_{1} < k_{2}, −u_{1} ∼ 1.

6. (α, x − π^{k/2}w), (π, x − e_{1}), (π, x − e_{2}),

if k ≡ 0 (mod 2), −e ∼ 1, k_{1} ≡ k_{2} ≡ 0 (mod 2) and either k_{1} = k_{2} or −u_{1} ≁ 1.

7. (π, x^{2} − π^{k}u), (α, x − e_{1}), (α, x − e_{2}),

if k ≡ 0 (mod 2), −e ≁ 1, k_{1} ≡ 1 (mod 2).

8. (π, x^{2} − π^{k}u), (α, x − e_{1}), (π, x − e_{1}),

if k ≡ 0 (mod 2), −e ≁ 1, k_{1} ≡ 0 (mod 2) and either k_{2} ≡ 1 (mod 2) or k_{1} < k_{2}, −u_{1} ∼ 1.

9. (π, x^{2} − π^{k}u), (π, x − e_{1}), (π, x − e_{2}),

if k ≡ 0 (mod 2), −e ≁ 1, k_{1} ≡ k_{2} ≡ 0 (mod 2) and either k_{1} = k_{2} or −u_{1} ≁ 1,

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1.

Let us prove firstly

Lemma 15 In case III the following algebras are unramified, nontrivial and not isomorphic to (π, α).

1. (α, x − e_{1}), (α, x − e_{2}), (α, (x − e_{1})(x − e_{2})), if k_{1} ≡ 1 (mod 2).

2. (α, x − e_{1}), (π, x − e_{1}), (απ, x − e_{1}), if k_{1} ≡ 0 (mod 2) and either k_{2} ≡ 1 (mod 2) or (k_{2} ≡ 0 (mod 2), k_{1} < k_{2} and −u_{1} ∼ 1).

3. (π, x − e_{1}), (π, x − e_{2}), (π, (x − e_{1})(x − e_{2})), if k_{1} ≡ k_{2} ≡ 0 (mod 2) and either k_{1} < k_{2} and −u_{1} ∼ 1 or k_{1} = k_{2}.

Proof of the lemma. In 1 we have: (α, x − e_{1})_{x−e_{2}} ≁ 1, (α, x − e_{2})_{x−e_{1}} ≁ 1, (α, (x − e_{1})(x − e_{2}))_{x−e_{2}} ∼ (α, x − e_{3})_{x−e_{2}} ≁ 1.

2. Let θ = e + π^{k_{1}}w. Then f(θ) ∼ w − u_{1}. If w satisfies the condition

\{ \begin{matrix}w_{1} ≁ 1, \\ w − u_{1} ∼ 1, \end{matrix}

then (π, x − e_{1})_{p} ≁ 1 and (πα, x − e_{1})_{p} ≁ 1, where p = Irr_{K|k}(θ). Furthermore, in case k_{2} ≡ 1 (mod 2) we have (α, x − e_{1})_{x−e_{3}} ≁ 1 and (α, x − e_{1})_{x−e−π^{k_{1}+1}} ∼ (α, π^{k_{1}+1}) ≁ 1 since f(e + π^{k_{1}+1}) ∼ −u_{1} ∼ 1, otherwise.

Finally, consider subcase 3. Let θ_{i} = e + π^{k_{1}}w_{i}, i = 1, 2, 3. Then f(θ_{i}) ∼ w_{i}(w_{i} − u_{1})(w_{i} − u_{2}), provided k_{1} = k_{2}. If w_{i} satisfies the conditions

\{ \begin{matrix}w_{1} ≁ 1, \\ w_{2} − u_{1} ≁ 1, \\ w_{3} − u_{2} ≁ 1, \\ (w_{1} − u_{1})(w_{1} − u_{2}) ≁ 1, \\ w_{2}(w_{2} − u_{2}) ≁ 1, \\ w_{3}(w_{3} − u_{3}) ≁ 1, \end{matrix}

then f(θ_{i}) ∼ 1 and (π, x − e_{i})_{p_{i}} ≁ 1, where p_{i} = Irr_{K|k}(θ_{i}). Moreover, (π, (x − e_{i})(x − e_{j}))_{p_{l}} ∼ (π, (x^{2} − π^{k}u)(x − e_{l}))_{p_{l}} ∼ (π, x − e_{l})_{p_{l}} ≁ 1, where {i, j, l} = {1, 2, 3}.

Let now k_{1} < k_{2} are even and −u_{1} ≁ 1. Let θ be as in 2. Then (π, x − e_{1})_{p} ≁ 1 and (π, (x − e_{1})(x − e_{2}))_{p} ≁ 1, where p is as above. We have also (π, x − e_{2})_{x−e_{1}} ∼ (π, −u_{1}) ≁ 1. The lemma is proved.

Now to prove the theorem it is enough to check that the algebra from lemma 12 is not isomorphic to all algebras from lemma 14 in the corresponding subcases.

Let firstly k ≡ 1 (mod 2). The extension of constants by \sqrt{α} leaves us in case 1 of lemma 12, and in cases 1 and 2 of lemma 13, therefore the algebra (euπ, x) does not split by this extension. On the other hand, all algebras from case 1 of lemma 14 split by it. And in case 2 of this lemma one only need to prove that (euπ, x) ≁ (π, x − e_{1}) and (euπ, x) ≁ (πα, x − e_{1}). But if w, θ and p are as in the proof of this case, then after the extension indicated above (euπ, x)_{p} ∼ (π, e) ∼ 1. On the other hand, (πα, x − e_{1})_{p} ∼ (π, x − e_{1})_{p} ≁ 1.

Let us take a unit η ∈ O_{K}^{∗} with the usual properties satisfying the condition η̄ ≠ ē, η ≁ 1, η − e ∼ 1. Then f(η) ∼ η − e ∼ 1 and (euπ, x)_{q} ≁ 1, where q = Irr_{K|k}(η). We have also (π, x − e_{i})_{q} ∼ 1 and (π, (x − e_{1})(x − e_{2}))_{q} ∼ 1. Thus case k ≡ 1 (mod 2) is considered.

If k is even and −e ∼ 1, then for any A ∈ {π, α, πα} (A, x − e_{i})_{x^{3}−π} ∼ (A, −e) ∼ 1, (A, (x − e_{1})(x − e_{1}))_{x^{3}−π} ∼ 1, but (α, x − π^{k/2}w)_{x^{3}−π} ≁ 1.

Finally, let k be even, −e ≁ 1, then (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1 and (α, x − e_{i})_{x−π^{k/2}w} ∼ 1, (α, (x − e_{1})(x − e_{2}))_{x−π^{k/2}w} ∼ 1. If p_{i} are the points from the proof of lemma 14 in which the corresponding algebras are nontrivial, then (π, x^{2} − π^{k}u)_{p_{i}} ∼ 1 and we are done.

Let us pass to the next two cases.

Theorem 12 Let f be as in cases IV or V. Then the group _{2}Br C is generated by the classes of the algebras (π, α), (π, x^{2} + ax + b), and classes of the following algebras.

• Case IV.

1. (π, x), (π, x − π^{k}u), if be ≁ 1.

2. (π, x), (α, x), if be ∼ 1.

• Case V.

1. (α, x), (α, x − π^{k_{1}}u_{1}), if k_{1} ≡ 1 (mod 2).

2. (α, x), (π, x), if k_{1} ≡ 0 (mod 2) and either k_{2} ≡ 1 (mod 2) or k_{1} < k_{2} and −bu_{1} ∼ 1.

3. (π, x), (π, x − π^{k_{1}}u_{1}), if k_{1} ≡ k_{2} ≡ 0 (mod 2) and either k_{1} = k_{2} or −bu_{1} ≁ 1.

Proof. In case IV (π, x) ≁ 1, (π, x − π^{k}u) ≁ 1 by lemma 3. For the same reason (π, x) ≁ 1 in V.2, V.3 and (π, x − π^{k_{1}}u_{1}) ≁ 1 in V.3. In IV if be ≁ 1, then for w ∈ O_{K}^{∗} we have f(π^{k}w) ∼ αw(w − e) and (π, x(x − π^{k}u))_{Irr_{K|k}(π^{k}w)} ∼ (π, w(w − e)) ≁ 1, provided w(w − e) ≁ 1. If be ∼ 1, then

(\frac{ᾱ , x}{k̄(C̃)})_{x} ∼ (\frac{ᾱ , x}{k̄⟨x⟩}) ≁ 1,

so (α, x) ≁ 1. Replacing ν = πα one can check that (πα, x) ≁ 1.

In case V if k_{1} ≡ 1 (mod 2), then (α, x)_{x−π^{k_{1}} u_{1}} ≁ 1 and (α, x − π^{k_{1}}u_{1})_{x} ≁ 1. Moreover, (α, x(x − π^{k_{1}}u_{1}))_{x−π^{k_{1}} u_{1}} ∼ (α, (x^{2} + ax + b)(x − π^{k_{2}}u_{2}))_{x−π^{k_{1}} u_{1}} ∼ (α, b(u_{1} − π^{k_{2}−k_{1}}u_{2})π^{k_{1}}) ≁ 1.

Now we are in subcase V.2. If k_{2} ≡ 1 (mod 2), then (α, x)_{x−π^{k_{2}} u_{2}} ≁ 1. In case k_{2} ≡ 0 (mod 2) we have f(π^{k_{1}+1}) ∼ −bu_{1} ∼ 1 and (α, x)_{x−π^{k_{1}+1}} ≁ 1. Replacing the prime element one can check that (απ, x) ≁ 1.

Consider the last subcase in V. We have f(π^{k_{1}}w) ∼ bw(w − u_{1})(w − π^{k_{2}−k_{1}}u_{2}) and (π, x(x − π^{k_{1}}u_{1}))_{Irr_{K|k}(π^{k_{1}} w)} ∼ (π, w(w − u_{1})) ≁ 1, provided that

\{ \begin{matrix}w(w − u_{1}) ≁ 1, \\ b(w − π^{k_{2}−k_{1}}u_{2}) ≁ 1. \end{matrix}

It remains to prove that the algebra (π, x^{2} + ax + b) is nontrivial and not isomorphic to the algebras listed above. In case IV we have (π, x^{2}+ax+b) ≁ 1, (π, x(x^{2}+ax+b)) ≁ 1, (π, (x−π^{k}u)(x^{2}+ax+b)) ≁ 1 by lemma 3. In IV.1 (π, x(x − π^{k}u)(x^{2} + ax + b)) ≁ 1 by the same reason. The extension of constants by \sqrt{π} leaves us in case IV.2, so that (π, x^{2} + ax + b) ≁ (α, x). After this extension we also have that (πα, x) ∼ (α, x) ≁ (π, x^{2} + ax + b).

Finally, let us compare the algebra (π, x^{2} + ax + b) with the others in case V. First of all, (π, x^{2} + ax + b) ≁ 1 and (π, (x^{2} + ax + b)x(x − π^{k_{1}}u_{1})) ≁ 1 by analogy with IV. After the extension of constants by \sqrt{α} we are not in the same case because the polynomial x^{2} + ax + b splits after this extension. But the algebra (π, x^{2} + ax + b) is still nontrivial in view of lemma 3. So this algebra is not isomorphic to (α, x), (α, x − π^{k_{1}}u_{1}) and (α, x(x − π^{k_{1}}u_{1})). To complete the proof one only need to check that (π, x^{2} + ax + b) is not isomorphic to (π, x), (πα, x) in V.2 and (π, x), (π, x − π^{k_{1}}u_{1})) in V.3. In V.2 we have f(π^{k_{1}}w) ∼ b(w −u_{1}), where w is as above. Let p = Irr_{K|k}(π^{k_{1}}w). Then (π, x)_{p} ∼ (πα, x)_{p} ∼ (π, w), (π, x^{2} + ax + b)_{p} ∼ (π, b). If we require

\{ \begin{matrix}b(w − u_{1}) ∼ 1, \\ bw ≁ 1, \end{matrix}

then (π, x)_{p} ∼ (πα, x)_{p} ≁ (π, x^{2} + ax + b)_{p}.

Consider subcase V.3. Let k_{1} = k_{2}. To prove that (π, x^{2} + ax + b) ≁ (π, x), (π, x − π^{k_{1}}u_{1}) we need to solve the systems

\{ \begin{matrix}bw_{1} ≁ 1, \\ (w_{1} − u_{1})(w_{1} − u_{2}) ≁ 1 \end{matrix} and \{ \begin{matrix}b(w_{2} − u_{1}) ≁ 1, \\ w_{2}(w_{2} − u_{2}) ≁ 1. \end{matrix}

By the same way one can prove (π, x^{2} + ax + b) ≁ (π, x), provided k_{1} < k_{2}. Finally, in this case (π, (x^{2} + ax + b)(x − π^{k_{1}}u_{1}))_{x} ∼ (π, −bu_{1}) ≁ 1. Theorem 13 Let f be as in cases V I or V II. Then the group _{2}Br C is generated by the classes of the algebra (π, α), and classes of the following algebras.

• Case V I. (π, x + e_{1}), (π, x + e_{2}).

• Case V II. (π, x + e) and (α, x + e) if −e ∼ 1, (π, x + e′) if −e ≁ 1. and

1. (α, x^{2} − π^{k}u) if m ≥ k/2, k ≡ 1 (mod 2).

2. (π, x^{2} − π^{k}u) if m ≥ k/2, k ≡ 0 (mod 4).

3. (πE, x − π^{k/2}w) if m ≥ k/2, k ≡ 2 (mod 4).

4. (πuve_{1}e_{2}, x) if m < k/2, k ≡ 1 (mod 2), m ≡ 0 (mod 2).

5. (πE, x − π^{k/2}w) if m < k/2, k ≡ 0 (mod 2), m ≡ 1 (mod 2).

6. (α, x − π^{m}v) if m < k/2, k ≡ m ≡ 1 (mod 2), uve_{1}e_{2} ∼ 1.

7. (α, x) if m < k/2, k ≡ m ≡ 1 (mod 2), uve_{1}e_{2} ≁ 1.

8. (π, x^{2} − π^{k}u) if m < k/2, k ≡ m ≡ 0 (mod 2), −ve_{1}e_{2} ≁ 1.

9. (α, x − π^{k/2}w) if m < k/2, k ≡ m ≡ 0 (mod 2), −ve_{1}e_{2} ∼ 1, where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1, E is chosen such that f(π^{k/2}w) ∼ πE, and in case V II the units e_{1} and e_{2} must be omitted in all algebras and conditions listed above. Proof. The algebras (π, x + e_{1}), (π, x + e_{2}), (π, (x + e_{1})(x + e_{2})) are nontrivial in case V I. The same is true for the algebras (π, x + e) and (π, x + e′) in case V II. In the last case we also have

(\frac{ᾱ , x + ē}{k̄(Ẽ)})_{x+ē} ∼ (\frac{ᾱ , x + ē}{k̄(\sqrt{x})})_{x+ē} ∼ (\frac{ᾱ , x + ē}{k̄⟨x + ē⟩(\sqrt{−ē})}) ≁ 1,

provided −e ∼ 1. Moreover, then the extension by \sqrt{α} does not split (π, x + e), so (πα, x + e) ≁ 1. If −e ≁ 1, then (π, (x + e)(x + e′))_{x+e} ∼ (π, (x^{2} − π^{k}u)(x − π^{m}v))_{x+e} ∼ (π, −e) ≁ 1.

Let us consider now the remaining algebra simultaneously in both cases.

1. (α, x^{2} − π^{k}u)_{x−π^{m}v} ≁ 1. Furthermore, (α, x^{2} − π^{k}u) ≁ (π, x + e_{i}), (π, (x + e_{1})(x + e_{1})) in case V I since the last two algebras do not split after the extension by \sqrt{α}. For the same reason (α, x^{2} − π^{k}u) ≁ (π, x + e), (π, x + e′), and (πα, x + e) in case V II. We have also (α, x + e)_{x−π^{m}v} ∼ 1, so that (α, x + e) ≁ (α, x^{2} − π^{k}u) and (π, (x + e)(x + e′))_{x−π^{m}v} ∼ (π, ee′) ∼ 1, so that (α, x^{2} − π^{k}u) ≁ (π, (x + e)(x + e′)).

2. Let w ∈ O_{K}^{∗} such that

\{ \begin{matrix}w^{2} − u ≁ 1, \\ a(w − vπ^{m−k/2}) ≁ 1, \end{matrix} where a = \{ \begin{matrix}e_{1}e_{2} in V I, \\ 1, in V II. \end{matrix}

Then f(π^{k/2}w) ∼ 1 and (π, x^{2} − π^{k}u)_{p} ≁ 1, where p = Irr_{K|k}(π^{k/2}w). Let now w_{0} ∈ O_{K}^{∗} such that

\{ \begin{matrix}w_{0}^{2} − u ∼ 1, \\ a(w_{0} − vπ^{m−k/2}) ∼ 1, \end{matrix}

Then f(π^{k/2}w) ∼ 1 and (π, x^{2} − π^{k}u)_{p_{0}} ∼ 1, where p_{0} = Irr_{K|k}(π^{k/2}w_{0}). In case e_{i} ∼ 1 we have (π, x^{2} −π^{k}u)_{p} ≁ (π, x+e_{i})_{p} and if e_{i} ≁ 1, then (π, x^{2} −π^{k}u)_{p_{0}} ≁ (π, x+e_{i})_{p_{0}}. In any case (π, x^{2} −π^{k}u) ≁ (π, x+e_{i}). By the same way one can check that (π, x^{2} −π^{k}u) ≁ (π, (x+e_{1})(x+e_{2})), (π, x+e), (π, x+e′) and (π, (x + e)(x + e′)). Finally, (π, x^{2} − π^{k}u) ≁ (πα, x + e), (α, x + e) since two last algebras do not split by \sqrt{π}. 3.

(πE, x − π^{k/2}w)_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨D(x^{2} − π^{k}u)⟩}) ≁ 1.

To finish the consideration of this subcase it is enough to observe that the completion in x^{2} − π^{k}u splits any other algebra from the list.

4. f(0) ∼ πuve_{1}e_{2}, so the algebra (πuve_{1}e_{2}, x) is unramified. Let w be from O_{K}^{∗} and q = Irr_{K|k}(π^{m}w). Then f(π^{m}w) ∼ e_{1}e_{2}(w − v) in V I or w − v in V II. Furthermore, (πuve_{1}e_{2}, x)_{q} ∼ (πuve_{1}e_{2}, w). If we require w ≁ 1 and e_{1}e_{2}(w − v) ∼ 1, then (πuve_{1}e_{2}, x)_{q} ≁ 1. Similarly, if we_{i} ≁ 1 and e_{1}e_{2}(w − v) ∼ 1, then (πuve_{1}e_{2}, x)_{q} ≁ (π, x + e_{i})_{q}. By analogy we can prove that (πuve_{1}e_{2}, x) is not isomorphic to the remaining algebras.

5. This case can be considered similarly to case 3.

6. f(0) ∼ uve_{1}e_{2} (or uv in V II) ∼ 1. Therefore (α, x − π^{m}v)_{x} ∼ (α, π^{m}) ≁ 1. The extension by \sqrt{α} does not split the algebras (π, x + e_{i}), (π, x + e), (π, x + e′), (π, (x + e_{1})(x + e_{2})), (πα, x + e). So it is sufficient to check that (α, x − π^{m}v) ≁ (α, x + e), (π, (x + e)(x + e′)). But this is true in view of (α, x + e)_{x} ∼ 1 and (π, (x + e)(x + e′))_{x} ∼ 1.

7. The algebra (α, x) is unramified since f(0) ∼ α. (α, x)_{x−π^{m}v} ∼ (α, π^{m}v) ≁ 1. In case V I the extension by \sqrt{α} does not split the algebras (π, x + e_{i}), (π, (x + e_{1})(x + e_{2})) since this extension leaves us in case V I. The same holds for the algebras (π, x + e), (πα, x + e) and (π, x + e′). Indeed, after this extension (πα, x + e) ∼ (π, x + e) ≁ 1 since we are in V II again. And (π, x + e′) is also nontrivial after the extension because of the units e and e′ are symmetrical. Finally, (α, x + e)_{x−π^{m}v} ∼ (α, e) ∼ 1 and (π, (x + e)(x + e′))_{x−π^{m}v} ∼ 1.

8. f(π^{k/2}w) ∼ −ve_{1}e_{2}(w^{2} − u) ∼ 1, provided w^{2} − u ≁ 1. Thus (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1. The algebras (π, (x^{2} − π^{k}u)(x + e_{i})), (π, (x^{2} − π^{k}u)(x + e)), (π, (x^{2} − π^{k}u)(x + e′)), (π, (x^{2} − π^{k}u)(x + e_{1})(x + e_{2})) are nontrivial by lemma 3. We only need to prove (π, x^{2} − π^{k}u) ≁ (π, (x + e)(x + e′)), (α, x + e), (πα, x + e). But two last algebras do not split after the extension by \sqrt{π} because this extension leaves us in case V II, −e ∼ 1. Finally, (π, (x + e)(x + e′))_{x−π^{k/2}w} ∼ 1.

9. f(π^{k/2}w) ∼ α, so the algebra (α, x − π^{k/2}w) is unramified. Let q = Irr_{k( ^{3} \sqrt{π})|k}(π^{m} ^{3} \sqrt{π}). Then f(π^{m} ^{3} \sqrt{π}) ∼ −ve_{1}e_{2} ∼ 1. Hence

(α, x − π^{k/2}w)_{q} ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨q⟩}) ≁ 1.

Since the extension by \sqrt{π} leaves us in the current subcase, in order to complete the proof one only need to check that (α, x − π^{k/2}w) ≁ (α, x + e). But this holds since (α, x + e)_{q} ∼ (α, e) ∼ 1.

Theorem 14 Let f be as in case V III. Then the group _{2}Br C is generated by the classes of the algebra (π, α), and the classes of the following algebras (l_{i} = x − π^{m_{i}}v_{i}).

• (π, l_{1}), (π, x + e) if e ≁ 1.

• (π, l_{1}), (α, l_{1}) if e ∼ 1.

and

1. (α, x^{2} − π^{k}u) if k/2 < m_{1}, e ∼ 1, k ≡ 1 (mod 2).

2. (π, x^{2} − π^{k}u) if k/2 < m_{1}, e ∼ 1, k ≡ 0 (mod 2) and −1 ∼ 1.

3. (α, x − π^{k/2}w) if k/2 < m_{1}, e ∼ 1, k ≡ 0 (mod 2) and −1 ≁ 1.

4. (π, x^{2} − π^{k}u) if k/2 < m_{1}, e ≁ 1, k ≡ 0 (mod 2) and −1 ≁ 1.

5. (α, l_{1}) if k/2 < m_{1}, e ≁ 1, k ≡ 0 (mod 2) and −1 ∼ 1.

6. (α, l_{1}) if k/2 < m_{1}, e ≁ 1, k ≡ 1 (mod 2).

7. (π, x^{2} − π^{k}u) if k/2 = m_{1} and either m_{1} < m_{2} or v̄_{1} ≠ v̄_{2}.

8. (π, x^{2} − π^{k}u) if k/2 = m_{1} = m_{2}, v̄_{1} = v̄_{2} and e(v_{1}^{2} − u) ≁ 1.

9. (α, l_{1}) if k/2 = m_{1} = m_{2}, v̄_{1} = v̄_{2} and e ≁ 1, v_{1}^{2} − u ≁ 1.

10. (α, x − π^{k/2}w) if k/2 = m_{1} = m_{2}, v̄_{1} = v̄_{2} and e ∼ 1, v_{1}^{2} − u ∼ 1.

11. (α, x^{2} − π^{k}u) if m_{1} < k/2 ≤ m_{2}, k ≡ 1 (mod 2).

12. (π, x^{2} − π^{k}u) if m_{1} < k/2 ≤ m_{2}, k ≡ 0 (mod 2) and m_{1} + k/2 ≡ 0 (mod 2).

13. (πE, x − π^{k/2}w) if m_{1} < k/2 ≤ m_{2}, k ≡ 0 (mod 2) and m_{1} + k/2 ≡ 1 (mod 2).

14. (πE, x − π^{k/2}w) if m_{2} < k/2, m_{1} + m_{2} ≡ 1 (mod 2), k ≡ 0 (mod 2).

15. (α, x^{2} − π^{k}u) if m_{2} < k/2, m_{1} + m_{2} ≡ 1 (mod 2), k ≡ 1 (mod 2) and −ev_{1}v_{2}u ∼ 1.

16. (α, x) if m_{2} < k/2, m_{1} + m_{2} ≡ 1 (mod 2), k ≡ 1 (mod 2) and −ev_{1}v_{2}u ≁ 1.

17. (πF, x) if m_{2} < k/2, m_{1} + m_{2} ≡ 0 (mod 2), k ≡ 1 (mod 2), where F ∈ O_{k}^{∗}, πF ∼ f(0).

18. (π, x^{2} − π^{k}u) if m_{2} < k/2, m_{1} + m_{2} ≡ 0 (mod 2), k ≡ 0 (mod 2), ev_{1}v_{2} ≁ 1.

19. (α, x − π^{k/2}w) if m_{2} < k/2, m_{1} + m_{2} ≡ 0 (mod 2), k ≡ 0 (mod 2), ev_{1}v_{2} ∼ 1, where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1 and E is chosen such that f(π^{k/2}w) ∼ πE. Proof. In view of lemma 3 the algebras (π, l_{1}) and (π, l_{1}(x+e)) are nontrivial. Furthermore, (π, x+e)_{l_{1}} ≁ 1, provided e ≁ 1. After the extension by \sqrt{α} we have (πα, l_{1}) ∼ (π, l_{1}) ≁ 1. Finally, f( ^{3} \sqrt{π}) ∼ e, so in case e ∼ 1

(α, l_{1})_{x^{3}−π} ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨x^{3} − π⟩}) ≁ 1.

Consider now the third generator.

1. The algebra (α, x^{2} − π^{k}u) is nontrivial since (α, x^{2} − π^{k}u)_{l_{1}} ≁ 1. (α, x^{2} − π^{k}u)_{x^{3}−π} ∼ 1, on the other hand (α, l_{1})_{x^{3}−π} ≁ 1, therefore these algebras are not isomorphic. The algebra (α, x^{2} − π^{k}u) is not isomorphic to (π, l_{1}), (πα, l_{1}) because the extension by \sqrt{α} does not split two last algebras.

2. (π, x^{2} − π^{k}u)_{l_{1}} ∼ (π, −u) ≁ 1. By lemma 3 (π, x^{2} − π^{k}u) ≁ (π, l_{1}). Since after the extension by \sqrt{π} one has (πα, l_{1}) ∼ (α, l_{1}) ≁ 1, it follows that (π, x^{2} − π^{k}u) is not isomorphic to these algebras.

3. Because of f(π^{k/2}w) ∼ α the algebra (α, x−π^{k/2}w)) is unramified. Moreover, (α, x−π^{k/2}w))_{x^{3}−π} ≁ 1. Let q = Irr_{k( ^{3} \sqrt{π})|k}(π^{s}), where s = k/2 + 1/3. Then f(π^{s}) ∼ 1, therefore

(α, x − π^{k/2}w)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2}}{k( ^{3} \sqrt{π})⟨q⟩} ) and (α, l_{1})_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{s}}{k( ^{3} \sqrt{π})⟨q⟩} ) ,

so (α, x − π^{k/2}w)_{q} ≁ (α, l_{1})_{q}. To complete the consideration of this subcase it is enough to observe that after the extension by \sqrt{π} we have (π, l_{1}) ∼ 1 ≁ (α, x − π^{k/2}w) and (πα, l_{1}) ∼ (α, l_{1}) ≁ (α, x − π^{k/2}w).

4. f(π^{k/2}w)) ∼ e(w^{2} −u) ∼ 1 and (π, x^{2} −π^{k}u)_{x−π^{k/2}w} ≁ 1. By lemma 3 (π, x^{2} −π^{k}u) ≁ (π, l_{1}), (π, l_{1}(x+ e)). Finally, (π, x^{2} − π^{k}u)_{l_{1}} ∼ 1 ≁ (π, x + e)_{l_{1}}.

5. Let

ε = \{ \begin{matrix}1, if k ≡ 0 (mod 4), \\ 2, if k ≡ 2 (mod 4), \end{matrix}

q = Irr_{k( ^{3} \sqrt{π})|k}(π^{k/2+ε/3}). Then f(π^{k/2+ε/3}) ∼ 1 and

(α, l_{1})_{q} ∼ (\frac{α , π^{k/2+ε/3}}{k( ^{3} \sqrt{π})⟨q⟩} ) ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨q⟩}) ≁ 1.

The algebra (α, l_{1}) is not isomorphic to the remaining algebras since the extension by \sqrt{π} leaves us in the current subcase and therefore it does not split this algebra.

6. Since the extension by \sqrt{α} leaves us in case V III, the algebras (π, l_{1}) and (π, l_{1}(x + e)) do not split by this extension, so they are not isomorphic to (α, l_{1}). Furthermore, if m_{1} < m_{2} or v̄_{1} ≠ v̄_{2}, then (α, l_{1})_{l_{2}} ∼ (α, π^{m_{1}}) and (α, l_{1})_{l_{1}} ∼ (α, (x^{2} − π^{k}u)l_{2}(x + e))_{l_{1}} ∼ (α, π^{m_{1}+1}). One of these two algebras is nontrivial. Thus (α, l_{1}) ≁ 1. Let now m_{1} = m_{2} and v_{2} = v_{1} + π^{s}τ, s > 0, τ ∈ O_{k}^{∗}. Then (α, l_{1})_{l_{2}} ∼ (α, π^{m_{1}+s}) and (α, l_{1})_{l_{1}} ∼ (α, π^{m_{1}+s+1}), so that the algebra (α, l_{1}) is nontrivial again. It remains to check that (α, l_{1}) ≁ (π, x + e). But (π, x + e)_{l_{1}} ≁ 1 and (π, x + e)_{l_{2}} ≁ 1. On the other hand, among the algebras (α, l_{1})_{l_{1}} and (α, l_{1})_{l_{2}} a trivial one exists.

7. f(π^{k/2}w) ∼ (w^{2} − u)(w − v_{1})(w − v_{2}π^{m_{2}−m_{1}}). If we require

\{ \begin{matrix}w^{2} − u ≁ 1, \\ (w − v_{1})(w − v_{2}π^{m_{2}−m_{1}}) ≁ 1, \end{matrix}

w ∈ O_{K}^{∗}, w̄ ≠ v̄_{1}, v̄_{2}, p = Irr_{K|k}(π^{k/2}w), then f(π^{k/2}w) ∼ 1 and (π, x^{2} − π^{k}u)_{p} ≁ 1. As we already had seen above, (π, x^{2} − π^{k}u) ≁ (π, l_{1}), (π, l_{1}(x + e)) by lemma 3. If

\{ \begin{matrix}w_{0}^{2} − u ∼ 1, \\ (w_{0} − v_{1})(w_{0} − v_{2}π^{m_{2}−m_{1}}) ∼ 1, \end{matrix}

w_{0} ∈ O_{K}^{∗}, w̄_{0} ≠ v̄_{1}, v̄_{2}, p_{0} = Irr_{K|k}(π^{k/2}w_{0}), then f(π^{k/2}w_{0}) ∼ 1 and (π, x^{2} − π^{k}u)_{p_{0}} ∼ 1. But (π, x + e)_{p_{0}} ∼ (π, e) ≁ 1, provided e ≁ 1. Thus in case e ≁ 1 we are done. Case e ∼ 1 can be considered by analogy with subcase 2.

8. Let e ≁ 1, v_{1}^{2} − u ∼ 1. There exists w ∈ O_{k}^{∗} such that w^{2} − u ≁ 1. So (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ∼ (π, w^{2} − u) ≁ 1 since f(π^{k/2}w) ∼ (w^{2} − u)e ∼ 1. If e ∼ 1, v_{1}^{2} − u ≁ 1, then (π, x^{2} − π^{k}u)_{x−π^{m_{1}} v_{1}} ∼ (π, v_{1}^{2} − u) ≁ 1. Let us check that the algebra (π, x^{2} − π^{k}u) is not isomorphic to the other ones. If e ≁ 1, then (π, x^{2} − π^{k}u) ≁ (π, l_{1}), (π, l_{1}(x + e)) as above. We also have (π, x^{2} − π^{k}u)_{x−π^{m_{1}} v_{1}} ∼ (π, v_{1}^{2} − u) ∼ 1 and (π, x + e)_{x−π^{m_{1}} v_{1}} ∼ (π, e) ≁ 1. Let now e ∼ 1. Then one can argue as in subcase 2.

9. Let

ε = \{ \begin{matrix}1, if k ≡ 0 (mod 4), \\ 2, if k ≡ 2 (mod 4), \end{matrix}

θ = π^{k/2}(v_{1} + ( ^{3} \sqrt{π})^{ε}), q = Irr_{k( ^{3} \sqrt{π})|k}(θ). Then f(θ) ∼ (v_{1}^{2} − u)e ∼ 1 and

(α, l_{1})_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m_{1}+ε}}{k( ^{3} \sqrt{π})⟨q⟩} ) ≁ 1.

(α, l_{1}) ≁ (π, l_{1}), (π, x + e), (π, l_{1}(x + e)) since the extension by \sqrt{π} leaves us in the current subcase and therefore it does not split the algebra (α, l_{1}).

10. The algebra (α, x − π^{k/2}w) is unramified. f( ^{3} \sqrt{π}) ∼ e ∼ 1 so (α, x − π^{k/2}w)_{x^{3}−π} ≁ 1. After the extension by \sqrt{π} we are in the current subcase again, therefore it is enough only to prove that (α, x − π^{k/2}w) ≁ (α, l_{1}). To check this fact assume θ_{0} = π^{k/2}(v_{1} + ^{3} \sqrt{π}), q_{0} = Irr_{k( ^{3} \sqrt{π})|k}(θ_{0}). Then f(θ_{0}) ∼ 1 and

(α, l_{1})_{q_{0}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2+1}}{k( ^{3} \sqrt{π})⟨q_{0}⟩} ) and (α, x − π^{k/2}w)_{q_{0}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2}}{k( ^{3} \sqrt{π})⟨q_{0}⟩} ) .

11. (α, x^{2} − π^{k}u)_{l_{2}} ∼ (α, π^{k}) ≁ 1. If e ∼ 1, then \sqrt{α} does not split (π, l_{1}) and (πα, l_{1}) so that these algebras are not isomorphic to (α, x^{2} −π^{k}u). Moreover, (α, x^{2} −π^{k}u)_{x^{3}−π} ≁ (α, l_{1})_{x^{3}−π}. Let e ≁ 1. Then we have (α, x^{2} − π^{k}u) ≁ (π, l_{1}), (π, l_{1}(x + e)). Finally, (α, x^{2} − π^{k}u)_{l_{1}} ∼ 1, but (π, x + e)_{l_{1}} ≁ 1.

12. Let w ∈ O_{K}^{∗} such that

\{ \begin{matrix}w^{2} − u ≁ 1, \\ −v_{1}e(w − v_{2}π^{m_{2}−k/2}) ≁ 1, \end{matrix}

then f(π^{k/2}w) ∼ 1 and (π, x^{2} −π^{k}u)_{Irr_{K|k}}(π^{k/2}w) ≁ 1. If e ≁ 1, then (π, x^{2} −π^{k}u) ≁ (π, l_{1}), (π, l_{1}(x+e)). We also have (π, x^{2} − π^{k}u)_{l_{1}} ∼ 1 and (π, x + e)_{l_{1}} ≁ 1. If e ∼ 1, then the extension by \sqrt{π} leaves us in subcase 12, so we can argue as in subcase 2.

13. The algebra (πE, x − π^{k/2}w) is unramified by the choice of E.

(πE, x − π^{k/2}w)_{x^{2}−π^{k}u} ∼ (\frac{πE , π^{k/2}(\sqrt{u} − w)}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(\sqrt{π(x^{2} − π^{k}u)})}) ∼

∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨\sqrt{π(x^{2} − π^{k}u)}⟩}) ≁ 1.

On the other hand,

(π, l_{1})_{x^{2}−π^{k}u} ∼ (\frac{π , −π^{m_{1}}v_{1}}{k(\sqrt{u})⟨\sqrt{π(x^{2} − π^{k}u)}⟩}) ∼ 1.

The same is true for the algebras (π, x + e), (π, l_{1}(x + e)), (α, l_{1}) and (πα, l_{1}).

14. This case is totally similar to the previous one.

15. f(0) ∼ 1, so (α, x^{2} − π^{k}u)_{x} ∼ (α, π^{k}) ≁ 1. The extension by \sqrt{α} does not split (π, l_{1}), (πα, l_{1}), (π, l_{1}(x + e)) in the corresponding cases, therefore (α, x^{2} − π^{k}u) is not isomorphic to them. If e ∼ 1, then (α, l_{1})_{x^{3}−π} ≁ (α, x^{2} − π^{k}u)_{x^{3}−π}. Finally, if e ≁ 1, then (π, x + e)_{l_{1}} ≁ (α, x^{2} − π^{k}u)_{l_{1}}.

16. f(0) ∼ −ev_{1}v_{2}u ∼ α. Thus the algebra (α, x) is unramified. We have also

(α, x)_{x^{2}−π^{k}u} ∼ (\frac{α , \sqrt{π^{k}u}}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{−α(x^{2} − π^{k}u)})}) ∼

∼ (\frac{α , \sqrt{πu}}{k(\sqrt{πu})⟨\sqrt{−α(x^{2} − π^{k}u)}⟩}) ≁ 1.

As it was already noticed, the other algebras are trivial after completion at x^{2} − π^{k}u.

17. The algebra (F π, x) is unramified by the choice of F. Let η ∈ O_{K}^{∗}, K be as above, η ≁ 1, η + e ∼ 1, and r = Irr_{K|k}(η). Then

(F π, x)_{r} ∼ (\frac{F π , η}{K⟨r⟩} ) ≁ 1,

since f(η) ∼ η+e ∼ 1. On the other hand, (α, l_{1})_{q} ∼ (α, η) ∼ 1 and (π, x+e)_{q} ∼ (π, η+e) ∼ 1. Let firstly m_{1} < m_{2} or v̄_{1} ≠ v̄_{2}. Then for w ∈ O_{K}^{∗}, w̄ ≠ v̄_{1}, v̄_{2} we have f(π^{m_{2}}w) ∼ −e(w − v_{2})(v_{1} − wπ^{m_{2}−m_{1}}). Let p = Irr_{K|k}(π^{m_{2}}w). Then (π, l_{1})_{p} ∼ (π, (−1)^{m_{1}+1}(v_{1} − wπ^{m_{2}−m_{1}})) and (F π, x)_{p} ∼ (π, (−F)^{m_{2}}w). We require

\{ \begin{matrix}−e(w − v_{2})(v_{1} − wπ^{m_{2}−m_{1}}) ∼ 1, \\ −F^{m_{2}}w(v_{1} − wπ^{m_{2}−m_{1}}) ≁ 1. \end{matrix}

Then (π, l_{1})_{p} ≁ (F π, x)_{p}. Similarly one can check (πα, l_{1}) ≁ (F π, x) and (π, l_{1}(x+e)) ≁ (F π, x). Consider now the case m_{1} = m_{2}, v_{2} = v_{1} + π^{r}μ, r > 0, μ ∈ O_{k}^{∗}. Suppose w ∈ O_{K}^{∗}, w̄ ≠ μ̄, s = Irr_{K|k}(π^{m_{2}}(v_{1} + π^{r}w)). Then f(π^{m_{2}}(v_{1} + π^{r}w)) ∼ ew(w − μ), (π, l_{1})_{s} ∼ (π, (−1)^{m_{1}+r}w) and (F π, x)_{s} ∼ (π, (−F)^{m_{1}}v_{1}). If we require

\{ \begin{matrix}ew(w − μ) ∼ 1, \\ (−1)^{r}F^{m_{1}}v_{1}w ≁ 1, \end{matrix}

then (π, l_{1})_{s} ≁ (F π, x)_{s}. For the remaining algebras the arguments are similar.

18. If w ∈ O_{k}^{∗}, w^{2} − u ≁ 1, then (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1 in view of f(π^{k/2}w) ∼ ev_{1}v_{2}(w^{2} − u) ∼ 1. Let η and r be from the previous case, then (π, l_{1})_{r} ≁ 1, (π, l_{1}(x + e))_{r} ≁ 1 and (πα, l_{1})_{r} ≁ 1, but (π, x^{2} − π^{k}u)_{r} ∼ 1. The extension by \sqrt{π} leaves us in the current case, so it does not split (α, l_{1}), provided e ∼ 1, but it splits (π, x^{2} − π^{k}u). Finally, (π, x + e)_{l_{1}} ≁ (π, x^{2} − π^{k}u)_{l_{1}} if e ≁ 1.

19. f(π^{k/2}w) ∼ ev_{1}v_{2}(w^{2} − u) ∼ α, hence the algebra (α, x − π^{k/2}w) is unramified. If

ε = \{ \begin{matrix}1, if m_{2} ≡ 0 (mod 2), \\ 2, if m_{2} ≡ 1 (mod 2), \end{matrix}

then f(π^{m_{2}+ε/3}) ∼ ev_{1}v_{2} ∼ 1. Assume p = Irr_{k( ^{3} \sqrt{π})|k}(π^{m_{2}+ε/3}). Then

(α, x − π^{k/2}w)_{p} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m_{2}+ε}}{k( ^{3} \sqrt{π})⟨p⟩} ) ≁ 1.

Since the extension by \sqrt{π} leaves us in case 19, it does not split the algebra (α, x − π^{k/2}w). Hence (α, x − π^{k/2}w) ≁ (π, l_{1}), (π, x + e), (π, l_{1}(x + e)). Since after this extension (πα, l_{1}) ∼ (α, l_{1}), it is enough to show that (α, x − π^{k/2}w) ≁ (α, l_{1}), provided e ∼ 1. Let p_{0} = Irr_{k( ^{3} \sqrt{π})|k}(π^{m_{2}+1/3}). Then f(π^{m_{2}+1/3}) ∼ 1 as above and

(α, x − π^{k/2}w)_{p_{0}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m_{2}+1}}{k( ^{3} \sqrt{π})⟨p_{0}⟩} ) , (α, l_{1})_{p_{0}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m_{1}}}{k( ^{3} \sqrt{π})⟨p_{0}⟩} ) .

Because of m_{1} ≡ m_{2} (mod 2) we have (α, x − π^{k/2}w)_{p_{0}} ≁ (α, l_{1})_{p_{0}} and the theorem is proved.

Without loss of generality one can assume that in case IX either 0 < k < 4 and k/2 ≤ m_{1} (case IX.1) or m_{1} = 1, k > 2 (case IX.2). Let l_{i} = x − π^{k_{i}}v_{i}, i = 1, 2, 3.

Theorem 15 Let f be as in case IX.1 and as soon as m_{1} = m_{2} = m_{3}, then v̄_{3} ≠ v̄_{i}, i = 1, 2. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

1. (α, x^{2} − π^{k}u), (α, l_{1}), (α, l_{2}), if

k ≡ 1 (mod 2), m_{1} ≡ 0 (mod 2).

2. (πE, p_{w}), (α, l_{1}), (α, l_{2}), if

k ≡ 0 (mod 2), m_{1} ≡ 1 (mod 2).

3. (πE, p_{w}), (α, l_{2}), (π, l_{2}), if

k ≡ m_{1} ≡ 0 (mod 2), m_{1} < m_{2}, uv_{1} ∼ 1.

4. (α, x^{2} − π^{k}u), (α, l_{2}), (π, l_{2}), if

k ≡ m_{1} ≡ 1 (mod 2), m_{1} < m_{2}, uv_{1} ∼ 1.

5. k ≡ m_{1} ≡ 1 (mod 2), and if m_{1} < m_{2}, then uv_{1} ≁ 1.

a) (α, x^{2} − π^{k}u), (π, l_{1}), (α, l_{1}), if m_{1} = m_{2}, v̄_{1} = v̄_{2} and ξ = −u(v_{1} − v_{3}π^{m_{3}−m_{1}}) ∼ 1.

b) (α, x^{2} − π^{k}u), (π, l_{1}), (π, l_{2}), otherwise.

6. k ≡ m_{1} ≡ 0 (mod 2), and if m_{1} < m_{2}, then uv_{1} ≁ 1.

a) (πE, p_{w}), (π, l_{1}), (α, l_{1}), if m_{1} = m_{2}, v̄_{1} = v̄_{2} and ξ ∼ 1.

b) (πE, p_{w}), (π, l_{1}), (π, l_{2}), otherwise,

where w ∈ O_{K}^{∗} satisfies the following conditions. If m_{1} > 1, then w ∈ O_{k}^{∗}, w^{2} − u ≁ 1, E = αw, and p_{w} = x − πw. If m_{1} = 1, then

• If 1 < m_{2}, then

\{ \begin{matrix}w^{2} − u ≁ 1, \\ w(w − v_{1}) ∼ 1. \end{matrix}

• If 1 = m_{2} < m_{3}, then

\{ \begin{matrix}(w^{2} − u)(w − v_{1})w ≁ 1, \\ (w^{2} − u)(w − v_{2})w ≁ 1, \\ (w^{2} − u)(w − v_{1})(w − v_{2}) ≁ 1. \end{matrix}

• If 1 = m_{2} = m_{3}, then

\{ \begin{matrix}(w^{2} − u)(w − v_{1})(w − v_{2}) ≁ 1, \\ (w^{2} − u)(w − v_{1})(w − v_{3}) ≁ 1, \\ (w^{2} − u)(w − v_{2})(w − v_{3}) ≁ 1 \end{matrix}

for a suitable unramified odd degree extension K|k, p_{w} = Irr_{K|k}(πw), E = N_{K|k}(E_{0}), and E_{0} ∈ O_{K}^{∗} is such that f(πw) ∼ πE_{0} in K.

It is convenient to divide the proof into several parts. We assume in the following lemmas that all conditions of the above theorem are satisfied.

Lemma 16 If k ≡ 1 (mod 2), then the algebra (α, x^{2} − π^{k}u) is unramified and nontrivial. Otherwise, the same is true for the algebra (πE, p_{w}), where w and p_{w} are from the above theorem.

We have (α, x^{2} − π^{k}u)_{l_{i}} ∼ (α, π^{k}) ≁ 1, since k < 2m_{i}, i = 1, 2, 3. If k = 2, m_{1} > 1, then f(πw) ∼ πw(w^{2} − u) ∼ πE and the algebra (πE, x − πw) is unramified. (πE, x − πw)_{l_{i}} ∼ (παw, −πw) ≁ 1. Let now k = 2, m_{1} = 1. Then

f(πw) ∼ π(w^{2} − u)(w − v_{1})(w − π^{m_{2}−1}v_{2})(w − π^{m_{3}−1}v_{3}).

Let w̄ ≠ v̄_{1} and if m_{i} = 1, then w̄ ≠ v̄_{i}. Then f(πw) ∼ πE_{0}, E_{0} ∈ O_{K}^{∗}, where

E_{0} = \{ \begin{matrix}(w^{2} − u)(w − v_{1}), if 1 < m_{2}, \\ (w^{2} − u)(w − v_{1})(w − v_{2})w, if 1 = m_{2} < m_{3}, \\ (w^{2} − u)(w − v_{1})(w − v_{2})(w − v_{3}), if 1 = m_{2} = m_{3}. \end{matrix}

We have

(πE, p_{w})_{p_{w}} ∼ (\frac{πE_{0} , p_{w}}{K(\sqrt{πE_{0}})⟨p_{w}⟩}) ∼ 1,

since E = N_{K|k}(E_{0}) ∼ E_{0} in K. So the algebra (πE, p_{w}) is unramified. Let us require (πE, p_{w})_{l_{i}} ≁ 1, i = 1, 2, 3. Then

(πE, p_{w})_{l_{i}} ∼ (π, N_{K|k}(E_{0}(w − v_{i}))),
(πE, p_{w})_{l_{2}} ∼ (π, N_{K|k}(E_{0}(w − v_{1}π^{m_{i}−1}))), i = 2, 3,

and these algebras are nontrivial iff E_{0}(w − v_{1}) ≁ 1 and E_{0}(w − v_{i}) ≁ 1, provided m_{i} = 1 or E_{0}w ≁ 1, provided m_{i} > 1, i = 2, 3. It just gives the conditions on w from the theorem. These conditions can always be satisfied by lemma 6. The lemma is proved.

Lemma 17 If k ≢ m_{1} (mod 2), then the classes of the algebras (π, α), (α, l_{1}), (α, l_{2}) together with the class of the algebra from lemma 16 give all generators of _{2}Br C.

Proof. Let firstly k/2 < m_{1}. Then (α, l_{1})_{l_{1}} ∼ (α, (x^{2} − π^{k}u)l_{2}l_{3})_{l_{1}} ∼ (α, π^{k+m_{1}}l_{2}(π^{m_{1}}v_{1})) ∼ (α, πl_{2}(π^{m_{1}}v_{1})) and (α, l_{1})_{l_{2}} ∼ (α, l_{2}(π^{m_{1}}v_{1})). Thus exactly one of the algebras (α, l_{1})_{l_{1}} and (α, l_{1})_{l_{2}} is nontrivial. Since the algebra from lemma 16 is nontrivial at l_{1}, l_{2}, the algebra (α, l_{1}) is not isomorphic to it. Let now k = 2, m_{1} = 1. Then (α, l_{1})_{l_{1}} ∼ (α, πl_{2}(π^{m_{1}}v_{1})) ≁ (α, l_{1})_{l_{2}} again.

Consider the algebras (α, l_{2}) and (α, l_{1}l_{2}). We have

(α, l_{2})_{l_{1}} ∼ (α, l_{2}(π^{m_{1}}v_{1})), (α, l_{2})_{l_{2}} ∼ (α, π^{k+m_{2}}l_{2}(π^{m_{1}}v_{1})), (α, l_{2})_{l_{3}} ∼ (α, π^{m_{2}}),
(α, l_{1}l_{2})_{l_{2}} ∼ (α, (x^{2} − π^{k}u)l_{3})_{l_{2}} ∼ (α, π^{k+m_{2}}), (α, l_{1}l_{2})_{l_{3}} ∼ (α, π^{m_{1}+m_{2}}).

Hence (α, l_{1}l_{2})_{l_{2}} ≁ (α, l_{1}l_{2})_{l_{3}}. This means that (α, l_{1}l_{2}) ≁ 1 and (α, l_{1}l_{2}) is not isomorphic to the algebra from lemma 16. Moreover, if m_{1} ≡ m_{2} (mod 2), then (α, l_{2})_{l_{2}} ∼ (α, πl_{2}(π^{m_{1}}v_{1})) ≁ (α, l_{2})_{l_{1}} and (α, l_{2})_{l_{2}} ∼ (α, l_{2}(π^{m_{1}}v_{1})) ∼ (α, π^{m_{1}}) ≁ (α, l_{2})_{l_{3}} otherwise, since in this case m_{1} < m_{2}. The lemma is proved.

Lemma 18 If k ≡ m_{1} (mod 2), m_{1} < m_{2}, uv_{1} ∼ 1, then the classes of the algebras (π, α), (α, l_{2}), (π, l_{2}) together with the class of the algebra from lemma 16 give all generators of _{2}Br C.

Proof. Let ε ∈ \{1, 2\}, τ ∈ O_{k}^{∗}, θ = π^{m_{1}}( ^{3} \sqrt{π})^{ε}τ, and p = Irr_{k( ^{3} \sqrt{π})|k}(θ). Then f(θ) ∼ π^{k+m_{1}}uv_{1} ∼ 1. Assume A ∈ {π, α, πα}. Then

(A, l_{2})_{p} ∼ (\frac{A , θ}{k( ^{3} \sqrt{π})⟨p⟩}) ∼ (\frac{A , ( ^{3} \sqrt{π})^{m_{1}+ε}τ}{k( ^{3} \sqrt{π})⟨p⟩} ) ,

and for any A we are always able to find ε and τ such that (A, l_{2})_{p} ≁ 1. Let now compare the algebras (A, l_{2}) with the algebra from lemma 16. Assume that ε and τ are such that (A, l_{2})_{p} ∼ 1. In view of k/2 < m_{1} + ε/3 and k/2 < m_{2} we have that the algebra from lemma 16 is nontrivial at p because it is nontrivial at l_{2}. This completes the proof of the lemma.

Lemma 19 Let k ≡ m_{1} (mod 2) and if m_{1} < m_{2}, then uv_{1} ≁ 1. Then the classes of the algebras (π, α), (π, l_{1}), the class of the algebra from lemma 16 and the class of the following algebra give all generators of _{2}Br C.

• (π, l_{2}), if m_{1} < m_{2} or v̄_{1} ≠ v̄_{2}.

• (π, l_{2}), if m_{1} = m_{2}, v̄_{1} = v̄_{2} and ξ ≁ 1.

• (α, l_{1}), if m_{1} = m_{2}, v̄_{1} = v̄_{2} and ξ ∼ 1.

Proof. In this case k/2 < m_{1}. If m_{1} < m_{2}, then (π, l_{1})_{l_{1}} ∼ (π, −π^{k}u) and (π, l_{1})_{l_{2}} ∼ (π, −π^{m_{1}}v_{1}) ∼ (π, −π^{k}v_{1}). Since uv_{1} ≁ 1 we have that among the algebras (π, l_{1})_{l_{1}} and (π, l_{1})_{l_{2}} there is just one trivial algebra. So (π, l_{1}) ≁ 1 and (π, l_{1}) is not isomorphic to the algebra from lemma 16. Let now m_{1} = m_{2} and μ_{i} ∈ O_{K}^{∗}. Then f(π^{m_{1}}μ_{i}) ∼ −u(μ_{i} − v_{1})(μ_{i} − v_{2})(μ_{i} − v_{3}π^{m_{3}−m_{1}}). If p_{i} = Irr_{K|k}(π^{m_{1}}μ_{i}), then (π, l_{1})_{p_{i}} ∼ (π, (−1)^{m_{1}}(μ_{i} − v_{1})). One can find K, μ_{1}, and μ_{2} such that f(π^{m_{1}}μ_{1}) ∼ f(π^{m_{1}}μ_{2}) ∼ 1 and (π, l_{1})_{p_{1}} ≁ 1, (π, l_{1})_{p_{2}} ∼ 1. Since the algebra from lemma 16 is not trivial at p_{2} we have again that (π, l_{1}) ≁ 1 and (π, l_{1}) is not isomorphic to this algebra.

Let us describe the last generator. Consider firstly the case m_{1} < m_{2}. Then f(π^{m_{1}}μ_{i}) ∼ −u(μ_{i} − v_{1}). If μ_{i} and p_{i} are such that −u(μ_{i} − v_{1}) ∼ 1, i = 1, 2, 3, 4, (−1)^{m_{1}}μ_{1} ∼ 1, (−1)^{m_{1}}μ_{2} ≁ 1, μ_{3} ≁ −u, and μ_{4} ∼ −u, then (π, l_{2})_{p_{1}} ∼ (π, (−1)^{m_{1}}μ) ∼ 1, (π, l_{2})_{p_{2}} ≁ 1, (π, l_{1}l_{2})_{p_{3}} ∼ (π, μ_{3}(μ_{3} − v_{1})) ∼ (π, −uμ_{3}) ≁ 1 and (π, l_{1}l_{2})_{p_{4}} ∼ 1. Thus (π, l_{2}) ≁ 1, (π, l_{2}) ≁ (π, l_{1}), (π, l_{2}) is not isomorphic to the algebra from lemma 16 and (π, l_{1}l_{2}) is not.

Let m_{1} = m_{2}. Then (π, l_{2}) ≁ 1 and (π, l_{2}) is not isomorphic to the algebra from lemma 16 in view of the symmetry for the algebras (π, l_{1}) and (π, l_{2}). Let v̄_{1} ≠ v̄_{2}. Then one can find μ_{1}, μ_{2} such that f(π^{m_{1}}μ_{1}) ∼ 1, i = 1, 2 and (μ_{1} −v_{1})(μ_{1} −v_{2}) ≁ 1, (μ_{2} −v_{1})(μ_{2} −v_{2}) ∼ 1. This means that (π, l_{1}l_{2})_{p_{1}} ≁ 1 and (π, l_{1}l_{2})_{p_{2}} ∼ 1.

Let now v_{2} = v_{1} + π^{s}τ, q = Irr_{K|k}(π^{m_{1}}(v_{1} + π^{s}η)), η ∈ O_{K}^{∗}. Then f(π^{m_{1}}(v_{1} + π^{s}η)) ∼ ξη(η − τ). If ξ ≁ 1, then we require η(η − τ) ≁ 1. In this case f(π^{m_{1}}(v_{1} + π^{s}η)) ∼ 1, (π, l_{1}l_{2})_{q} ∼ (π, η(η − τ)) ≁ 1. Moreover, (π, l_{1}l_{2})_{l_{3}} ∼ (π, (π^{m_{3}−m_{1}}v_{3} − v_{1})(π^{m_{3}−m_{1}}v_{3} − v_{2})) ∼ 1.

Finally, let ξ ∼ 1. After the replacement of prime element ν = πα one can check that the algebra (πα, l_{1}) is nontrivial and not isomorphic to the algebra from lemma 16. To complete the proof it is sufficient to show that the same is true for the algebra (α, l_{1}). If θ_{i} = π^{m_{1}}(v_{1} + ( ^{3} \sqrt{π})^{i}), i = 1, 2, t_{i} = Irr_{k( ^{3} \sqrt{π})|k}(θ_{i}), then f(θ_{i}) ∼ ξ ∼ 1, and

(α, l_{1})_{t_{i}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m_{1}+i}}{k( ^{3} \sqrt{π})⟨t_{i}⟩} ) .

So among the algebras (α, l_{1})_{t_{i}}, i = 1, 2 there is exactly one nontrivial algebra. Since the algebra from lemma 16 is nontrivial at t_{1} and t_{2}, the lemma is proved.

Now the proof of the theorem follows immediately from lemmas 16–19.

Let us consider the remaining subcase of case IX.1. In this subcase one can assume that f(x) = (x^{2} − π^{k}u)(x − π^{m}v)(x − π^{m}v_{1})(x − π^{m}v_{2}), v_{i} = v + π^{s_{i}}τ_{i}, τ_{i} ∈ O_{k}^{∗}, i = 1, 2, 0 < s_{1} ≤ s_{2} and if s_{1} = s_{2}, then τ̄_{1} ≠ τ̄_{2}. Let l = x − π^{m}v, l_{i} = x − π^{m}v_{i}, and

ξ = \{ \begin{matrix}uτ_{1}, if k/2 < m_{1}, \\ (u − v^{2})τ_{1}, otherwise. \end{matrix}

Theorem 16 Let f be as in case IX.1, m_{1} = m_{2} = m_{3}, v̄_{1} = v̄_{2} = v̄_{3}. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

• A = (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2).

• A = (πE, x − πw), if k = 2,

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1,

E = \{ \begin{matrix}αw, if m > 1, \\ α(w − v), if m = 1, \end{matrix}

and

1. (α, l), (α, l_{1}), if

k + m + s_{1} ≡ 1 (mod 2).

2. (π, l), (π, l_{1}), if

k + m + s_{1} ≡ 0 (mod 2), s_{1} = s_{2}.

3. (π, l), (π, l_{2}), if

k + m + s_{1} ≡ 0 (mod 2), s_{1} < s_{2}, ξ ≁ 1.

4. (π, l), (α, l), if

k + m + s_{1} ≡ 0 (mod 2), s_{1} < s_{2}, ξ ∼ 1,

Proof. If k is odd, then (α, x^{2} −π^{k}u)_{l} ≁ 1 and (α, x^{2} −π^{k}u)_{l_{i}} ≁ 1, i = 1, 2. Let k = 2. Then f(πw) ∼ πE and the algebra A is unramified. We also have (πE, x − πw)_{l} ∼ (π, w^{2} − u) ≁ 1. By the same argument (πE, x − πw)_{l_{i}} ≁ 1. So A is nontrivial at l, l_{1}, l_{2} in any case.

Observe that (α, l)_{l} ∼ (α, π^{k+s_{1}+s_{2}}), (α, l)_{l_{1}} ∼ (α, π^{m+s_{1}}), (α, l)_{l_{2}} ∼ (α, π^{m+s_{2}}), (α, l_{1})_{l} ∼ (α, π^{m+s_{1}}), (α, l_{1})_{l_{1}} ∼ (α, π^{k}), (α, l_{1})_{l_{2}} ∼ (α, π^{m+s_{1}}). Let k + m + s_{1} ≡ 1 (mod 2). Then among the algebras (α, l)_{l} and (α, l)_{l_{2}} there exists exactly one nontrivial algebra, so that (α, l) ≁ 1 and (α, l) ≁ A. The same is true for the algebras (α, l_{1})_{l_{1}} and (α, l_{1})_{l_{2}}. Finally, (α, ll_{1})_{l} ∼ (α, π^{s_{1}+s_{2}+1}) and (α, ll_{1})_{l_{2}} ∼ (α, π^{s_{1}+s_{2}}).

Let now k + m + s_{1} ≡ 0 (mod 2), θ_{i} = π^{m}(v + π^{s_{1}}η_{i}), η_{i} ∈ O_{K}^{∗}, p_{i} = Irr_{K|k}(θ_{i}),

δ = \{ \begin{matrix}−u, if k/2 < m, \\ v^{2} − u, otherwise. \end{matrix}

Then

f(θ_{i}) ∼ \{ \begin{matrix}δ(η_{i} − τ_{1}), if s_{1} < s_{2}, \\ δη_{i}(η_{i} − τ_{1})(η_{i} − τ_{2}) otherwise. \end{matrix}

Let us require f(θ_{i}) ∼ 1, then (π, l)_{p_{i}} ∼ (π, (−1)^{k}η_{i}). One can check that A_{p_{i}} ≁ 1. There exist η_{1}, η_{2} such that (−1)^{k}η_{1} ≁ 1, (−1)^{k}η_{2} ∼ 1. Then (π, l)_{p_{1}} ≁ 1 and (π, l)_{p_{2}} ≁ A_{p_{2}}. Consider the case s_{1} = s_{2}. We have (π, l_{1})_{p_{i}} ∼ (π, (−1)^{k}(η_{i} − τ_{1})). One can always find an element η_{i} such that

\{ \begin{matrix}(−1)^{k}(η_{1} − τ_{1}) ∼ 1, \\ δ(−1)^{k}η_{1}(η_{1} − τ_{2}) ∼ 1 \end{matrix} ⟹ (π, l_{1})_{p_{1}} ≁ A_{p_{1}}.

\{ \begin{matrix}(−1)^{k}(η_{2} − τ_{1}) ≁ 1, \\ δ(−1)^{k}η_{2}(η_{2} − τ_{2}) ≁ 1 \end{matrix} ⟹ (π, l_{1})_{p_{2}} ≁ 1.

\{ \begin{matrix}η_{3}(η_{3} − τ_{1}) ≁ 1, \\ δ(η_{3} − τ_{2}) ≁ 1 \end{matrix} ⟹ (π, l_{1})_{p_{3}} ≁ (π, l_{1})_{p_{3}}.

\{ \begin{matrix}η_{4}(η_{4} − τ_{1}) ∼ 1, \\ δ(η_{4} − τ_{2}) ∼ 1 \end{matrix} ⟹ (π, l_{1})_{p_{4}} ≁ A_{p_{4}} ⊗ (π, l_{1})_{p_{4}}.

Let now k + m + s_{1} ≡ 0 (mod 2), s_{1} < s_{2}, ξ ≁ 1. Consider the algebra (π, l_{2}). If v′ = v + π^{s_{2}}τ_{2}, then f(x) = (x^{2} − π^{k}u)(x − π^{m}v′)(x − π^{m}(v′ + π^{s_{1}}τ′_{1}))(x − π^{m}(v′ + π^{s_{2}}τ′_{2})). So (π, l_{2}) = (π, l′) ≁ 1, A since we are in the case k + m + s_{1} ≡ 0 (mod 2) again. Furthermore, (π, ll_{2})_{l_{1}} ∼ (π, (π^{m+s_{1}}τ_{1})^{2}) ∼ 1, therefore (π, ll_{2}) ≁ A. We have also (π, ll_{2})_{l} ∼ (π, ξ) ≁ 1.

Finally, let k + m + s_{1} ≡ 0 (mod 2), s_{1} < s_{2}, ξ ∼ 1. If θ_{i} = π^{m}(v + π^{s_{1}}( ^{3} \sqrt{π})^{i}), q_{i} = Irr_{k( ^{3} \sqrt{π})|k}(θ_{i}), i = 1, 2, then f(θ) ∼ ξ ∼ 1 and

(α, l)_{q_{i}} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m+s_{1}+i}}{k( ^{3} \sqrt{π})⟨q_{i}⟩} ) , A_{q_{i}} ≁ 1.

Thus (α, l) ≁ 1, A. To finish the proof one only need to check that (πα, l) ≁ 1, A. But it is evident after the replacing the prime element by ν = απ.

Theorem 17 Let f be as in case IX.2 Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

1. k/2 ≤ m_{2}. Two generators are (α, l_{1}), (α, l_{2}). The third one is a) (π, l_{2}) if k ≡ 1 (mod 2),

b) (πE, x − π^{k/2}w) otherwise.

2. m_{2} < k/2 ≤ m_{3}. Two generators are (α, l_{1}), (α, l_{2}). The third one is a) (α, x^{2} − π^{k}u) if k ≡ 1 (mod 2),

b) (π, l_{3}) if k ≡ 0 (mod 2), m_{2} + k/2 ≡ 1 (mod 2),

c) (πE, x − π^{k/2}w) if k ≡ 0 (mod 2), m_{2} + k/2 ≡ 0 (mod 2),

3. m_{3} < k/2 and if m_{2} = m_{3} = 1,then v̄_{3} ≠ v̄_{1}, v̄_{3} ≠ v̄_{2}. Two generators are (α, l_{1}), (α, l_{2}). The third one is

a) (πE, x − π^{k/2}w) if m_{2} + m_{3} ≡ k ≡ 0 (mod 2),

b) (α, x − π^{k/2}w) if m_{2} + m_{3} ≡ 1 (mod 2), k ≡ 0 (mod 2) and −v_{1}v_{2}v_{3} ∼ 1,

c) (π, x^{2} − π^{k}u) if m_{2} + m_{3} ≡ 1 (mod 2), k ≡ 0 (mod 2) and −v_{1}v_{2}v_{3} ≁ 1,

d) (α, x^{2} − π^{k}u) if m_{2} + m_{3} ≡ 0 (mod 2), k ≡ 1 (mod 2) and uv_{1}v_{2}v_{3} ∼ 1,

e) (f(0), x), otherwise.

4. m_{2} = m_{3} = 1 < k/2, v_{i} = v_{1} +π^{s_{i}}τ_{i}, i = 2, 3, s_{2} ≤ s_{3} and if s_{1} = s_{2}, then τ̄_{2} ≠ τ̄_{3}. Two generators are

a) (α, l_{1}), (α, l_{2}) if s_{2} ≡ 0 (mod 2),

b) (π, l_{1}), (π, l_{2}) if s_{2} = s_{3} ≡ 1 (mod 2),

c) (π, l_{1}), (α, l_{1}) if s_{2} ≡ 1 (mod 2), s_{2} < s_{3} and −τ_{2} ∼ 1,

d) (π, l_{1}), (π, l_{3}) if s_{2} ≡ 1 (mod 2), s_{2} < s_{3} and −τ_{2} ≁ 1. The third one is α) (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2), uv_{1} ∼ 1, β) (α, x), if k ≡ 1 (mod 2), uv_{1} ≁ 1, γ) (πE, x − π^{k/2}w), if k ≡ 0 (mod 2),

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1, and E ∈ O_{k}^{∗} can be found from the condition f(π^{k/2}w) ∼ πE.

Proof.

1. We have (α, l_{1})_{l_{2}} ∼ (α, −πv_{1}) ≁ 1, (α, l_{2})_{l_{1}} ∼ (α, πv_{1}) ≁ 1 and (α, l_{1}l_{2})_{l_{1}} ∼ (α, (x^{2} − π^{k}u)l_{3})_{l_{1}} ∼ (α, (πv_{1})^{2}πv_{1}) ≁ 1.

Assume firstly k ≡ 1 (mod 2). Suppose ε ∈ \{1, 2\}, ε ≡ m_{2} (mod 2), θ = α( ^{3} \sqrt{π})^{3m_{2}−ε} and p = Irr_{k( ^{3} \sqrt{π})|k}(θ). Then 3k/2 < v_{k( ^{3} \sqrt{π})}(θ) < 3m_{2}. Let uv_{1} ∼ 1. Then we have f(θ) ∼ uv_{1} ∼ 1. Thus

(π, l_{2})_{p} ∼ (\frac{α , θ_{1}}{k( ^{3} \sqrt{π})⟨p⟩}) ∼ (\frac{^{3} \sqrt{π} , αd^{2}}{k( ^{3} \sqrt{π})⟨p⟩}) ≁ 1.

So (π, l_{2}) ≁ 1. If now uv_{1} is arbitrary, then after the extension by \sqrt{α} we are still in subcase 1.a and uv_{1} ∼ 1, therefore the algebra (π, l_{2}) is nontrivial again and it does not split by \sqrt{α}. Hence (π, l_{2}) ≁ (α, l_{1}), (α, l_{2}), (α, l_{1}l_{2}).

Let now k ≡ 0 (mod 2). We have

(πE, x − π^{k/2}w)_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})}) ≁ 1.

Since (α, l_{i})_{x^{2}−π^{k}u} ∼ 1 and (α, l_{1}l_{2})_{x^{2}−π^{k}u} ∼ 1, we are done.

2. (α, l_{1})_{l_{3}} ≁ 1, (α, l_{1}l_{2})_{l_{1}} ≁ 1. If m_{2} > 1, then (α, l_{2})_{l_{1}} ≁ 1. If m_{2} = 1, then (α, l_{2})_{l_{3}} ≁ 1. Let us find the last algebra. Consider the case k ≡ 1 (mod 2). (α, x^{2} − π^{k}u)_{l_{3}} ≁ 1. We have also (α, x^{2} − π^{k}u)_{l_{1}} ∼ 1, (α, l_{1}l_{1})_{l_{1}} ≁ 1. If either m_{2} > 1 or v̄_{1} ≠ v̄_{2}, then (α, l_{1})_{l_{2}} ≁ (α, x^{2}−π^{k}u)_{l_{2}} and (α, l_{2})_{l_{1}} ≁ (α, x^{2}−π^{k}u)_{l_{1}}. Let m_{2} = 1 and v̄_{1} = v̄_{2}. Suppose θ = π ^{3} \sqrt{π}. Then f(θ) ∼ v_{1}v_{2} ∼ 1. So

(α, l_{1})_{x^{3}−π^{4}} ∼ (\frac{α , −πv_{1}}{k( ^{3} \sqrt{π})⟨x^{3} − π^{4}⟩}) ≁ 1, (α, l_{2})_{x^{3}−π^{4}} ≁ 1

but

(α, x^{2} − π^{k}u)_{x^{3}−π^{4}} ∼ (α, θ^{2}) ∼ 1.

Let now k ≡ 0 (mod 2) and m_{2} + k/2 ≡ 0 (mod 2). Then

(πE, x − π^{k/2}w)_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩}) ≁ 1.

To complete the consideration of subcase 2.c it is enough to observe that all other algebras split at x^{2} − π^{k}u.

Finally, if m_{2} + k/2 ≡ 1 (mod 2), then f(π^{k/2}w_{i}) ∼ v_{1}v_{2}(w_{i}^{2} − u)(w − π^{m_{3}−k/2}v_{3}), where w_{i} ∈ O_{K}^{∗} and w̄_{i} ≠ v̄_{3}, provided m_{3} = k/2. For p_{i} = Irr_{K|k}(π^{k/2}w_{i}) we have (π, l_{3})_{p_{i}} ∼ (π, (−1)^{k/2}(w − π^{m_{3}−k/2}v_{3})). Let us require

\{ \begin{matrix}(−1)^{k/2}(w_{1} − π^{m_{3}−k/2}v_{3}) ≁ 1, \\ (−1)^{k/2}v_{1}v_{2}(w_{1}^{2} − u) ≁ 1, \end{matrix} \{ \begin{matrix}(−1)^{k/2}(w_{2} − π^{m_{3}−k/2}v_{3}) ∼ 1, \\ (−1)^{k/2}v_{1}v_{2}(w_{2}^{2} − u) ∼ 1. \end{matrix}

Then (π, l_{3})_{p_{1}} ≁ 1, (π, l_{3})_{p_{2}} ∼ 1. Furthermore, the algebras (π, l_{1})_{p_{1}} and (π, l_{1})_{p_{2}} are either both trivial or both nontrivial. The same is true for the algebras (π, l_{2})_{p_{1}}, (π, l_{2})_{p_{2}} and for the algebras (π, l_{1}l_{2})_{p_{1}}, (π, l_{1}l_{2})_{p_{2}}. So (π, l_{3}) is not isomorphic to (π, l_{i}) and (π, l_{1}l_{2}).

3. We have (α, l_{1})_{l_{3}} ≁ 1, (α, l_{2})_{l_{3}} ≁ 1 and (α, l_{1}l_{2})_{l_{1}} ≁ 1. If m_{2} + m_{3} and k are even, then one can argue as in subcase 2.c.

Let m_{2} + m_{3} ≡ 1 (mod 2), k ≡ 0 (mod 2) and −v_{1}v_{2}v_{3} ∼ 1. Then f(π^{k/2}w) ∼ −v_{1}v_{2}v_{3}(w^{2} − u) ∼ α, so the algebra (α, x − π^{k/2}w) is unramified. Then f(θ_{i}) ∼ θ_{i}^{2} ∼ 1 and

(α, x − π^{k/2}w)_{p_{i}} ∼ (\frac{α , θ_{i}}{k( ^{3} \sqrt{π})⟨p_{i}⟩}) ,

therefore there is just one nontrivial algebra among (α, x − π^{k/2}w)_{p_{i}}, i = 1, 2. In since the algebras (α, l_{1})_{p_{1}}, (α, l_{1})_{p_{2}} are either simultaneously trivial or simultaneously nontrivial, we have (α, x − π^{k/2}w) ≁ (α, l_{1}). By analogy (α, x − π^{k/2}w) ≁ (α, l_{2}), (α, l_{1}l_{2}).

Let now −v_{1}v_{2}v_{3} ≁ 1. We have f(π^{k/2}w) ∼ −v_{1}v_{2}v_{3}(w^{2} − u) ∼ 1 and (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ∼ (π, w^{2} − u) ≁ 1. For any i (π, x^{2} − π^{k}u)_{l_{i}} ∼ 1. On the other hand, (α, l_{1})_{l_{3}} ≁ 1, (α, l_{1}l_{2})_{l_{3}} ≁ 1 and either (α, l_{2})_{l_{1}} ≁ 1 or (α, l_{2})_{l_{3}} ≁ 1.

The case of even k is done. Let k be odd. Then f(0) ∼ uv_{1}v_{2}v_{3}π^{m_{2}+m_{3}}. If we are in subcase 3.d, then f(0) ∼ 1 and (α, x^{2} − π^{k}u)_{x} ≁ 1. Moreover, (α, x^{2} − π^{k}u) ≁ (α, l_{i}), (α, l_{1}l_{2}) as in the previous subcase.

Finally, in 3.e the algebra (f(0), x) is unramified. Assume firstly that m_{2} + m_{3} ≡ 0 (mod 2), so f(0) ∼ α,

(α, x)_{x^{2}−π^{k}u} ∼ (\frac{α , \sqrt{πu}}{k(\sqrt{πu})⟨−α(x^{2} − π^{k}u)⟩}) ≁ 1.

On the other hand,

(α, l_{1})_{x^{2}−π^{k}u} ∼ (\frac{α , −πv_{1}}{k(\sqrt{πu})⟨−α(x^{2} − π^{k}u)⟩}) ∼ 1 and (α, l_{2})_{x^{2}−π^{k}u} ∼ (α, l_{1}l_{2})_{x^{2}−π^{k}u} ∼ 1.

Let now m_{2} + m_{3} ≡ 1 (mod 2). To finish the consideration of case 3 it is sufficient to prove that the algebra (πuv_{1}v_{2}v_{3}, x) does not split after the extension by \sqrt{α}. This extension leaves us in subcase 3.e, so we can assume that −1 ∼ u ∼ v_{i} ∼ 1. If θ = π^{m_{3}} ^{3} \sqrt{π}w and p = Irr_{k( ^{3} \sqrt{π})|k}(θ), then

(π, x)_{p} ∼ (\frac{^{3} \sqrt{π} , w}{k( ^{3} \sqrt{π})⟨p⟩}) ≁ 1,

since f(θ) ∼ −v_{1}v_{2}v_{3} ∼ 1.

4. Consider case s_{2} ≡ 0 (mod 2). We have (α, l_{1})_{l_{2}} ∼ (α, π^{s_{2}+1}) ≁ 1, (α, l_{2})_{l_{1}} ∼ (α, π^{s_{2}+1}) ≁ 1, and (α, l_{1}l_{2})_{l_{2}} ∼ (α, l_{3}(x^{2} − π^{k}u))_{l_{2}} ∼ (α, π^{s_{2}+1}) ≁ 1.

If s_{2} = s_{3} ≡ 1 (mod 2), let θ_{i} = π(v_{1} + π^{s_{2}}μ_{i}), μ_{i} ∈ O_{K}^{∗}, p_{i} = Irr_{K|k}(θ_{i}), then f(θ_{i}) ∼ μ_{i}(μ_{i} − τ_{2})(μ_{i} − τ_{3}),

(π, l_{1})_{p_{i}} ∼ (\frac{π , μ_{i}}{K(\sqrt{f(θ_{i})})⟨p_{i}⟩}) , (π, l_{2})_{p_{i}} ∼ (\frac{π , μ_{i} − τ_{2}}{K(\sqrt{f(θ_{i})})⟨p_{i}⟩}) ,

(π, l_{1}l_{2})_{p_{i}} ∼ (\frac{π , μ_{i}(μ_{i} − τ_{2})}{K(\sqrt{f(θ_{i})})⟨p_{i}⟩}) .

If we require

\{ \begin{matrix}μ_{1} ≁ 1, \\ (μ_{1} − τ_{2})(μ_{1} − τ_{3}) ≁ 1, \end{matrix} \{ \begin{matrix}μ_{2} − τ_{2} ≁ 1, \\ μ_{2}(μ_{2} − τ_{3}) ≁ 1, \end{matrix} \{ \begin{matrix}μ_{3}(μ_{3} − τ_{2}) ≁ 1, \\ μ_{3} − τ_{3} ≁ 1, \end{matrix}

then (π, l_{1})_{p_{1}} ≁ 1, (π, l_{2})_{p_{2}} ≁ 1, and (π, l_{1}l_{2})_{p_{3}} ≁ 1.

Let s_{1} be odd, s_{1} < s_{2} and θ = π(v_{1} + π^{s_{2}}μ), μ ∈ O_{K}^{∗}, p = Irr_{K|k}(θ), then f(θ) ∼ μ − τ_{2}, (π, l_{1})_{p} ∼ (π, μ). If

\{ \begin{matrix}μ ≁ 1, \\ μ − τ_{2} ≁ 1, \end{matrix}

then (π, l_{1})_{p} ≁ 1. Suppose η = π(v_{1} + π^{s_{2}} ^{3} \sqrt{π}), q = Irr_{k( ^{3} \sqrt{(}π))|k)}(η). In this case f(η) ∼ −τ_{2} and

(α, l_{1})_{q} ∼ (\frac{α , π^{s_{2}+1} ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})|k} ) ≁ 1,

provided −τ_{2} ∼ 1. After replacing the prime element ν = πα we have −τ′_{2} = −τ_{2}α^{−s_{2}−1} ∼ 1, therefore we are in the same subcase, hence (πα, l_{1}) = (ν, l_{1}) ≁ 1. Let now −τ_{2} ≁ 1. The replacement v′_{1} = v_{1} + π^{s_{3}}τ_{3} leaves us in the current subcase, so (π, l_{3}) ∼ (π, l′_{1}) ≁ 1. (π, l_{1}l_{3})_{l_{1}} ∼ (π, l_{2})_{l_{1}} ∼ (π, −τ_{2}) ≁ 1.

So, two generators are found. To find the third one we can argue just by the same way as in subcase 3. To complete the proof it remains only to show that the third generator does not belong to the group generated by the first two ones. This is true for the algebra from α), because it is trivial everywhere in this proof, where the algebras from the group, generated by algebras a)–d) are nontrivial. Finally, the algebras from β and γ) are nontrivial at x^{2} − π^{k}u, but the algebras from a)–d) are trivial there. The theorem is proved.

4 Splitting type (1,2,2).

This section is devoted to case deg f_{1} = deg f_{2} = 2, deg f_{3} = 1 and the reduction is bad.

The evident list of all cases under consideration according to the reduction type of f(x) is following (f(x) = (x^{2} − π^{k}u)((x − a)^{2} − δ)(x − e), k ≥ 0, u ∈ O_{k}^{∗}, a, δ, e ∈ O_{k}, π^{k}u, δ ≁ 1).

I. k = 0, a ∈ M_{k}, δ ∈ O_{k}^{∗}, ū = δ̄.

II. k > 0, δ ∈ O_{k}^{∗}, e ∈ O_{k}^{∗}.

III. k > 0, δ ∈ O_{k}^{∗}, e ∈ M_{k}.

IV. k > 0, δ ∈ M_{k}, a ∈ O_{k}^{∗}, e ∈ O_{k}^{∗}, ā ≠ ē, k ≤ v(δ).

V. k > 0, δ ∈ M_{k}, a ∈ O_{k}^{∗}, e ∈ M_{k}.

V I. k > 0, δ ∈ M_{k}, a ∈ M_{k}, e ∈ O_{k}^{∗}, k ≤ v(δ).

V II. k > 0, δ ∈ M_{k}, a ∈ M_{k}, e ∈ M_{k}, k ≤ v(δ).

Lemma 20

| _{2}Br C| = 8.

Theorem 18 Let f be as in case I. Then the group _{2}Br C is generated by the classes of the algebras (π, α), (π, x^{2} − u) and the class of the following algebra.

1. (π, x − e), if either e ∈ O_{k}^{∗}, e^{2} − u ≁ 1 or e ∈ M_{k}, −1 ∼ 1.

2. (α(x^{2} − u), x − w), otherwise,

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1.

Remark 1 The algebra (α(x^{2} − u), x − w) does not split by any quadratic extension of scalars.

Proof of the theorem. The algebra (π, x^{2} − u) is nontrivial by lemma 3. We have also

(π, x − e)_{x^{2}−u} ∼ (\frac{π , \sqrt{u} − e}{k(\sqrt{u})⟨x^{2} − u⟩(\sqrt{D(x^{2} − u)})}) , D ∈ k^{∗}.

The last algebra is nontrivial iff \sqrt{u} − e ≁ 1 in k(\sqrt{u}). And this is true if either e ∈ O_{k}^{∗}, e^{2} − u ≁ 1 or e ∈ M_{k}, −1 ∼ 1. The algebra (π, (x^{2} − u)(x − e)) is nontrivial by lemma 3 again.

Consider now the algebra A = (α(x^{2}−u), x−w). It is unramified because A_{x−w} ∼ (α(w^{2}−u), x−w) ∼ 1. Let θ = (1 + ^{3} \sqrt{π})\sqrt{u} and p = Irr_{k( ^{3} \sqrt{π},\sqrt{u})|k}(θ). Then f(θ) ∼ \sqrt{u} − e ∼ 1, provided we are in subcase 2. So

A_{p} ∼ (\frac{(1 + ^{3} \sqrt{π})^{2}u − u , \sqrt{u} − w}{k( ^{3} \sqrt{π}, \sqrt{u})⟨p⟩} ) ∼ (\frac{^{3} \sqrt{π} , \sqrt{u} − w}{k( ^{3} \sqrt{π}, \sqrt{u})⟨p⟩}) ≁ 1.

On the other hand,

(π, x^{2} − u)_{p} ∼ (\frac{^{3} \sqrt{π} , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π}, \sqrt{u})⟨p⟩}) ∼ 1.

The theorem is proved.

Theorem 19 Let f be as in cases II or III. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

• Case I. The first algebra is (π, x − e) and the second one is

1. (eξuπ, x), if k ≡ 1 (mod 2).

2. (π, x^{2} − π^{k}u), if k ≡ 0 (mod 2) and eξ ≁ 1.

3. (α, x − π^{k/2}w), if k ≡ 0 (mod 2) and eξ ∼ 1.

• Case II (e = π^{m}v).

1. (π, x − π^{m}v), (π, x^{2} − π^{k}u), if m ≤ k/2, m, k ≡ 0 (mod 2) and either ξv ≁ 1 or m = k/2.

2. (π, x − π^{m}v), (α, x − π^{k/2}w), if m ≤ k/2, m, k ≡ 0 (mod 2), ξv ∼ 1 and m < k/2.

3. (π, (x − a)^{2} − δ), (αξηπ, x − π^{k/2}w), if m ≤ k/2, k ≡ 0 (mod 2), m ≡ 1 (mod 2),

η = \{ \begin{matrix}v, if m < k/2, \\ v − w, if m = k/2. \end{matrix}

4. (π, x − π^{m}v), (α, x^{2} − π^{k}u), if m ≤ k/2, m, k ≡ 1 (mod 2), −ξuv ∼ 1.
5. (π, x − π^{m}v), (α, x), if m ≤ k/2, m, k ≡ 1 (mod 2), −ξuv ≁ 1.
6. (π, x − π^{m}v), (−ξuvπ, x), if m ≤ k/2, k ≡ 1 (mod 2), m ≡ 0 (mod 2).
7. (π, x − π^{m}v), (α, x^{2} − π^{k}u), if m > k/2, k ≡ 1 (mod 2).
8. (π, x − π^{m}v), (π, x^{2} − π^{k}u), if m > k/2, k ≡ 0 (mod 4).
9. (π, (x − a)^{2} − δ), (−ξwαπ, x − π^{k/2}w), if m > k/2, k ≡ 2 (mod 4).

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1 and

ξ = \{ \begin{matrix}δ, if a ∈ M_{k}, \\ δ − a^{2}, otherwise. \end{matrix}

Proof. In II and III (π, x − e) ≁ 1 by lemma 3. In II.1 we have f(0) ∼ −eξuπ, so the algebra (−eξuπ, x) is unramified. Let τ ∈ O_{K}^{∗} such that K|k is an unramified odd degree extension and the following conditions hold:

\{ \begin{matrix}τ ≁ 1, \\ τ − e ∼ 1, \\ (τ − a)^{2} − δ ∼ 1. \end{matrix}

Such K and τ exist in view of lemmas 5 and 6. Let p = Irr_{K|k}(τ). Then f(τ) ∼ 1. Thus

(π, x − e)_{p} ∼ (\frac{π , τ − e}{K⟨p⟩}) ∼ 1 and (−eξuπ, x)_{p} ∼ (\frac{π , τ}{K⟨p⟩}) ≁ 1.

Consider subcase II.2. We have f(wπ^{k/2}) ∼ eξ(w^{2} − u) ∼ 1, so (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1. Finally, (π, x − e) ≁ (π, x^{2} − π^{k}u) by lemma 3. If we are in II.3, then f(wπ^{k/2}) ∼ α, so that the algebra (α, x−wπ^{k/2}) is unramified. Let ε ∈ {1, 2}, ε ≡ k/2 (mod 2), θ = π^{k/2−1}( ^{3} \sqrt{π})^{ε} and p = Irr_{k( ^{3} \sqrt{π})|k}(θ). Then f(θ) ∼ eξ ∼ 1 and

(α, x − wπ^{k/2})_{p} ∼ (\frac{α , π^{k/2−1}( ^{3} \sqrt{π})^{ε}}{k( ^{3} \sqrt{π})⟨p⟩} ) ≁ 1.

To prove (π, x − e) ≁ (α, x − wπ^{k/2}) it is enough to observe that the extension by \sqrt{π} leaves us in subcase II.3, hence this extension does not split the algebra (α, x − wπ^{k/2}). In III the algebras (π, x − π^{m}v) and (π, (x − a)^{2} − δ) are nontrivial. III.1. Since f(π^{k/2}w) ∼ ξv ∼ 1, provided m < k/2, then (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1. Let now k/2 = m. Then we require

\{ \begin{matrix}τ^{2} − u ∼ α, \\ v − τ ∼ αξ, \end{matrix}

τ ∈ O_{K}^{∗}, K|k is unramified of odd degree, then f(π^{k/2}τ) ∼ 1 and (π, x^{2}−π^{k}u)_{Irr(π^{k/2}τ)} ≁ 1. Furthermore, (π, x^{2} − π^{k}u) ⊗ (π, x − π^{m}v) ∼ (π, (x − a)^{2} − δ) ≁ 1. III.2. f(π^{k/2}w) ∼ α, so that (α, x − π^{k/2}w) is unramified. Now we can argue as in II.3. III.3. In view of f(π^{k/2}w) ∼ ξη(w^{2} − u)π the algebra A = (ξη(w^{2} − u)π, x − π^{k/2}w) is unramified.

A_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ≁ 1, (D ∈ k^{∗}).

On the other hand,

(π, (x − a)^{2} − δ)_{x^{2}−π^{k}u} ∼ (\frac{π , −ξ}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ∼ 1.

III.4. We have f(0) ∼ −ξuv. Thus (α, x^{2} − π^{k}u)_{x} ≁ 1. Moreover, although the extension by \sqrt{α} leads us to case (1, 1, 1, 2), it does not split the algebra (π, x − π^{m}v). Therefore (α, x^{2} − π^{k}u) ≁ (π, x − π^{m}v). III.5. f(0) ∼ α, so the algebra (α, x) is unramified. (α, x)_{x−π^{m}v} ≁ 1. The further arguments are similar to the previous subcase. III.6. The algebra A = (−ξuvπ, x) is unramified. Let θ = π^{m}τ, τ ∈ O_{K}^{∗}, p = Irr_{K|k}(θ). Then f(θ) ∼ ξ(v − τ) and A_{p} ∼ (π, τ). If we require

\{ \begin{matrix}τ ≁ 1, \\ ξ(v − τ) ∼ 1, \end{matrix}

then A_{p} ≁ 1. And if we require

\{ \begin{matrix}τ ≁ −ξ, \\ ξ(v − τ) ∼ 1, \end{matrix}

then A_{p} ≁ (π, x − π^{m}v)_{p} since in this case (π, x − π^{m}v)_{p} ∼ (π, −ξ). III.7. (α, x^{2} − π^{k}u)_{x−π^{m}v} ≁ 1. Moreover, the extension by \sqrt{α} does not split the algebra (π, x − π^{m}v). III.8. We have f(π^{k/2}τ) ∼ −ξτ(τ^{2} − u). If τ^{2} − u ≁ 1 and −ξτ ≁ 1, then (π, x^{2} − π^{k}u)_{Irr(π^{k/2}τ)} ≁ 1. III.9. f(π^{k/2}w) ∼ −ξwαπ, so the algebra A = (−ξwαπ, x − π^{k/2}w) is unramified. Finally,

A_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(D(x^{2} − π^{k}u))}) ≁ 1,

but

(π, (x − a)^{2} − δ)_{x^{2}−π^{k}u} ∼ (\frac{π , −ξ}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(D(x^{2} − π^{k}u))}) ∼ 1.

The theorem is proved.

Theorem 20 Let f be as in case IV. f(x) = (x − e)(x^{2} − π^{k}u)((x − v)^{2} − π^{m}γ), e, u, v, γ ∈ O_{k}^{∗}, v̄ ≠ γ̄, 0 < k ≤ m. Then the group _{2}Br C is generated by the classes of the algebra (π, α) and the classes of the following algebras. The first algebra is

1. (π, x − e), if k ≡ 0 (mod 2), −e ≁ 1.

2. (α, x − π^{k/2}w), if k ≡ 0 (mod 2), −e ∼ 1.

3. (πue, x), if k ≡ 1 (mod 2).

The second one is

a. (π, (x − v)^{2} − π^{m}γ), if m ≡ 0 (mod 2), v − e ≁ 1.

b. (α, x − v − π^{m/2}τ), if m ≡ 0 (mod 2), v − e ∼ 1.

c. (π(e − v)γ, x − v), if m ≡ 1 (mod 2),

where w, τ ∈ O_{k}^{∗}, w^{2} − u ≁ 1, τ^{2} − γ ≁ 1.

Proof. If k ≡ 0 (mod 2), then f(π^{k/2}w) ∼ −eα. In case 1 we have f(π^{k/2}w) ∼ 1 and (π, x−e)_{x−π^{k/2}w} ∼ (π, −e) ≁ 1. In addition let us consider the algebra (π, x^{2}−π^{k}u). It is also nontrivial at this place. Indeed, (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ∼ (π, w^{2} − u) ≁ 1.

In case 2 f(π^{k/2}w) ∼ α and the algebra (α, x − π^{k/2}w) is unramified. Now one can prove (α, x − π^{k/2}w) ≁ 1 by analogy with the theorem 19, subcase II.3.

Let now k ≡ 1 (mod 2), μ ∈ O_{K}^{∗}, K be as above and

\{ \begin{matrix}μ ∼ αue, \\ u − μ ∼ αu. \end{matrix}

Assume θ = \sqrt{π^{k}μ} ∈ K(\sqrt{πμ}) and q = Irr_{K(\sqrt{πμ})|k}(θ). Then f(θ) ∼ e(u − μ)π ∼ eμ(u − μ) ∼ 1 and

(πue, x)_{q} ∼ (\frac{πue , θ}{K(\sqrt{πμ})⟨q⟩}) ∼ (\frac{ueτ , \sqrt{πμ}}{K(\sqrt{πμ})⟨q⟩}) ≁ 1.

After replacing x′ = x − v we have

f(x′) = (x′ − (e − v))((x′)^{2} − π^{m}γ)((x′ − (−v))^{2} − π^{k}u).

The above arguments gives that the algebras from items a, b, c are unramified and nontrivial. (Because nowhere in this arguments we used that k/2 < m.) To finish the proof it is enough only to check, that these algebras are not isomorphic to the algebras from items 1, 2, 3 in the corresponding subcases. 1.a. We have (π, x − e) ⊗ (π, (x − v)^{2} − π^{m}γ) ∼ (π, x^{2} − π^{k}u) ≁ 1. 2.b. If A is the algebra from 2 and A′ is the one from b, then A_{p} ≁ 1 and A′_{p} ∼ (α, −v) ∼ 1, where p is constructed by analogy with the proof of theorem 19, subcase II.3.. 1.b, 2.a. In view of symmetry one need to consider only one case, say, 2.a. In this case for p as above A_{p} ≁ 1, but (π, (x − v)^{2} − π^{m}γ)_{p} ∼ (π, v^{2}) ∼ 1. 3.a. (πue, x)_{q} ≁ 1, but (π, (x − v)^{2} − π^{m}γ)_{q} ∼ 1. 3.b. A′_{q} ∼ (α, −v) ∼ 1. 3.c. Finally, we have

(π(e − v)γ, x − v)_{q} ∼ (\frac{π(e − v)γ , −v}{K(\sqrt{πμ})⟨q⟩} ) ∼ (\frac{τ(e − v)γ , −v}{K(\sqrt{πμ})⟨q⟩} ) ∼ 1,

and this completes the proof.

Theorem 21 Let f be as in case V. f(x) = ((x−v)^{2} −π^{m}γ)(x−π^{r}δ)(x^{2} −π^{k}u) v, γ.δ, u ∈ O_{k}^{∗}, m, r, k > 0, π^{m}γ, π^{k}u ≁ 1. Then the group _{2}Br C is generated by the classes of the algebra (π, α) and the classes of the following algebras. The first algebra is

i) (π, (x − v)^{2} − π^{m}γ), if m ≡ 0 (mod 2), v ≁ 1.

ii) (α, x − v − π^{m/2}w), if m ≡ 0 (mod 2), v ∼ 1.

iii) (−vγπ, x − v), if m ≡ 1 (mod 2).

The second one is

a. (α, x − π^{r}δ), if k/2 < r, k ≡ 1 (mod 2).

b. (π, x − π^{r}δ), if k/2 < r, k ≡ 0 (mod 4).

c. (τ(τ^{2} − u)π, x − π^{k/2}τ), if k/2 < r, k ≡ 2 (mod 4).

d. (πδu, x), if k/2 > r, k ≡ 1 (mod 2), r ≡ 0 (mod 2).

e. (π, x − π^{r}δ), if k ≡ 0 (mod 2), r ≡ 0 (mod 2) and either k/2 = r or k/2 > r, −δ ≁ 1.

f. (α, x − π^{k/2}τ), if k ≡ 0 (mod 2), r ≡ 0 (mod 2), k/2 > r and −δ ∼ 1.

g. (ξαπ, x − π^{k/2}τ), if k ≡ 0 (mod 2), r ≡ 1 (mod 2), k/2 ≥ r,

ξ = \{ \begin{matrix}−δ, if k/2 > r, \\ τ − δ − a^{2}, if k/2 = r. \end{matrix}

h. (α, x − π^{r}δ), if k ≡ 1 (mod 2), r ≡ 1 (mod 2), k/2 > r and δu ∼ 1.

k. (α, x), if k ≡ 1 (mod 2), r ≡ 1 (mod 2), k/2 > r and δu ≁ 1,

where w, τ ∈ O_{k}^{∗}, w^{2} − γ ≁ 1, τ^{2} − u ≁ 1.

Proof. i. f(π^{m/2}w + v) ∼ v(w^{2} − u) ∼ 1, so (π, (x − v)^{2} − π^{m}γ)_{x−v−π^{m/2w}} ≁ 1. ii. f(π^{m/2}w + v) ∼ v(w^{2} − u) ∼ α, so the algebra (α, x − v − π^{m/2}w) is unramified. We have also f(v + ^{3} \sqrt{π}) ∼ 1, therefore

(α, x − v − π^{m/2}w)_{(x−v)^{3}−π} ∼ (\frac{α , ^{3} \sqrt{π}}{k( ^{3} \sqrt{π})⟨(x − v)^{3} − π⟩}) ≁ 1.

iii. f(v) ∼ −vγπ, (−vγπ, x − v) is unramified. As usual, let χ ∈ O_{K}^{∗} satisfy the conditions

\{ \begin{matrix}χ ∼ −αvγ, \\ χ − γ ∼ −αγ. \end{matrix}

Let θ = v + \sqrt{π^{m}χ}, q = Irr_{K(\sqrt{πχ})|k}(θ). Then f(θ) ∼ χv(χ − γ) ∼ 1, hence

(−vγπ, x − v)_{q} ∼ (\frac{−vγχ , \sqrt{πχ}}{K(\sqrt{πχ})⟨q⟩}) ≁ 1.

a. (α, x − π^{r}δ)_{x−π^{r}δ} ∼ (α, ((x − v)^{2} − π^{m}γ)(x^{2} − π^{k}u)_{x−π^{r}δ}) ∼ (α, −π^{k}u) ≁ 1. b. We have f(μπ^{k/2}) ∼ −μ(μ^{2} − u) for some μ ∈ O_{K}^{∗}. Thus (π, x − π^{r}δ)_{Irr(μπ^{k/2})} ∼ (π, μ) ≁ 1, provided

\{ \begin{matrix}μ ≁ 1, \\ u − μ^{2} ≁ 1. \end{matrix}

c. f(π^{k/2}τ) ∼ ταπ. Furthermore,

(ταπ, x − π^{k/2}τ)_{x^{2}−π^{k}u} ∼ (\frac{π , τ − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩}) ≁ 1.

d. Let ν ∈ O_{K}^{∗} satisfy the conditions

\{ \begin{matrix}ν ∼ αδu, \\ ν − u ∼ −αu, \end{matrix}

and t = Irr_{K(\sqrt{πν})|k}(\sqrt{π^{k}ν}). Then f(\sqrt{π^{k}ν}) ∼ −δν(ν − u) ∼ 1, so that

(πδu, x)_{t} ∼ (\frac{νδu , \sqrt{πν}}{K(\sqrt{πν})⟨t⟩}) ≁ 1.

e. If k/2 = r, then f(π^{k/2}λ) ∼ (λ − δ)(λ^{2} − u) for some λ ∈ O_{K}^{∗}. Moreover, if s = Irr_{K|k}(π^{k/2}λ), then

(π, x − π^{r}δ)_{s} ∼ (\frac{π , λ − δ}{K(\sqrt{(λ − δ)(λ^{2} − u)})⟨s⟩}) .

If we require

\{ \begin{matrix}λ − δ ≁ 1, \\ λ^{2} − u ≁ 1, \end{matrix}

then (π, x − π^{r}δ)_{s} ≁ 1.

Let now k/2 > r. We have f(π^{k/2}τ) ∼ −δ(τ^{2} − u) ∼ 1. So, (π, x − π^{r}δ)_{x−π^{k/2}τ} ∼ (π, −δ) ≁ 1. f. f(π^{k/2}τ) ∼ −δ(τ^{2} − u) ∼ α and the algebra (α, x − π^{k/2}τ) is unramified. To prove (α, x − π^{k/2}τ) ≁ 1 one can follow the proof of theorem 19, subcase II.3. g. By analogy with subcase c we have (ξαπ, x − π^{k/2}τ)_{x^{2}−π^{k}u} ≁ 1. h. f(0) ∼ π^{k+r}δu ∼ 1, therefore (α, x − π^{r}δ)_{x} ≁ 1. k. f(0) ∼ π^{k+r}δu ∼ α, hence the algebra (α, x) is unramified. Finally, (α, x)_{x−π^{r}δ} ≁ 1.

Now, to complete the proof one only need to check that the algebra from i)–iii) is not isomorphic to the algebra from a)–k). Note, that in all subcases a)–k) the nontriviality of the corresponding algebra is showed by the completion of this algebra at the place with prime element l ∈ k[x] such that l = Irr(θ), θ ∈ M_{k_{alg}}. Thus, in i) and ii) we have (π, (x−v)^{2}−π^{m}γ)_{l} ∼ (π, v^{2}) ∼ 1 and (α, x−v−π^{m/2}w)_{l} ∼ (α, −v) ∼ 1. The nontriviality of the algebra (−vγπ, x) from iii is proved by the completion at q = Irr_{K(\sqrt{πχ})|k}(θ), θ = v + \sqrt{π^{m}χ}. Any algebra from a)–k) completed in q can be obtained from some algebra over k by extending of scalars. So it is trivial at q, since the extension k(\sqrt{πχ}) splits any algebra over k. The theorem is proved.

Consider the next case.

Theorem 22 Let f be as in case V I. f(x) = (x − e)(x^{2} − π^{k}u)((x − π^{m}v)^{2} − π^{r}δ), u, v, δ, e ∈ O_{k}^{∗}, 0 < m, 0 < k ≤ r, π^{k}u, π^{r}δ ≁ 1. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

Let if k, r ≡ 1 (mod 2), then 2m ≤ k, and if k = r ≡ 0 (mod 2), 2m > k, then ū ≠ δ̄. Then the first algebra is

i. (πE, x), if k ≡ 1 (mod 2).

ii. (π, x^{2} − π^{k}u), if k ≡ 0 (mod 2) and −e ≁ 1.

iii. (α, x − π^{k/2}w), if k ≡ 0 (mod 2), either k < r or k = r > 2m, and −e ∼ 1.

iv. (α, Irr_{K|k}(π^{k/2}β)), if k ≡ 0 (mod 2), k = r ≤ 2m, −e ∼ 1, where β ∈ O_{K}^{∗} and

(β^{2} − u) × \{ \begin{matrix}β^{2} − δ, if k < 2m, \\ (β − v)^{2} − δ, if k = 2m. \end{matrix}

The second one is

a. (πF, x − π^{m}v), if r ≡ 1 (mod 2).

b. (π, (x − π^{m}v)^{2} − π^{r}δ) if r ≡ 0 (mod 2) and

• k = r and either k ≤ 2m or −e ≁ 1,

• 2m < k < r and −e ≁ 1,

• 2m = k < r and −e(v^{2} − u) ≁ 1,

• k < r, k < 2m, k ≡ 0 (mod 2), and ue ≁ 1.

c. (α, x − π^{m}v − π^{r/2}τ) if r ≡ 0 (mod 2) and

• k = r > 2m, −e ∼ 1,

• 2m < k < r and −e ∼ 1,

• 2m = k < r and −e(v^{2} − u) ∼ 1,

• k < r, k < 2m, k ≡ 0 (mod 2), and ue ∼ 1.

d. (ueN_{K|k}(γ^{2} − δ)π, Irr_{K|k}(π^{m}v + π^{r/2}γ)), if r ≡ 0 (mod 2), k < r, k < 2m and k ≡ 1 (mod 2), where γ ∈ O_{K}^{∗}, K|k is of odd degree and unramified,

γ^{2} − δ ≁ \{ \begin{matrix}1 , if 2m < r, \\ v^{2} − δ , if 2m = r, \\ −δ , if 2m > r. \end{matrix}

Otherwise we have 1. (π, x^{2} − π^{k}u), (α(x^{2} − π^{k}u), π^{k/2}(x − π^{k/2}w)), if k = r ≡ 0 (mod 2), 2m > k, ū = δ̄.

2. The first algebra: (euπ, x), the second one: either (α, x^{2} − π^{k}u), provided −ueδ ∼ 1, or (α, x − π^{m}v) otherwise, if k, r ≡ 1 (mod 2), k < 2m < r.

3. The first algebra: either (α, x^{2} − π^{k}u), provided −ueδ ∼ 1, or (α, x) otherwise, the second one: (πue, x − π^{s}), k/2 < s < r/2, if k, r ≡ 1 (mod 2), k < r < 2m.

4. The first algebra from 3, and (x, −πu(x^{2} − π^{k}u)), if k = r ≡ 1 (mod 2), k < 2m.

where E, F, w, tau ∈ O_{k}^{∗} can be found from the conditions f(0) ∼ πE, f(π^{m}v) ∼ πF, w^{2} − u ≁ 1, τ^{2} − δ ≁ 1.

Remark 2 The algebra (α(x^{2} −π^{k}u), π^{k/2}(x−π^{k/2}w)) from 1. does not split by any quadratic extension of scalars.

Proof of the theorem. i. If μ ∈ O_{K}^{∗}, then

f(\sqrt{π^{k}μ}) ∼ \{ \begin{matrix}−e(μ − u)(μ − δ) , if k = r < 2m, \\ −eμ(μ − u) , otherwise. \end{matrix}

Let t = Irr_{K(\sqrt{πμ})|k}(\sqrt{π^{k}μ}). If Eμ ≁ 1 in K and f(\sqrt{π^{k}μ}) ∼ 1 in K, then

(πE, x)_{t} ∼ (\frac{Eμ , \sqrt{πμ}}{K(\sqrt{πμ})⟨t⟩}) ≁ 1.

The conditions on K and μ always can be satisfied, provided that in case k = r < 2m we have ū ≠ δ̄. ii. We have

f(π^{k/2}ξ) ∼ \{ \begin{matrix}−e(ξ^{2} − u),  if k < r or k = r > 2m, \\ −e(ξ^{2} − u)(ξ^{2} − δ),  if k = r < 2m, \\ −e(ξ^{2} − u)((ξ − v)^{2} − δ), if k = r = 2m. \end{matrix}

Thus if k < r or k = r > 2m, then (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1. Let now k = r ≤ 2m. Then (π, x^{2} − π^{k}u)_{Irr_{K|k}(π^{k/2}ξ)} ≁ 1, provided ξ^{2} − u ≁ 1 and ξ^{2} − δ ∼ 1 if k < 2m, (ξ − v)^{2} − δ ∼ 1 if k = 2m. iii. To prove (α, x − π^{k/2}τ) ≁ 1 one can follow the proof of theorem 19, subcase II.3. iv. From the conditions on β we have the algebra to be unramified. The nontriviality can be proved as in iii. a. Similarly as in subcase i for t = Irr_{K(\sqrt{πμ})|k}(θ), θ = π^{m}v + \sqrt{π^{r}μ} we can find K and μ such that (πF, x − π^{m}v)_{t} ≁ 1. Indeed,

f(θ) ∼ \{ \begin{matrix}−eμ(μ − δ),  if 2m < k, \\ ueμ(μ − δ),  if 2m, r > k, \\ −eμ(μ − δ)(v^{2} − u), if k = 2m < r, \end{matrix}

and the conditions f(θ) ∼ 1 and F μ ≁ 1 can be satisfied in some unramified extension K|k of odd degree. b,c. By replacing x′ = x − π^{m}v case k = r is reduced to ii and iii. Furthermore,

f(π^{m}v + π^{r/2}τ) ∼ −eα × \{ \begin{matrix}1, if 2m < k, \\ v^{2} − u, if 2m = k, \\ −uπ^{k}, if 2m < k. \end{matrix}

So, in b the algebra (π, (x − π^{m}v)^{2} − π^{r}δ)_{x−π^{m}v+π^{r/2}τ} is nontrivial, and the algebra from c is unramified. The nontriviality of the last algebra can be checked by analogy with the proof of theorem 19, subcase II.3. Indeed, this algebra is nontrivial at Irr_{k( ^{3} \sqrt{π})|k}(θ), where θ = π^{m}v + π^{r/2−1}( ^{3} \sqrt{π})^{ε}, ε ∈ \{1, 2\}, ε ≡ r/2 (mod 2). d. We have f(π^{m}v + π^{r/2}γ) ∼ ueπ(γ^{2} − δ). Since K|k is of odd degree γ^{2} − δ ∼ N_{K|k}(γ^{2} − δ) in K, so that the algebra A = (ueN_{K|k}(γ^{2} − δ)π, Irr_{K|k}(π^{m}v + π^{r/2}γ)) is unramified. Let μ ∈ O_{K′}^{∗}, K′|k is of odd degree and unramified such that

\{ \begin{matrix}μ ≁ ue(γ^{2} − δ), \\ μ − u ≁ −u(γ^{2} − δ). \end{matrix}

Let t = Irr_{K′(\sqrt{πμ})|k}(\sqrt{π^{k}μ}). Then f(\sqrt{π^{k}μ}) ∼ −eμ(μ − u) ∼ 1 and

A_{t} ⊗ K ∼ (\frac{ue(γ^{2} − δ)μ , ∏_{σ∈GK|k}(\sqrt{π^{k}μ} − π^{m}v − π^{r/2}γ)}{KK′(\sqrt{πμ})⟨t⟩}) ∼ (\frac{α , \sqrt{πμ}}{KK′(\sqrt{πμ})⟨t⟩}) ≁ 1.

Let us check now that the algebra A from i − iv is not isomorphic to the algebra B from a)–d).

Let firstly k and r be odd. Then 2m < k, A = (πE, x), and A_{t} ∼ 1. On the other hand,

B_{t} ∼ (πF, x − π^{m}v)_{t} ∼ (\frac{πF , −π^{m}v}{K(\sqrt{πμ})⟨t⟩} ) ∼ 1.

Let now k be odd, r be even. We have

B_{t} ∼ (π, (x − π^{m}v)^{2} − π^{r}δ)_{t} ∼ \{ \begin{matrix}(\frac{π , π^{2m}v^{2}}{K(\sqrt{πμ})⟨t⟩}) , if 2m < k, \\ (\frac{π , π^{k}μ}{K(\sqrt{πμ})⟨t⟩}) , otherwise \end{matrix} ∼ 1.

In c we have B = (α, x − π^{m}v − π^{r/2}δ) and 2m < k. So

B_{t} ∼ (\frac{α , −π^{m}v}{K(\sqrt{πμ})⟨t⟩}) ∼ 1.

Finally, if case d takes place, then 2m > k and

B_{t} ⊗ K′ ∼ (\frac{ue(γ^{2} − δ)π , \sqrt{π^{k}μ}}{KK′(\sqrt{πμ})⟨t⟩}) ∼ (\frac{αueμ , \sqrt{πμ}}{KK′(\sqrt{πμ})⟨t⟩}) ∼ 1,

since from the proof of i) we have αueμ ∼ 1.

Conversely, let k be even and r be odd. From the proof of a) B_{t} ≁ 1, t = Irr_{K\sqrt{πμ}|k}(θ), θ = π^{m}v + \sqrt{π^{r}μ} . On the other hand, if A = (π, x^{2} − π^{k}u), then

A_{t} ∼ (\frac{π , π^{2m}v^{2} − π^{k}u}{K(\sqrt{πμ})⟨t⟩}) ∼ 1

in view of min{2m, r} < min{r, m + r/2}. For A = (α, x − π^{k/2}w)

A_{t} ∼ (\frac{α , π^{m}v − π^{k/2}w}{K(\sqrt{πμ})⟨t⟩} ) ∼ 1,

provided that if k = 2m, then w̄ ≠ v̄.

Finally, let k and r be even. The extension by \sqrt{π} leaves us in case V I. Moreover, if the algebra (α, x−π^{k/2}w) is nontrivial (subcase iii)), then it does not split by this extension, since we are still in iii). The same is true for the algebras from iv) and c). So it is enough to prove that in ii), b) (π, x − e) ≁ 1, and in iii), c) (α, x − π^{k/2}w) ≁ (α, x − π^{m}v − π^{r/2}τ). Case iv), c) is impossible.

In ii), b) we have (π, x − e)_{Irr(π^{k/2}ξ)} ∼ (π, −e) ≁ 1. Consider now iii), c). Then −e ∼ 1. If k = r, then k > 2m. For θ from the proof of iii) satisfying the conditions θ = π^{k/2−1}( ^{3} \sqrt{π})^{ε}, ε ∈ \{1, 2\}, ε ≡ m + k/2 (mod 2) we have (q = Irr_{k( ^{3} \sqrt{π})|k}(θ))

(α, x − π^{k/2}w)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m−1}}{k( ^{3} \sqrt{π})⟨q⟩} ) and (α, x − π^{m}v − π^{r/2}τ)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{m}}{k( ^{3} \sqrt{π})⟨q⟩} ) ,

so A_{q} ≁ B_{q}.

Let now k < r. Case 2m < k is similar to the previous one. In case 2m = k consider θ = π^{m}v + π^{r/2−1}( ^{3} \sqrt{π})^{ε}, ε ∈ \{1, 2\}, ε ≡ k/2 + r/2 (mod 2). Then

(α, x − π^{k/2}w)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2}}{k( ^{3} \sqrt{π})⟨q⟩} ) and (α, x − π^{m}v − π^{r/2}τ)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2−1}}{k( ^{3} \sqrt{π})⟨q⟩} ) .

Finally, in case 2m > k let θ = π^{k/2}( ^{3} \sqrt{π}). We have

(α, x − π^{k/2}w)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2}}{k( ^{3} \sqrt{π})⟨q⟩} ) and (α, x − π^{m}v − π^{r/2}τ)_{q} ∼ (\frac{α , ( ^{3} \sqrt{π})^{k/2+1}}{k( ^{3} \sqrt{π})⟨q⟩} ) .

Consider now cases 1–4 step by step.

1. There exists a ∈ O_{k}^{∗} such that a^{2} − 4u ≁ 1. If p = Irr_{k(\sqrt{u})|k}(π^{k/2}(a + \sqrt{u})), then

(π, x^{2} − π^{k}u)_{p} ∼ (\frac{π , a + 2\sqrt{u}}{k(\sqrt{u})⟨p⟩}) ,

since f(π^{k/2}(a + \sqrt{u})) ∼ −e(a^{2} + 2a\sqrt{u})^{2} ∼ 1 in k(\sqrt{u}). So, (π, x^{2} − π^{k}u)_{p} ≁ 1.

Furthermore, (α(x^{2} − π^{k}u), π^{k/2}(x − π^{k/2}w))_{x−π^{k/2}w} ∼ (α(w^{2} − u), π^{k/2}(x − π^{k/2}w)) ∼ 1, so that this algebra is unramified. Let us prove the nontriviality. Suppose θ = (1 + ^{3} \sqrt{π})π^{k/2}\sqrt{u} ∈ k(\sqrt{u}, ^{3} \sqrt{π}) and q = Irr(θ). Then θ^{2} − π^{k}u ∼ (θ − π^{m}v)^{2} − π^{k}δ ∼ ^{3} \sqrt{π} in k(\sqrt{u}). Therefore f(θ) ∼ 1 and

(α(x^{2} − π^{k}u), π^{k/2}(x − π^{k/2}w))_{q} ∼ (\frac{α ^{3} \sqrt{π} , π^{k/2}((1 + ^{3} \sqrt{π})π^{k/2}\sqrt{u} − π^{k/2}w)}{k(\sqrt{u}, ^{3} \sqrt{π})⟨q⟩}) ∼

∼ (\frac{^{3} \sqrt{π} , \sqrt{u} − w}{k(\sqrt{u}, ^{3} \sqrt{π})⟨q⟩}) ≁ 1.

Thus \sqrt{α} does not split the algebra. To check that \sqrt{π} (and \sqrt{πα}) does not split it, one only need to observe that this extension leaves us in subcase V I.1.

2. The first algebra is unramified and nontrivial by analogy with subcase i). We have f(π^{m}v) ∼ −ueδ. If −ueδ ∼ 1, then (α, x^{2} − π^{k}u)_{x−π^{m}v} ≁ 1. Otherwise, the algebra (α, x − π^{m}v) is unramified. Moreover, (α, x)_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1. Finally, the extension by \sqrt{α} leaves us in subcase V I.2, therefore it does not split the first algebra.

3. If f(0) ∼ −ueδ ∼ 1, then (α, x^{2} −π^{k}u)_{x} ≁ 1. If −ueδ ≁ 1, then (α, x) is unramified and (α, x)_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1. Let θ = (β − \sqrt{α}) ^{3} \sqrt{π}, where β^{2} − α ≁ 1, q = Irr_{k(\sqrt{α}, ^{3} \sqrt{π})|k}(θ). Then f(θ) ∼ 1 and

(πue, x − π^{s})_{q} ∼ ( \frac{^{3} \sqrt{π} , (β − \sqrt{α}) ^{3} \sqrt{π}}{k(\sqrt{α}, ^{3} \sqrt{π})⟨q⟩} ) ≁ 1.

4. (x, −πu(x^{2} − π^{k}u))_{x} ∼ (x, π^{k+1}u^{2}) ∼ 1, so the second algebra B is unramified. To prove that B is nontrivial let us do it for the algebra B′ = B ⊗ k(C)(\sqrt{α}). If θ = π^{k/2}μ ∈ K(\sqrt{πμ}), μ ∈ O_{K}^{∗}, K is of odd degree and unramified over k(\sqrt{α}), then f(θ) ∼ (μ − u)(μ − δ) and

B′_{Irr(θ)} ∼ (\frac{\sqrt{πμ} , μ − u}{K(\sqrt{πμ})(\sqrt{f(θ)})⟨Irr(θ)⟩}) .

Anyway, we can find such K and μ that μ − u ∼ μ − δ ≁ 1, ū ≠ μ̄, δ̄ ≠ μ̄. Then B′_{Irr(θ)} ≁ 1. Case V I is considered.

It is convenient to divide the last case into several parts.

Lemma 21 Let case V II take place, i.e.

f(x) = (x − π^{s}e)(x^{2} − π^{k}u)((x − π^{m}v)^{2} − π^{r}δ),

where e, u, v, δ ∈ O_{k}^{∗}, s, k, m, r > 0, π^{k}u, π^{r}δ ≁ 1, k < r. Then the following subcases do not intersect and cover case V II.

V II_{1}. s = 1, k > 1 and if m = 1, then v̄ ≠ ē.

V II′_{1}. s = 1, k > 1, m = 1, and v̄ = ē.

V II_{2}. s ≥ 2, k ≥ 4, m = 1.

V II_{3}. 1 ≤ k ≤ 3 and k < 2s. Proof. If s > 1, m > 1, and k > 3, then one can use replacement x = x′π^{2}, y = y′π^{5}. Theorem 23 Let f be as in case V II_{1}. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

Let if k ≡ 1 (mod 2), then 2m < k, and if k = r ≡ 0 (mod 2), 2m > k, then ū ≠ δ̄. Then the first algebra is i. (−παN_{K|k}(e − π^{k/2−1}τ), p), if k ≡ 0 (mod 2),

where τ ∈ O_{K}^{∗} such that K|k is of odd degree and unramified, p = Irr_{K|k}(π^{k/2}τ) and τ satisfies the conditions: (τ^{2} − u) ≁ 1, ϕ(τ) ∼ 1,

ϕ(x) = \{ \begin{matrix}1, if k < r or 2m < r, \\ x^{2} − δ, if k = r and 2m > r, \\ (x − v)^{2} − δ, if k = r = 2m. \end{matrix}

ii. (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2) and ueψ ∼ 1, where

ψ = \{ \begin{matrix}1, if 2m < r, \\ v^{2} − δ, if 2m = r, \\ −δ, if 2m > r, r is even. \end{matrix}

iii. (α, x), if k ≡ 1 (mod 2) and ueψ ≁ 1. The second one is a. (−π(μ^{2} − δ)χ(μ)N_{L|k}(e − vπ^{m−1} − μπ^{r/2−1}), q), if r ≡ 0 (mod 2),

where μ ∈ O_{L}^{∗} such that L|k is of odd degree and unramified, q = Irr_{L|k}(π^{m}v+π^{r/2}μ) and μ satisfies the condition: (μ^{2} − δ) ≁ 1 if k is even, and (μ^{2} − δ)χ(μ) ≁ 1, otherwise,

χ(x) = \{ \begin{matrix}1,  if 2m < k, \\ v^{2} − u, if 2m = k < r, \\ −u,  if 2m > k, k < r, \\ x^{2} − u , if 2m > k = r, \\ (x + v)^{2} − u , if k = r = 2m. \end{matrix}

b. (α, (x − π^{m}v)^{2} − π^{r}δ), if r ≡ 1 (mod 2) and δ(e − π^{m−1}v)χ(μ) ∼ 1. c. (α, x − π^{m}v), if r ≡ 1 (mod 2) and δ(e − π^{m−1}v)χ(μ) ≁ 1.

Otherwise we have

1. The first algebra is from ii) or iii), and the second one is (−πueδ, x − π^{m}v), if k, r ≡ 1 (mod 2), k < 2m < r.

2. The first algebra is (−πueδ, x), the second one is either (α, x^{2} − π^{k}u), provided ue ∼ 1 or (α, x − π^{(k+1)/2}), provided ue ≁ 1, if k, r ≡ 1 (mod 2), k < r < 2m.

3. If k ≡ 1 (mod 2), r ≡ 0 (mod 2), and 2m > k, then the first algebra is from ii) or iii), and the second one is

- (π, (x − π^{m}v)^{2} − π^{r}δ) in the following cases:

1. 2m < r and ue ≁ 1.
2. 2m = r, ue ≁ 1, and v^{2} − δ ∼ 1.
3. 2m > r, ue ≁ 1, and −1 ≁ 1.

- (π, t) in cases

1. 2m = r, ue ∼ 1, and v^{2} − δ ≁ 1.

2. 2m > r, ue ∼ 1, and −1 ∼ 1,

where t = Irr_{K(\sqrt{πw})|k}(\sqrt{π^{k}w}), K is as above, w ∈ O_{K}^{∗} \ (O_{K}^{∗})^{2}, and −e(w − u) ≁ 1 in K.

- (α, Irr_{K|k}(π^{m}v + π^{r/2}λ)) in cases

1. 2m < r, ue ∼ 1. v^{2} − δ ≁ 1.

2. 2m > r, ue ≁ −1,

where λ ∈ O_{K}^{∗}, and λ^{2} − δ ≁ 1 in case 2m < r, λ^{2} − δ ∼ −1 in case 2m > r. Note that if 2m < r, then one can suppose K = k, and if 2m > r, then this is true as soon as k̄ ≠ F_{3}.

- (α, Irr_{K}(π^{m}η)) if 2m = r and ue ∼ v^{2} − δ, where η ∈ O_{K}^{∗}, ue((η − v)^{2} − δ) ≁ 1.

4. (−π(e−π^{k−2}w), x−π^{k/2}w), (α(x^{2} −π^{k}u), π^{k/2}(x−π^{k/2}w)), w^{2} −u ≁ 1, if k = r ≡ 0 (mod 2), 2m > k, ū = δ̄ and if k = 2, then e^{2} − u ∼ 1.

5. (π, x − πe), (π, x^{2} − π^{k}u), if k = r = 2, m > 1, ū = δ̄ and e^{2} − u ≁ 1.

6. If k = r ≡ 1 (mod 2), 2m > k, then the first algebra is (x, −πu(x^{2} − π^{k}u)) and the second one is

- (−πueδ, x), if ū ≠ δ̄.

- (α, x^{2} − π^{k}u), if ū = δ̄ and either −ue ∼ 1 or 2m = k + 1.

- (α, Irr_{k(\sqrt{πu})|k}(θ)), θ = \sqrt{π^{k}u}(1 + \sqrt{πu}), if ū = δ̄, −ue ≁ 1, and 2m > k + 1.

Remark 3 The algebra (α(x^{2} − π^{k}u), π^{k/2}(x − π^{k/2}w)) from 4 does not split by any quadratic extension of scalars.

Proof of the theorem. i). In view of f(π^{k/2}τ) ∼ −πα(e − π^{k/2}τ) in K the algebra is unramified. We also have (−παN_{K|k}(e − π^{k/2−1}τ), p)_{x−πe} ∼ (−παN(e − π^{k/2−1}τ), ∏_{σ}(πe − π^{k/2}τ^{σ}))(−παN(e − π^{k/2−1}τ), πN(πe − π^{k/2}τ)) ≁ 1. ii), iii). f(0) ∼ ueψ, so in iii) the algebra (α, x) is unramified. Moreover, in ii) we have (α, x^{2}−π^{k}u)_{x} ≁ 1 and in iii) (α, x)_{x−πe} ≁ 1. a). f(π^{m}v + π^{r/2}μ) ∼ −π(e − π^{m−1}v − π^{r/2−1}μ)(μ^{2} − δ)χ(μ), so the algebra is unramified. If k is even, then

(−παχ(μ)N_{L|k}(e − vπ^{m−1} − π^{r/2−1}μ), q)_{x^{2}−π^{k}u} ∼ (\frac{π , ∏_{λ}(π^{k/2}\sqrt{u} − π^{m}v − π^{r/2}μ^{λ})}{k(\sqrt{α})⟨x^{2} − π^{k}u⟩(\sqrt{C(x^{2} − π^{k}u)})}) .

The latter algebra is nontrivial if and only if χ(μ) is not a square in L. We also have (−π(μ^{2} − δ)χ(μ)N_{L|k}(e − vπ^{m−1} − π^{r/2−1}μ), q)_{x−πe} ∼ (−π(μ^{2} − δ)χ(μ)N(e − vπ^{m−1} − π^{r/2−1}μ), ∏_{λ}(πe − π^{m}v − π^{r/2}μ^{λ})) ∼ (π, (μ^{2} − δ)χ(μ)), and this algebra is nontrivial iff (μ^{2} − δ)χ(μ) is not a square in L. Anyway, the algebra from a) is nontrivial. b), c). We have f(π^{m}v) ∼ δ(e − π^{k−2}v)χ(μ) and (α, x − π^{m}v)_{x−πe} ≁ 1.

Let us check now that the algebra from i), ii), iii) is not isomorphic to the algebra from a), b), c) in the corresponding subcases. 1). Let firstly k, r ≡ 1 (mod 2). We have (α, x^{2} − π^{k}u)_{x−π^{m}v} ∼ 1 ≁ (α, (x − π^{m}v)^{2} − π^{r}δ)_{x−π^{m}v}. Furthermore,

(α, x)_{x^{2}−π^{k}u} ∼ (\frac{α , \sqrt{πu}}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ≁ 1,

but

(α, x − π^{m}v)_{x^{2}−π^{k}u} ∼ (\frac{α , −π^{m}u}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ∼ 1.

Similarly one can check that (α, (x − π^{m}v)^{2} − π^{r}δ)_{x^{2}−π^{k}u} ∼ 1. Finally, we obtain

(α, x − π^{m}v)_{(x−π^{m}v)^{2}−π^{r}δ} ∼ (α, \sqrt{πδ}) ≁ 1, and (α, x^{2} − π^{k}u)_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1.

2). k ≡ 0 (mod 2), r ≡ 1 (mod 2). Then the algebra from i is nontrivial at x − πe, but (α, (x − π^{m}v)^{2} − π^{r}δ)_{x−πe} ∼ 1. Moreover,

(−παN_{K|k}(e − π^{k/2−1}τ), p)_{(x−π^{m}v)^{2}−π^{r}δ} ∼

(\frac{−δαN(e − π^{k/2−1}τ) , ∏_{σ}(π^{m}v + \sqrt{π^{r}δ} − π^{k/2}τ^{σ})}{k(\sqrt{πδ})⟨(x − π^{m}v)^{2} − π^{r}δ⟩(\sqrt{C((x − π^{m}v)^{2} − π^{r}δ)})}) .

One can choose τ such that if 2m = k, then τ̄ ≠ v̄. For this τ the latter algebra is trivial in view of v_{K}(π^{m}v − π^{k/2}τ^{σ}) < r/2. On the other hand, (α, x − π^{m}v)_{(x−π^{m}v)^{2}−π^{r}δ} ≁ 1. 3). k ≡ 1 (mod 2), r ≡ 0 (mod 2). The algebra from a) is nontrivial at x − πe, but (α, x^{2} − π^{k}u) is trivial at this place. We also have (α, x)_{x^{2}−π^{k}u} ≁ 1 and

(−παN_{L|k}(e − vπ^{m−1} − π^{r/2−1}μ), q)_{x^{2}−π^{k}u} ∼

(\frac{−αu(e − vπ^{m−1}) , ∏_{λ}(\sqrt{π^{k}u} − π^{m}v − π^{r/2}μ^{λ})}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ∼

(\frac{−αu(e − vπ^{m−1}) , −π^{m}v}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ∼ 1.

4). k, r ≡ 0 (mod 2). We have

(−παN_{K|k}(e − π^{k/2−1}), p)_{x^{2}−π^{k}u} ∼ (\frac{π , ∏_{σ}(τ^{σ} − \sqrt{u})}{k(\sqrt{α})⟨x^{2} − π^{k}u⟩(\sqrt{C(x^{2} − π^{k}u)})}) .

This algebra is nontrivial since τ^{2} − u ≁ 1 in K. So the algebra from i) is nontrivial either at x − πe or at x^{2} − π^{k}u. But the algebra from a) is nontrivial at one of these places, and trivial at the other. Therefore, these algebras are not isomorphic.

Consider the remaining cases.

1. Let w be a unit from a sufficiently large unramified extension K of odd degree over k, such that

\{ \begin{matrix}w ≁ −ueδ, \\ w − u ∼ −e, \end{matrix}

and t = Irr_{K(\sqrt{πw})|k}(\sqrt{π^{k}w}). Then f(\sqrt{π^{k}w}) ∼ −e(w − u) ∼ 1 and

(−πueδ, x − π^{m}v)_{t} ∼ (\frac{−wueδ , \sqrt{πw}}{K(\sqrt{πw})⟨t⟩} ) ≁ 1.

To finish the proof of this subcase it is enough to check that the extension of scalars by \sqrt{α} does not split algebra (−πueδ, x − π^{m}v). Indeed, this extension leaves us in subcase V II.7.

2. All algebras are unramified. Since f(π^{(k+1)/2}) ∼ ue we obtain (x^{2} − π^{k}u)_{x−π^{(k+1)/2}} ≁ 1. Furthermore,

(α, x − π^{(k+1)/2})_{x^{2}−π^{k}u} ∼ (\frac{α , \sqrt{πu}}{k(\sqrt{πu})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ≁ 1.

The nontriviality of the algebra (−πueδ, x) can be checked similarly as in the previous subcase. Finally, the extension by \sqrt{α} leaves us in the same subcase, so it does not split the latter algebra.

3. The nontriviality of algebras from ii) and iii) was already shown. Let firstly 2m < r. Then f(π^{m}v + π^{r/2}λ) ∼ ue(λ^{2} − δ), so if ue ∼ 1, λ^{2} − δ ≁ 1, then (α, Irr(π^{m}v + π^{r/2}δ)) is unramified. In this case the first algebra is (α, x^{−}π^{k}u), it is trivial at x − πe, but (α, Irr(π^{m}v + π^{r/2}δ))_{x−πe} ≁ 1.

If ue ≁ 1, then f(π^{m}v+π^{r/2}λ) ∼ 1, provided λ^{2}−δ ≁ 1. Therefore (π, (x−π^{v})^{2}−π^{r}δ)_{Irr(π^{m}v+π^{r/2}λ)} ≁ 1. The first algebra is (α, x), (α, x)_{x−πe} ≁ 1. On the other hand, (π, (x − π^{v})^{2} − π^{r}δ)_{x−πe} ∼ 1.

Let now 2m = r. We have f(π^{m}η) ∼ ue((η − v)^{2} − δ), so algebra (α, Irr(π^{m}η)) is unramified. It is nontrivial at x − πe. Since ue(v^{2} − δ) ∼ 1, the first algebra is trivial at x − πe.

If ue ≁ 1, v^{2} −δ ∼ 1, then (π, (x−π^{m}v)^{2} −π^{r}δ)_{Irr(π^{m}η)} ≁ 1. In this case the first algebra is nontrivial at x − πe, and this is not true for the second one.

Suppose ue ∼ 1 and v^{2} − δ ≁ 1. We have f(\sqrt{π^{k}w}) ∼ −e(w − u) ∼ α in K(\sqrt{πw}). So

(π, t)_{t} ∼ (\frac{w , t}{K(\sqrt{πw}, \sqrt{α})}) ⟨t⟩ ∼ 1,

and algebra (π, t) is unramified. It is not isomorphic to (α, x) at x − πe. It is easy to check that f(π^{m−1/3}) ∼ ue ∼ 1. Hence (π, t)_{Irr_{k( ^{3} \sqrt{π})|k}(π^{m−1/3})} ∼ ( ^{3} \sqrt{π}, w) ≁ 1.

Finally, let 2m > r. One can consider this case by analogy to the previous ones.

4. We have f(π^{k/2}w) ∼ −π(e − π^{k−2}w) and α((π^{k/2}w)^{2} − π^{k}u) ∼ 1, so the both algebras are unramified. (−π(e − π^{k−2}), x − π^{k/2}w)_{x^{2}−π^{k}u} ∼ (π, w − \sqrt{u}) ≁ 1. Let us check that the second algebra is nontrivial, not isomorphic to the first one, and does not split by any quadratic extension of the scalars. To prove this it is sufficient to show that

B = (α(x^{2} − π^{k}u), π^{k/2}(x − π^{k/2}w)) ⊗_{k(C)} k(\sqrt{π}, \sqrt{α})(C) ≁ 1.

Assume ν = \sqrt{π}, θ = ν^{k}\sqrt{u}(1 + ν). Then f(θ) ∼ 1 in view of δ = u + ν^{2l}μ, l > 0 and if k = 2, then e − \sqrt{u} ≁ 1. Thus

B_{x−θ} ∼ (\frac{ν , w − \sqrt{u}}{k(\sqrt{π}, \sqrt{α})⟨x − θ⟩}) ≁ 1.

5. In this case (π, x − πe)_{x^{2}−π^{2}u} ∼ (π, e − \sqrt{u}) ≁ 1 and (π, x^{2} − π^{2}u)_{x−πe} ∼ (π, e^{2} − u) ≁ 1. Moreover, ((π, x − πe)_{⊗}(π, x^{2} − π^{2}u))_{x−πe} ∼ (π, (x − π^{m}v)^{2} − π^{r}δ)_{x−πe} ∼ (π, e^{2} − δ) ≁ 1.

6. We have (x, −πu(x^{2}−π^{k}u))_{x} ∼ (x, π^{k+1}u^{2}) ∼ 1, therefore the algebra (x, −πu(x^{2}−π^{k}u)) is unramified. Let K, w, and t be as in subcase 1, but

\{ \begin{matrix}uew(w − δ) ≁ 1, \\ −u(w − u) ≁ 1. \end{matrix}

Then f(\sqrt{π^{k}w}) ∼ −ew(w − u)(w − δ) ∼ 1 and

(x, −πu(x^{2} − π^{k}u))_{t} ∼ (\frac{\sqrt{πw} , −u(w − u)}{K(\sqrt{πw}⟨t⟩)}) ≁ 1.

Consider the second algebra. For ū ≠ δ̄ one always can find w and K such that

\{ \begin{matrix}−ueδw ≁ 1, \\ −ew(w − u)(w − δ) ∼ 1. \end{matrix}

Then f(\sqrt{π^{k}w}) ∼ 1 and (−πueδ, x)_{t} ∼ (−ueδw, \sqrt{πw}) ≁ 1. So the second algebra is nontrivial. We also can find such K, w, and t that

\{ \begin{matrix}w ∼ −ueδ, \\ w − u ≁ −u, \\ w − δ ≁ −δ. \end{matrix}

For such w the first algebra is nontrivial at t and the second one is trivial at t.

Let now ū = δ̄ and θ_{w} = \sqrt{π^{k}u}(1 + \sqrt{πu}w). Then f(θ_{w}) ∼ −ew(uw − π^{m−(k+1)/2}) (if 2m = k + 1 let ūw̄ ≠ v̄). In case 2m > k + 1 we have f(θ_{w}) ∼ −ue ∼ 1 for any w, and otherwise we can find w such that f(θ_{w}) ∼ 1. Thus for q = Irr_{K(\sqrt{πw})|k}(θ_{w})

(α, x^{2} − π^{k}u)_{q} ∼ (\frac{α , 2w\sqrt{πu}}{K(\sqrt{πw})⟨q⟩}) ≁ 1.

Let ū = δ̄, 2m > k + 1, and −ue ≁ 1. Then for θ = θ_{w}, K = k, w = 1 we find f(θ) ∼ −ue ∼ α and the algebra (α, q) is unramified.

(α, q)_{x^{2}−π^{k}u} ∼ (\frac{α , (\sqrt{π^{k}u} − \sqrt{π^{k}u}(1 + \sqrt{πu}))(\sqrt{π^{k}u} + \sqrt{π^{k}u}(1 − \sqrt{πu}))}{k(\sqrt{πu})⟨\sqrt{D(x^{2} − π^{k}u)}⟩}) ≁ 1.

Both algebras (α, x^{2} − π^{k}u) and (α, q) are not isomorphic to the first algebra from the list. Indeed, the extension by \sqrt{α} leaves us in the current case and subcase. So it does not split the first algebra.

The theorem is proved.

Theorem 24 Let f be as in case V II′_{1}, i.e.

f(x) = (x − πv(1 + π^{l}τ))(x^{2} − π^{k}u)((x − πv)^{2} − π^{r}δ), 1 < k ≤ r, π^{k}u, π^{r}δ ≁ 1.

Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras. The first algebra A is

i . (πE(w), x − π^{k/2}w), if k ≡ 0 (mod 2),

ii . (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2) and uv ∼ 1,

iii .(α, x), if k ≡ 1 (mod 2) and uv ≁ 1,

where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1, and

E(w) = α \{ \begin{matrix}−v, if k > 2, \\ w − v, if k = 2. \end{matrix} \} \{ \begin{matrix}1, if r > 2, \\ (w − v)^{2} − δ, if r = 2. \end{matrix} \} .

The second algebra B is

a . (α, x − πv),

if r ≡ 1 (mod 2), l ≡ 0 (mod 2), τvδξ ≁ 1, where

ξ = \{ \begin{matrix}1, if k > 2, \\ v^{2} − u if k = 2. \end{matrix}

b . (α, (x − πv)^{2} − π^{r}δ),

if r ≡ 1 (mod 2), l ≡ 0 (mod 2), τvδξ ∼ 1.

c . (πvδτξ, x − πv),

if r, l ≡ 1 (mod 2), 2l + 2 < r.

d . (παN_{K|k}(η(γ)ζ(γ)), p),

if either r, l ≡ 0 (mod 2), 2l + 2 ≤ r, or r ≡ 2 (mod 4), 2l + 2 > r, where

η(x) = \{ \begin{matrix}1, if k > 2, \\ v^{2} − u if k = 2 < r, \\ (x + v)^{2} − δ, if k = r = 2, \end{matrix} ζ(x) = \{ \begin{matrix}−vτ if 2l + 2 < r, \\ x − vτ, if 2l + 2 = r, \\ x, if 2l + 2 > r, \end{matrix}

γ ∈ O_{K}^{∗}, K|k is of odd degree and unramified, p = Irr_{K|k}(πv + π^{r/2}γ), γ^{2} − δ ≁ 1, and if r = 2,

then (γ + v)^{2} − u ∼ 1.

e . (π, (x − πv)^{2} − π^{r}δ),

if r ≡ 0 (mod 2), l ≡ 1 (mod 2), and either 2l + 2 = r or 2l + 2 < r and −τvξ ≁ 1.

f . (α, x − πv − π^{r/2}μ),

if r ≡ 0 (mod 2), l ≡ 1 (mod 2), 2l + 2 < r, and −τvξ ∼ 1, where μ ∈ O_{k}^{∗} and μ^{2} − δ ≁ 1.

g . (π, (x − πv)^{2} − π^{r}δ),

if r ≡ 0 (mod 4), 2l + 2 > r.

h . (α, (x − πv)^{2} − π^{r}δ),

if r, l ≡ 1 (mod 2), 2l + 2 > r.

Note that in d) if r > 2, then one can assume K = k, p = x − πv − π^{r/2}γ.

Proof. i). We have f(π^{k/2}w) ∼ πE(w), so A is unramified.

A_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(\sqrt{D(x^{2} − π^{k}u)})}) ≁ 1.

ii),iii). Then f(0) ∼ uv, so (α, x^{2} − π^{k}u)_{x} ≁ 1 in case uv ∼ 1, and (α, x) is unramified otherwise. Finally, (α, x)_{x^{2}−π^{k}u} ≁ 1.

It easy to check that f(πv) ∼ π^{r+l+1}τvδξ and for even r and arbitrary μ ∈ O_{K}^{∗} f(π^{v} + π^{r/2}μ) ∼ (μ^{2} − δ)η(μ)ζ(μ)π^{min{l+1,r/2}}. a. We have B_{(x−πv)^{2}−π^{r}δ} ≁ 1 and

(α, x)_{(x−πv)^{2}−π^{r}δ} ∼ (\frac{α , πv}{k(\sqrt{πδ})⟨…⟩}) ∼ 1,

(α, x^{2} − π^{k}u)_{(x−πv)^{2}−π^{r}δ} ∼ (\frac{α , ξ}{k(\sqrt{πδ})⟨…⟩}) ∼ 1,

(πE(w), x − π^{k/2}w)_{(x−πv)^{2}−π^{r}δ} ∼ (\frac{δE(w) , πv − π^{k/2}w}{k(\sqrt{πδ})⟨…⟩} ) ∼ 1,

so A ≁ B. b. f(πv) ∼ 1, therefore B_{x−πv} ≁ 1. On the other hand, (α, x^{2} − π^{k}u)_{x−πv} ∼ 1. For the remaining A we have A_{x^{2}−π^{k}u} ≁ 1, but B_{x^{2}−π^{k}u} ∼ (α, (πμ)^{2}) ∼ 1. c. Let θ = πv + \sqrt{π^{r}λ}. Then f(θ) ∼ −vτξλ(λ − δ) in k(\sqrt{πλ}). B_{Irr(θ)} ∼ (vδτξλ, \sqrt{πλ}). The conditions −vτξλ(λ − δ) ∼ 1 and vδτξλ ≁ 1 can be satisfied in a sufficiently large unramified extension of odd degree of k. Thus B ≁ 1. We have also (α, x)_{Irr(θ)} ∼ (α, πv) ∼ 1, (α, x^{2} − π^{k}u)_{Irr(θ)} ∼ (α, (πv)^{2}) ∼ 1, and (πE(w))_{Irr(θ)} ∼ (λE(w), πv − π^{k/2}w) ∼ 1. So A ≁ B. d. B is unramified since f(πv + π^{r/2}γ) ∼ αη(γ)ζ(γ)π.

B_{(x−πv)^{2}−π^{r}δ} ∼ (\frac{π , ∏_{σ}(γ^{σ} − \sqrt{δ})}{k(\sqrt{δ})⟨…⟩}) ≁ 1.

The algebras (α, x), (α, x^{2} − π^{k}u) split by \sqrt{δ}, so they are not isomorphic to B. Finally, (πE(w), x − π^{k/2}w)_{(x−πv)^{2}−π^{r}δ} ∼ (π, πv − π^{k/2}w + π^{r/2}\sqrt{δ}). If r > 2, then the latter algebra is trivial, and A ≁ B again. Let k = r = 2. Then A_{x^{2}−π^{2}u} ≁ 1 and B_{x^{2}−π^{2}u} ∼ (π, ∏_{σ}(γ^{σ} + v − \sqrt{u})) ∼ 1 in view of (γ + v)^{2} − u ∼ 1 in K. e. In e) and f) we have

f(πv + π^{r/2}μ) ∼ (μ^{2} − δ)ημζμ ∼ (μ^{2} − δ)ξ \{ \begin{matrix}−vτ, if 2l + 2 < r, \\ μ − vτ, if 2l + 2 = r. \end{matrix} \} .

If either 2l + 2 = r or −τvξ ≁ 1, then there exists μ ∈ O_{K}^{∗} such that μ^{2} − δ ≁ 1 and f(πv + π^{r/2}μ) ∼ 1. For such μ we have B_{Irr(πv+π^{r/2}μ)} ∼ (π, μ^{2} − δ) ≁ 1. Since r ≥ 2l + 2 ≥ 4, then B_{x} ∼ 1 and B_{x^{2}−π^{k}u} ∼ 1. So B ≁ A. f. In this case f(πv + π^{r/2}μ) ∼ α, so B is unramified. Let θ = πv(1 + π^{l+1/3}) ∈ k(Π), Π = ^{3} \sqrt{π}. Then f(θ) ∼ −τvξ(π^{2l+8/3}v^{2} − π^{r}δ) ∼ 1.

B_{Irr(θ)} ∼ (\frac{α , π^{l+4/3}v − π^{r/2}μ}{k(Π)⟨Irr(θ)⟩}) ∼ (\frac{α , Π^{3l+4}}{k(Π)⟨Irr(θ)⟩}) ≁ 1.

Moreover, A ≁ B. Indeed, if k is even, then B_{x^{2}−π^{k}u} ∼ 1 since B splits by \sqrt{u}, and if k is odd, then (α, x^{2} − π^{k}u)_{Irr(θ)} ∼ 1. Finally, for odd k (α, x)_{x^{2}−π^{k}u} ≁ 1 and B_{x^{2}−π^{k}u} ∼ (α, −πv) ∼ 1. g. Let μ ∈ O_{K}^{∗} such that μ^{2} − δ ≁ 1, μη(μ) ≁ 1. Then f(πv + π^{r/2}μ) ∼ 1 and B_{Irr(πv+π^{r/2}μ)} ≁ 1. But B_{x} ∼ 1, so B ≁ (α, x^{2} − π^{k}u). The remaining algebra A is not trivial at x^{2} − π^{k}u, but B_{x^{2}−π^{k}u} ∼ (π, (\sqrt{π^{k}u} − πv)^{2} − π^{r}δ) ∼ 1 since r > 2. h. B_{x−πv(1+π^{l}τ)} ∼ (α, π^{2l+2}v^{2}τ^{2}) − π^{r}δ ≁ 1. We have also B_{x} ∼ (α, π^{2}v^{2}) ∼ 1. If k is odd, then B_{x^{2}−π^{k}u} ∼ (α, (\sqrt{π^{k}u} − πv)^{2} − π^{r}δ) ∼ 1 since r > 2. Finally, if k is even, then B_{x^{2}−π^{k}u} ∼ 1 in view of B splits by \sqrt{u}.

Consider the next case.

Theorem 25 Let f be as in case V II_{2}, i.e.

f(x) = (x − π^{s}e)(x^{2} − π^{k}u)((x − πv)^{2} − π^{r}δ), s ≥ 2, 4 ≤ k ≤ r, π^{k}u, π^{r}δ ≁ 1.

Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras.

If 2s < k, then the first algebra A is

1 . (α, x), if k, s ≡ 1 (mod 2) and ue ≁ 1.

2 . (α, x^{2} − π^{k}u), if k, s ≡ 1 (mod 2) and ue ∼ 1.

3 . (−παe, x − π^{k/2}w), if k ≡ 0 (mod 2), s ≡ 1 (mod 2), where w ∈ O_{k}^{∗}, w^{2} − u ≁ 1.

4 . (π, x^{2} − π^{k}u), if k, s ≡ 0 (mod 2) and −e ≁ 1.

5 . (α, x − π^{k/2}w), if k, s ≡ 0 (mod 2) and −e ∼ 1, w^{2} − u ≁ 1.

6 . (ueπ, x), if k ≡ 1 (mod 2), s ≡ 0 (mod 2).

If 2s ≥ k, then the first algebra A is

7 . (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2).

8 . (αξ(w)π, x − π^{k/2}w), if k ≡ 2 (mod 4), w^{2} − u ≁ 1,

ξ(x) = \{ \begin{matrix}w, if 2s > k, \\ w − e, if 2s = k, w̄ ≠ ē. \end{matrix}

9 . (π, x^{2} − π^{k}u), if k ≡ 0 (mod 4).

And the second algebra B is

a . (α, x − πv), if r ≡ 1 (mod 2), −vδ ≁ 1.

b . (α, (x − πv)^{2} − π^{r}δ), if r ≡ 1 (mod 2), −vδ ∼ 1.

c . (παv, x − πv − π^{r/2}μ), if r ≡ 0 (mod 2), where μ ∈ O_{k}^{∗}, μ^{2} − δ ≁ 1.

Proof. Let 2s < k. We have f(0) ∼ ueπ^{k+s}, f(πv) ∼ −vδπ^{r+1}. If k is even and w^{2} − u ≁ 1, then f(π^{k/2}w) ∼ −eαπ^{s}. If r is even and μ^{2} − δ ≁ 1, then f(πv + π^{r/2}μ) ∼ vαπ.

1. (α, x)_{x−π^{s}e} ≁ 1.

2. (α, x^{2} − π^{k}u)_{x} ≁ 1.

3. (−παe, x − π^{k/2}w)_{x−π^{s}e} ≁ 1.

4. (π, x^{2} − π^{k}u)_{x−π^{k/2}w} ≁ 1.

5. f(π^{s+1}) ∼ −π^{s}e ∼ 1, so (α, x − π^{k/2}w)_{x−π^{s+1}} ∼ (α, π^{s+1}) ≁ 1.

6. f(\sqrt{π^{k}γ}) ∼ −eγ(γ − u) in k(\sqrt{πγ}). So (ueπ, x)_{Irr(\sqrt{π^{k}γ})} ∼ (ueγ, \sqrt{πγ}) ≁ 1, provided γ ∈ O_{K}^{∗} satisfies the condition γ ≁ ue, γ − u ≁ −u.

Let now k ≤ 2s. If k is even and w^{2} − u ≁ 1, then f(π^{k/2}w) ∼ π^{k/2}αξ(w).

7. (α, x^{2} − π^{k}u)_{x−π^{s}e} ≁ 1.

8. (αξ(w)π, x − π^{k/2}w)_{x^{2}−π^{k}u} ≁ 1.

9. Let γ ∈ O_{K}^{∗}, γ^{2} − u ≁ 1, and ξ(γ) ≁ 1. Then f(π^{k/2}γ) ∼ 1 and (π, x^{2} − π^{k}u)_{Irr(π^{k/2}γ)} ≁ 1. a. (α, x − πv)_{(x−πv)^{2}−π^{r}δ} ≁ 1. On the other hand, A_{(x−πv)^{2}−π^{r}δ} ∼ 1. b. (α, (x − πv)^{2} − π^{δ})_{x−πv} ≁ 1. A ≁ B since B is trivial at all the places from 1–9, at which A is nontrivial. c. (παv, x − πv − π^{r/2}μ)_{(x−πv)^{2}−π^{r}δ} ∼ (π, μ − \sqrt{δ}) ≁ 1. On the other hand, A_{(x−πv)^{2}−π^{r}δ} ∼ 1.

The theorem is proved.

Let us pass to the last case.

Theorem 26 Let f be as in case V II_{3}, i.e.

f(x) = (x^{2} − π^{k}u)(x − π^{s}e)((x − π^{m}v)^{2} − π^{r}δ),

where k ∈ {1, 2, 3}, k < 2s, k ≤ r, π^{k}u, π^{r}δ ≁ 1. Then the group _{2}Br C is generated by the class of the algebra (π, α) and the classes of the following algebras. The first algebra A is

i. (α, x^{2} − π^{k}u), if k ≡ 1 (mod 2).

ii. (πE(w), x − πw), if k = 2, where w^{2} − u ≁ 1,

E(x) = \{ \begin{matrix}αx if r > 2, \\ αx(x^{2} − δ) if r = 2, m > 1, ū ≠ δ̄ \\ x, if r = 2, m > 1, ū = δ̄ \\ αx((x − v)^{2} − δ), if r = 2, m = 1. \end{matrix}

1. If r < 2s, r ≤ 2m, then the second algebra B is

a. (α, (x − π^{m}v)^{2} − π^{r}δ), if r ≡ 1 (mod 2).

b. (π, (x − π^{m}v)^{2} − π^{r}δ), if r ≡ 0 (mod 2), k + r/2 ≡ 0 (mod 2).

c. (πN_{K|k}(S(μ)), Irr_{K|k}(π^{m}v + π^{r/2}μ)), if r ≡ 0 (mod 2), k + r/2 ≡ 1 (mod 2), and if k = r = 2, m > 1, then ū ≠ δ̄. Here μ ∈ O_{K}^{∗}, K|k is of odd degree and unramified, μ^{2} − δ ≁ 1. If k = r = 2, then the additional condition on μ is

\{ \begin{matrix}μ^{2} − u ∼ 1, if m > 1, \\ (μ + v)^{2} − u ∼ 1, if m = 1. \end{matrix}

Note that if k = r = 2 does not hold, then one can assume K = k.

S(x) = α \{ \begin{matrix}x, if 2m > r, \\ x + v, if 2m = r, \end{matrix} \} \{ \begin{matrix}−u, if k < r, \\ x^{2} − u, if k = r < 2m, \\ (x + v)^{2} − u, if k = r = 2m, \end{matrix} \}

2. If k ≤ 2m < r, m < s, then the second algebra B is

a. (αvξπ, x − π^{m}v − π^{r/2}μ), if m + k ≡ 1 (mod 2), r ≡ 0 (mod 2), where μ^{2} − δ ≁ 1 and

ξ = \{ \begin{matrix}−u, if 2m > k, \\ v^{2} − u, if 2m = k. \end{matrix}

b. (−δvξπ, x − π^{m}v), if m + k ≡ 0 (mod 2), r ≡ 1 (mod 2).

c. (α, (x − π^{m}v)^{2} − π^{r}δ), if m + k, r ≡ 1 (mod 2) and −δvξ ∼ 1.

d. (α, x − π^{m}v), if m + k, r ≡ 1 (mod 2) and −δvξ ≁ 1.

e. (π, (x − π^{m}v)^{2} − π^{r}δ), if m + k, r ≡ 0 (mod 2), vξ = −uv ≁ 1.

f. (α, x − π^{m}v − π^{r/2}μ), if m + k, r ≡ 0 (mod 2), vξ = −uv ∼ 1, μ^{2} − δ ≁ 1.

3. Let 2s ≤ r, s ≤ m, and if s = m, then v̄ ≠ ē. The second algebra B is

a. (uδψπ, x − π^{m}v), if k + s ≡ 0 (mod 2), r ≡ 1 (mod 2).

b. (α, x − π^{m}v), if k + s, r ≡ 1 (mod 2), ψuδ ≁ 1.

c. (α, (x − π^{m}v)^{2} − π^{r}δ), if k + s, r ≡ 1 (mod 2), ψuδ ∼ 1.

d. (−πuαψ(μ), x − π^{m}v − π^{r/2}μ), if k + s ≡ 1 (mod 2), r ≡ 0 (mod 2), where μ^{2} − δ ≁ 1 and

ψ(x) = \{ \begin{matrix}−e, if s < m, 2s < r, \\ v − e, if s = m, 2s < r, \\ x − e, if s < m, 2s = r, \\ x + v − e, if s = m, 2s = r, \end{matrix}

if 2s < r, then ψ(x) = ψ does not depend on x.

e. (π, (x − π^{m}v)^{2} − π^{r}δ), if k + s, r ≡ 0 (mod 2), 2s < r, −uψ ≁ 1.

f. (α, x − π^{m}v − π^{r/2}μ), if k + s, r ≡ 0 (mod 2), 2s < r, −uψ ∼ 1, μ^{2} − δ ≁ 1.

g. (π, x − π^{s}e), if k + s, r ≡ 0 (mod 2), 2s = r.

4. If 2m < k (i.e. m = 1, k = 3, s ≥ 2), then the second algebra B is

a . (παv, x − πv − π^{r/2}μ), r ≡ 0 (mod 2), μ^{2} − δ ≁ 1.

b. (α, (x − πv)^{2} − π^{r}δ), r ≡ 1 (mod 2), −vδ ∼ 1.

c. (α, x − πv), r ≡ 1 (mod 2), −vδ ≁ 1.

5. If k < 2m = 2s < r, e = v(1 + π^{l}τ), l > 0, then the second algebra B is

a. (α, x − π^{m}v), if k + m + l, r ≡ 1 (mod 2), −uvτδ ≁ 1.

b. (α, (x − π^{m}v)^{2} − π^{r}δ), if k + m + l, r ≡ 1 (mod 2), −uvτδ ∼ 1.

c. (π, (x − π^{m}v)^{2} − π^{r}δ), if m + l ≤ r/2, k + m + l, r ≡ 0 (mod 2), and either m + l < r/2, uvτ ≁ 1, or m + l = r/2.

d. (α, x − π^{m}v − π^{r/2}μ), if m + l < r/2, k + m + l, r ≡ 0 (mod 2), and uvτ ∼ 1. μ^{2} − δ ≁ 1.

e. (−πuvτδ, x − π^{m}v), if m + l < r/2, k + m + l ≡ 0 (mod 2), r ≡ 1 (mod 2).

f. (πT (μ), x−π^{m}v −π^{r/2}μ), if m+l ≤ r/2, k +m+l ≡ 1 (mod 2), r ≡ 0 (mod 2), where μ^{2} −δ ≁ 1,

and

T (x) = \{ \begin{matrix}αvτu, if m + l < r/2, \\ −αu(x − vτ), if m + l = r/2. \end{matrix}

g. (−πuμα, x − π^{m}v − π^{r/2}μ), if m + l > r/2, k + r/2 ≡ 1 (mod 2), r ≡ 0 (mod 2).

h. (π, (x − π^{m}v)^{2} − π^{r}δ), if m + l > r/2, k + r/2, r ≡ 0 (mod 2).

j. (α, (x − π^{m}v)^{2} − π^{r}δ), if m + l > r/2, r ≡ 1 (mod 2), k + l + m ≡ 0 (mod 2).

6. If k = r = 2, m, s ≥ 2, ū = δ̄, then the second algebra B is

a. (π, x^{2} − π^{k}u), if −1 ∼ 1.

b. (α(x^{2} − π^{k}u), x − πw), if −1 ≁ 1, where w^{2} − u ≁ 1.

Proof. i),ii). (α, x^{2} − π^{k}u)_{x−π^{s}e} ≁ 1, provided k is odd. Otherwise we have f(πw) ∼ πE(w), so A is unramified. Moreover,

(πE(w), x − πw)_{x^{2}−π^{k}u} ∼ (\frac{π , w − \sqrt{u}}{k(\sqrt{u})⟨x^{2} − π^{k}u⟩(D(x^{2} − π^{k}u))}) ≁ 1.

1.a. B_{x−π^{s}e} ∼ (α, π^{r}) ≁ 1. If k is odd, then (A ⊗ B)_{x^{2}−π^{k}u} ∼ (x − π^{s}e)_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1, so A ≁ B. Let k is even. Then B_{x^{2}−π^{k}u} ∼ 1, and A ≁ B again. 1.b. If r is even, then f(π^{m}v + π^{r/2}μ) ∼ π^{k+r/2}S(μ). In case 1.b) one can find K, μ such that S(μ) ∼ 1 and μ^{2} − δ ≁ 1. Therefore B_{Irr(π^{m}v+π^{r/2}μ)} ∼ (π, μ^{2} − δ) ≁ 1. For odd k A_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1, but B_{x^{2}−π^{k}u} ∼ 1. Let k be even. Then r/2 is even, so r > k and B_{x^{2}−π^{k}u} ∼ 1. Anyway, A ≁ B. 1.c. We have B_{(x−π^{m}v)^{2}−π^{r}δ} ≁ 1. On the other hand, for odd k A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1. Let k is even. If r > 2, then A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ (π, −πw) ∼ 1. Let k = r = 2.

B_{x^{2}−π^{k}u} ∼ (\frac{π , ∏_{σ}((μ^{σ} + π^{m−1}v) − \sqrt{u})}{k(\sqrt{u})⟨…⟩}) ∼ 1.

Thus, A ≁ B.

It is easy to check that in 2) f(π^{m}v) ∼ −πm + k + rδξv and for even r f(π^{m}v + π^{r/2}μ) ∼ (μ^{2} − δ)ξvπ^{m+k}. Moreover, E(w) ∼ αw, since k < r. So in 2) A_{x−π^{s}e} ≁ 1 and A_{x^{2}−π^{k}u} in case i) as well as in case ii). 2.a. Let firstly m be even and k be odd. Then B_{x^{2}−π^{k}u} ∼ (\sqrt{πu}, −αv) and B_{x−π^{s}e} ∼ (π, −v). There is exactly one nontrivial algebra among two last algebras. So B ≁ 1 and A ≁ B.

Let now m be odd and k be even. Then B_{x−π^{s}e} ∼ (αvξπ, −πv) ∼ (π, αξ). Furthermore, the algebra B_{x^{2}−π^{k}u} is nontrivial iff the algebra (π, ξ) is nontrivial over k. Thus, B ≁ 1, A ≁ B again. 2.b. Assume θ = π^{m}v + \sqrt{π^{r}λ}, λ ∈ O_{K}^{∗}, p = Irr_{K(\sqrt{πλ})|k}(θ). We have f(θ) ∼ vξλ(λ − δ) in K(\sqrt{πλ}).

B_{p} ∼ (\frac{\sqrt{πλ} , −δvξλ}{K(\sqrt{πλ})⟨p⟩}) .

One always can find K, λ such that B_{p} ≁ 1. If k is odd, then A_{p} ∼ (α, π) ∼ 1 over K(\sqrt{πλ}) and A ≁ B. Let k be even. A_{p} ∼ (αλw, π^{m}v + \sqrt{π^{r}λ} − πw) ∼ 1. 2.c. B_{x−π^{m}v} ≁ 1. In addition, B_{x−π^{s}e} ∼ 1, so A ≁ B. 2.d. If k is odd and m is even, then B_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1 and B_{x−π^{s}e} ∼ (α, π^{m}) ∼ 1. Otherwise, B_{x−π^{s}e} ≁ 1 and B_{x^{2}−π^{k}u} ∼ 1. 2.e. Let μ^{2} − δ ≁ 1, then f(π^{m}v + π^{r/2}μ) ∼ 1 and B_{Irr(π^{m}v+π^{r/2}μ)} ≁ 1. B_{x−π^{s}e} ≁ 1. 2.f. Assume Π = ^{3} \sqrt{π}, θ = π^{m}v + π^{m}Π, and p = Irr_{k(Π)|k}(θ). We have f(θ) ∼ 1, B_{p} ∼ (α, Ππ^{m}), B_{x−π^{s}e} ∼ (α, π^{m}). There is exactly one nontrivial algebra among the two last algebras. On the other hand, A_{x−π^{s}e} ≁ 1 and A_{p} ≁ 1. Indeed, if k is odd, then A_{p} ∼ (α, −π^{k}u) and A_{p} ∼ (αwπ, −πw). So B ≁ 1 and B ≁ A.

In 3) f(π^{m}v) ∼ π^{k+r+s}uδψ and f(π^{m}v + π^{r/2}μ) ∼ −π^{k+s}uψ(μ)(μ^{2} − δ), provided r is even. If k is odd, then A_{x^{2}−π^{k}u} ∼ (α, \sqrt{πu}) ≁ 1. If k is even, then A_{x−π^{s}e} ∼ (παw, −πw) ≁ 1. Anyway, A is nontrivial at two these places. On the other hand, A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1. 3.a. This case can be considered by analogy with 2.b). 3.b. We have B_{(x−π^{m}v)^{2}−π^{r}δ} ≁ 1 and A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1. 3.c. B ≁ 1 since B_{x−π^{m}v} ≁ 1, and B ≁ A in view of B_{x−π^{s}e} ∼ 1. 3.d. B_{(x−π^{m}v)^{2}−π^{r}δ} ∼ (π, μ − \sqrt{δ}) ≁ 1, provided μ^{2} − δ ≁ 1. So B ≁ 1, A. 3.e. For μ^{2} − δ ≁ 1 f(π^{m}v + π^{r/2}μ) ∼ 1 and B_{x−π^{m}v−π^{r/2}μ} ≁ 1. B_{x−π^{s}e} ∼ 1. 3.f. Let θ_{i} = π^{m}v + π^{s}Π^{i} ∈ k(Π), Π = ^{3} \sqrt{π}, p_{i} = Irr_{k(Π)|k}(θ_{i}), i = 1, 2. Then f(θ_{i}) ∼ −π^{k}uπ^{s}ψ(π^{2s}Π^{2i} − π^{r}δ) ∼ −uψ ∼ 1 and B_{p_{i}} ∼ (α, π^{s}Π^{i}) over k(Π). So there is exactly one nontrivial algebra among B_{p_{i}}, i = 1, 2. On the other hand, A_{p_{i}} ≁ 1. Therefore, B ≁ 1, A. 3.g. Let η_{i} = π^{s}γ_{i}, γ_{i} ∈ O_{K}^{∗}, γ̄_{i} ≠ ē, q_{i} = Irr_{K|k}(η_{i}), i = 1, 2. Then

f(η_{i}) ∼ −u(γ_{i} − e) \{ \begin{matrix}γ_{i}^{2} − δ, if m > s, \\ (γ_{i} − v)^{2} − δ, if m = s. \end{matrix} \}

We require γ_{1} − e ≁ γ_{2} − e and f(η_{i}) ∼ 1, i = 1, 2. Then B_{q_{i}} ∼ (π, (−1)^{s}(γ_{i} − e)) and A_{q_{i}} ∼ 1, i = 1, 2. Since there is exactly one nontrivial algebra among B_{q_{i}}, i = 1, 2, B ≁ 1, A.

In 4) we have f(πv) ∼ −vδπ^{r+1} and f(πv + π^{r/2}μ) ∼ v(μ^{2} − δ)π, provided r is even. In this case A = (α, x^{2} − π^{3}u) and A_{x−π^{s}e} ≁ 1, A_{x^{2}−π^{3}u} ≁ 1. 4.a. B_{x−π^{s}e} ∼ (αvπ, −πv) ≁ 1, B_{x^{2}−π^{k}u} ∼ 1, so that B ≁ 1, A. 4.b. B ≁ 1 in view of B_{x−πv} ≁ 1, and B ≁ A since B_{x−π^{s}e} ∼ 1. 4.c. Finally, in this case B_{x−π^{s}e} ≁ 1 and B_{x^{2}−π^{3}u} ∼ 1.

In 5) f(π^{m}v) ∼ −π^{k+m+l+r}uvτδ and if r is even, then

f(π^{m}v + π^{r/2}μ) ∼ (μ^{2} − δ) \{ \begin{matrix}π^{k+l+m}uvτ, if m + l < r/2, \\ −π^{k+l+m}u(μ − vτ), if m + l = r/2, \\ −π^{k+r/2}uμ, if m + l > r/2. \end{matrix} \}

We have also A_{x−π^{s}e} ≁ 1 and A_{x^{2}−π^{k}u} ≁ 1. 5.a. B_{(x−π^{m}v)^{2}−π^{r}δ} ≁ 1 and A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1. 5.b. B_{x−π^{m}v} ≁ 1, B_{x^{2}−π^{k}u} ∼ 1. 5.c. If m + l < r/2, then f(π^{m}v + π^{r/2}μ) ∼ 1, provided μ^{2} − δ ≁ 1. Otherwise, f(π^{m}v + π^{r/2}μ) ∼ (μ^{2} − δ)(vτ − μ), and we always can find such K, μ that f(π^{m}v + π^{r/2}μ) ∼ 1 and μ^{2} − δ ≁ 1 again. Anyway, B_{Irr(π^{m}v+π^{r/2}μ)} ≁ 1. B ≁ A in view of B_{x^{2}−π^{k}u} ∼ 1. 5.d. Let θ_{i} = π^{m}v + π^{(m+l)}Π^{i}, p_{i} = Irr_{k(Π)|k}(θ_{i}), i = 1, 2. Then f(θ_{i}) ∼ uvτ ∼ 1 and B_{p_{i}} ∼ (α, Π^{m+l+i}). So there is exactly one nontrivial algebra among B_{p_{i}}, i = 1, 2. On the other hand, A_{p_{i}} ≁ 1. Therefore, B ≁ 1, A. 5.e. Let θ = π^{m}v + \sqrt{π^{r}λ} ∈ K(\sqrt{πλ}) and p = Irr(θ). Then f(θ) ∼ uvτλ(λ − δ) and

B_{p} ∼ (\frac{\sqrt{πλ} , −λuvτδ}{K(\sqrt{πλ})⟨p⟩} ) .

One always can choose K and λ such that B_{p} ≁ 1. Furthermore, B ≁ A since A_{p} ∼ 1. 5.f. f(π^{m}v+π^{r/2}μ) ∼ πT (μ), so B is unramified. A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1, but B_{(x−π^{m}v)^{2}−π^{r}δ} ∼ (π, μ−\sqrt{δ}) ≁ 1. 5.g. B_{(x−π^{m}v)^{2}−π^{r}δ} ≁ 1 and A_{(x−π^{m}v)^{2}−π^{r}δ} ∼ 1. 5.h. In view of f(π^{m}v + π^{r/2}μ) ∼ −uμ(μ^{2} − δ) we can find μ such that f(π^{m}v + π^{r/2}μ) ∼ 1 and (μ^{2} − δ) ≁ 1. Then B_{Irr(π^{m}v+π^{r/2}μ)} ≁ 1. Moreover, B_{x^{2}−π^{k}u} ∼ 1. 5.j. B_{x−π^{s}e} ∼ (α, π^{r}) ≁ 1. Finally, B ≁ A since A_{x^{2}−π^{k}u} ∼ 1.

6. A = (πw, x − πw), w^{2} − u ≁ 1. So, A_{x^{2}−π^{k}u} ≁ 1 and A_{x−π^{s}e} ∼ (πw, −πw) ∼ 1. Let −1 ∼ 1. Then B_{x−π^{s}e} ∼ (π, −u) ≁ 1. Thus, B ≁ 1, A. Let now −1 ≁ 1. In this case B_{x−πw} ∼ (α(w^{2} − u), x − πw) ∼ 1, so B is unramified. Finally, B_{x−π^{s}e} ∼ (−uα, −πw) ≁ 1.

The theorem is proved.

5 Splitting type (1,4).

This section is devoted to case deg f_{1} = 4, deg f_{2} = 1 and the reduction is bad.

In our further considerations we will need lemmas from the previous sections.

Lemma 22 | _{2}Br C| = 4.

So it is enough to find one unramified and nontrivial algebra that is not isomorphic to scalar algebra (π, α). This algebra is not isomorphic to (π, α) by lemma 12.

Lemma 23 By an appropriate replacing of the variable x the polynomial f can be reduced to the form f(x) = g(x)(x − π^{l}e), l ≥ 0, e ∈ O_{k}^{∗}, g is irreducible, and either I. g(x) = (x^{2} − u)^{2} + π^{m}vx + π^{n}δ, or II. g(x) = x^{4} + π^{k}ux^{2} + π^{m}vx + π^{r}δ,

where k, m, n > 0, u, v, δ ∈ O_{k}^{∗}, and in case I u ∉ (O_{k}^{∗})^{2}.

Proof. Since g(x) is irreducible, then by Hensel’s lemma we have only the following two possibilities for ḡ(x). I. ḡ(x) = (x^{2} + px + q)^{2}, p, q ∈ k̄, p^{2} − 4q ∉ k̄^{2}, II. ḡ(x) = (x − γ)^{4}, γ ∈ k̄.

In the second case one can assume that γ = 0, i.e. g(x) = x^{4} + ax^{3} + bx^{2} + cx + d, a, b, c, d ∈ M_{k}. Moreover, by replacing x ↦ x − a/4 we can remove the monomial ax^{3}.

In the first case the same replacement gives (x^{2} + px + q)^{2} = x^{4} + b̄x^{2} + c̄x + d̄, so p = 0, b̄ = 2q, d̄= q^{2}. Suppose b = −2u, ū = −q, then d = u^{2} + λ, λ ∈ M_{k}, and g(x) = (x^{2} − u)^{2} + cx + λ. The lemma is proved.

Theorem 27 Let the curve C correspond to case I from lemma 23. Then the following algebra A is nontrivial, unramified, and not isomorphic to the scalar one. (We define Δ = g(\sqrt{u}) = π^{m}v\sqrt{u} + π^{n}δ ∈ k(\sqrt{u})).

1. v(Δ) ≡ 1 (2), πΔξ(\sqrt{u}) ∼ 1 in k(\sqrt{u}), where

ξ(x) = \{ \begin{matrix}x, if l > 0, \\ x − e, if l = 0. \end{matrix}

A = (π, x^{2} − u).

2. v(Δ) ≡ 1 (2), πΔξ(\sqrt{u}) ≁ 1.

A = (α(x^{2} − u), π(x − β)), β ∈ O_{k}^{∗}, β^{2} − u ≁ 1.

3. v(Δ) ≡ 0 (2), ξ(\sqrt{u}) ≁ 1 in k(\sqrt{u}).

A = (π, x − π^{l}e).

4. v(Δ) ≡ 0 (2), ξ(\sqrt{u}) ∼ 1.

A = (αp(x), x − β), where p = Irr_{k(\sqrt{u}|k)}(θ), θ = \sqrt{u}(1 + π^{s}w), w ∈ O_{k(\sqrt{u})}^{∗} such that μ + 4u^{2}w^{2} ∈ O_{k(\sqrt{u})}^{∗} \ (O_{k(\sqrt{u})}^{∗})^{2}, Δ = π^{2s}μ.

Proof.

1. We have f(\sqrt{u}) ∼ Δξ(\sqrt{u}), so A is unramified. Let K|k be unramified extension of odd degree and η ∈ O_{K}^{∗} such that

\{ \begin{matrix}ξ(η) ∼ 1, \\ η^{2} − u ≁ 1. \end{matrix}

Then f(η) ∼ ξ(η) ∼ 1 and

A_{Irr_{K|k}(η)} ∼ (\frac{π , η^{2} − u}{K⟨…⟩}) ≁ 1.

2. We have A_{x−β} ∼ (α^{2}, π(x − β)) ∼ 1 and

A_{x^{2}−u} ∼ (\frac{x^{2} − u , π(β − \sqrt{u})}{k(\sqrt{u})(\sqrt{Δξ(\sqrt{u})})⟨x^{2} − u⟩}) ∼ 1,

since Δξ(\sqrt{u}) ∼ π(β − \sqrt{u}) in k(\sqrt{u}).

Thus, A is unramified. If η ∈ O_{K}^{∗} satisfies the conditions

\{ \begin{matrix}ξ(η) ∼ 1, \\ η^{2} − u ∼ 1, \end{matrix}

then f(η) ∼ ξ(η) ∼ 1 and

A_{Irr_{K|k}(η)} ∼ (\frac{α(η^{2} − u) , π(η − β)}{K⟨…⟩}) ∼ (α, π) ≁ 1.

3. Let μ, s, θ, p be from item 4 of this theorem. Then g(θ) ∼ μ + 4u^{2}w^{2} ≁ 1, and f(θ) ∼ g(θ)ξ(\sqrt{u}) ∼ 1. So

A_{p} ∼ (\frac{π , θ − π^{l}e}{k(\sqrt{u})⟨p⟩}) ∼ (\frac{π , ξ(\sqrt{u})}{k(\sqrt{u})⟨p⟩}) ≁ 1.

4. In view of αp(β) ∼ α(β − \sqrt{u} − π^{s}\sqrt{u}w)(β + \sqrt{u} + π^{s}\sqrt{u}w^{σ}) ∼ α(β^{2} − u) ∼ 1 we have A_{x−β} ∼ 1. Furthermore, f(θ) ∼ μ + 4u^{2}w^{2} ≁ 1, therefore f(θ) ∼ β − \sqrt{u} in k(\sqrt{u}) and

A_{p} ∼ (\frac{p , β − \sqrt{u}}{k(\sqrt{u})(\sqrt{f(θ)})⟨p⟩}) ∼ 1.

Thus A is unramified. We prove its nontriviality. Assume that ζ = \sqrt{u}(1 + Π) ∈ k(\sqrt{u}, Π), Π = ^{3} \sqrt{π}. Then f(ζ) ∼ (u^{2}(2 + Π)^{2}Π^{2} + π^{2s}μ)ξ(\sqrt{u}) ∼ 1 and

A_{Irr_{k(\sqrt{u},Π)}}(ζ) ∼ (\frac{αp(ζ) , ζ − β}{k(\sqrt{u}, Π)⟨…⟩}) ∼ (\frac{p(ζ) , β − \sqrt{u}}{k(\sqrt{u}, Π)⟨…⟩}) ≁ 1,

since p(ζ) ∼ (ζ − θ)(ζ − θ^{σ}) ∼ 2\sqrt{u}(Π\sqrt{u} − π^{s}w\sqrt{u}) ∼ Π.

The theorem is proved.

Let us pass to case II. It is convenient to introduce the following notations.

d = min \{\frac{k}{2} , \frac{m}{3}, \frac{n}{4}\} ∈ \frac{1}{4}Z ⋃ \frac{1}{3}Z, h(t) = t^{4} + π^{k−2d}ut^{2} + π^{m−3d}vt + π^{n−4d}δ ∈ k(π^{d})[t].

Then for a suitable unramified K|k of odd degree and w ∈ O_{K}^{∗} we have

g(π^{d}w) ∼ π^{4d}h(w) ∼ h(w) ∈ O_{K(π^{d})}^{∗}.

In addition, h̄(t) ∈ k̄[t] is a monic polynomial which is of degree 4 and does not coincide with t^{4}.

Theorem 28 Assume we are in case II, d ≤ l, and h̄(t) ∉ k̄[t]^{2}. Then the following algebra A is unramified, nontrivial, and not isomorphic to the scalar one.

1. d ∈ 2Z, A = (π, x − π^{l}e).

2. d ∈ 1 + 2Z, A = (π, Irr_{K|k}(π^{d}w)), w ∈ O_{K}^{∗} such that

\{ \begin{matrix}ξ(w) ≁ 1, \\ h(w) ≁ 1, \end{matrix} where ξ(x) = \{ \begin{matrix}x, if d < l \\ x − e, if d = l. \end{matrix}

3. d ∈ 2/3 + Z, A = (π, x − π^{l}e).

4. d ∈ 1/3 + Z, A = (π, Irr_{K( ^{3} \sqrt{π})|k}(π^{d}w)), K and w are from 2.

5. d ∈ 1/4 + Z ∪ 3/4 + Z, A = (α, x − π^{l}e).

6. d ∈ 1/2 + Z, A = (−πIrr_{K(\sqrt{π})|k}(π^{d}w), αx), w ∈ O_{K}^{∗}, h(w) ≁ 1.

Proof. Since h̄(t) ∉ k̄[t]^{2} the equation y^{2} = ah̄(t), a ∈ k̄^{∗} gives a variety in A^{2}(k̄_{s}).

Therefore by Leng-Weil theorem for any N ∈ N there exists a sufficiently large extension K̃|k̄ such that |V (K̃)| > N. So there exists w̃_{i} ∈ K̃, i = 1, 2 such that h̄(w̃_{1}) ∼ 1, h̄(w̃_{2}) ≁ 1 in K̃. After lifting K̃ to K and w̃_{i} to w_{i} ∈ O_{K}^{∗} (K̄ = K̃, w̄_{i} = w̃_{i}) we have h(w_{1}) ∼ 1, h(w_{2}) ≁ 1. If w is required to have some additional property (for example, ξ(w) ≁ 1), then one can consider a variety V in A^{3}(k̄_{s})

\{ \begin{matrix}y^{2} = ah̄(t), \\ z^{2} = ᾱξ̄(t) (lemma 6). \end{matrix}

1. Let K, w be as in item 2 of this theorem. Then f(π^{d}w) ∼ h(w)ξ(w) ∼ 1, so

A_{Irr(π^{d}w)} ∼ (\frac{π , ξ(w)}{K⟨Irr(π^{d}w)⟩}) ≁ 1.

2. f(π^{d}w) ∼ h(w)ξ(w)π ∼ π, so A is unramified. We have also

A_{x−π^{l}e} ∼ (π, ∏_{σ∈G(K|k)}(π^{l}e − π^{d}w^{σ})) ∼ (π, N_{K|k}(ξ(w))) ≁ 1.

3. f(π^{d}w) ∼ wh(w) in K( ^{3} \sqrt{π}). Let w, h(w) ≁ 1, then

A_{Irr(π^{d}w)} ∼ (\frac{^{3} \sqrt{π} , w}{K( ^{3} \sqrt{π})⟨…⟩}) ≁ 1.

4. f(π^{d}w) ∼ πwh(w) ∼ π and A is unramified.

A_{x−π^{l}e} ∼ (π, ∏_{σ∈G(K|k)}(x^{3} − π^{3d}(w^{σ})^{3}))_{x−π^{l}e} ∼ (π, −πN_{K|k}(w)) ≁ 1.

5. In this case n ≡ 1 (2) and d = n/4 < min{k/2, m/3, l}. Let d < d′ < min{k/2, m/3, l}, d′ ∈ Q such that d′ = p/q, p, q ∈ N, (p, q) = 1, p, q ≡ 1 (2). Under these assumptions f(π^{d′}δ) ∼ π^{(p+nq)/q} ∼ 1 in k(π^{1/q}). Therefore

A_{Irr(π^{d′} δ)} ∼ (α, π^{d′}δ) ∼ (\frac{α , π^{1/q}}{k(π^{1/q})⟨…⟩}) ≁ 1.

6. We have

A_{x} ∼ (−π ∏_{σ} −π^{2d}(w^{σ})^{2}, αx) ∼ 1.

f(π^{d}w) ∼ π^{d}wh(w) ∼ αwπ^{d} in k(\sqrt{π}), hence

A_{Irr(π^{d}w)} ∼ (\frac{−πIrr(π^{d}w) , αwπ^{d}}{K(\sqrt{π})(\sqrt{f(π^{d}w)})⟨…⟩}) ∼ 1.

Thus, A is unramified. It is nontrivial in view of f(π^{2/7}) ∼ π^{2/7} − π^{l}e ∼ 1 in k(π^{1/7}), so

A_{x^{7}−π^{2}} ∼ (\frac{−π ∏_{σ}(π^{4/7} − π^{2d}(w^{σ})^{2}) , απ^{2/7}}{k(π^{1/7})⟨x^{7} − π^{2}⟩} ) ∼ (\frac{−π^{1/7} , α}{k(π^{1/7})⟨x^{7} − π^{2}⟩}) ≁ 1.

The theorem is proved.

Theorem 29 Let case II take place and d > l, h̄(t) ∉ k̄[t]^{2}. Then the following algebra A is unramified, nontrivial, and not isomorphic to the scalar one.

1. l ≡ 0 (2), d ∈ (1/3)Z, −e ∼ 1.

A = (α, Irr_{K|k}(π^{d}w)), h(w) ≁ 1.

2. l ≡ 0 (2), d ∈ (1/3)Z, −e ≁ 1.

A = (α, x^{q} − π^{p}), where l < p/q < d, p, q ∈ N, (p, q) = 1, p, q ≡ 1 (2).

3. l ≡ 0 (2), n ≡ 1 (2). (This is always true if d ∈ 1/4 + Z ∪ 3/4 + Z).

A = (−πδe, x).

4. l ≡ 0 (2), n ≡ 0 (2), d ∈ 1/2 + Z, −δe ≁ 1.

A = (α, x).

5. l ≡ 0 (2), n ≡ 0 (2), d ∈ 1/2 + Z, −δe ∼ 1.

A = (α, Irr_{K(\sqrt{π})|k}(π^{d}w)), h(w) ≁ 1.

6. l ≡ 1 (2), n ≡ 1 (2), −δe ∼ 1.

A = (α, x − π^{l}e).

7. l ≡ 1 (2), n ≡ 1 (2), −δe ≁ 1.

A = (α, x).

8. l ≡ 1 (2), n ≡ 0 (2), d ∈ (1/3)Z.

A = (−αeπ, Irr_{K(π^{d})|k}(π^{d}w)), h(w) ≁ 1.

9. l ≡ 1 (2), n ≡ 0 (2), d ∈ 1/2 + Z.

A = (−πδe, x). Proof.

1. f(π^{d}w) ∼ h(w)(−π^{l}e) ∼ α, so A is unramified. f(π^{p/q}) ∼ −π^{l}e ∼ 1. Then

A_{x^{p}−π^{q}} ∼ (\frac{α , ∏_{σ}(π^{p/q} − π^{d}w^{σ})}{k(π^{1/q})⟨…⟩}) ∼ (\frac{α , π^{1/q}}{k(π^{1/q})⟨…⟩}) ≁ 1.

2. f(π^{p/q}) ∼ −e, so A is unramified. Moreover, f(π^{d}w) ∼ −eh(w) ∼ 1, so that

A_{Irr(π^{d}w)} ∼ (\frac{α , −π^{p}}{k(π^{d})⟨…⟩}) ≁ 1.

3. f(0) ∼ −πδe, i.e. A is unramified. Let τ ∈ O_{K}^{∗}, τ ≁ 1, τ − e ∼ 1. Then f(π^{l}τ) ∼ τ − e ∼ 1 and A_{Irr(π^{l}τ)} ∼ (−πδe, π^{l}τ) ∼ (π, τ) ≁ 1.

4. f(0) ∼ −δe, so that A is unramified. If w satisfies h(w) ≁ δ, then f(π^{d}w) ∼ −eh(w) ∼ 1. We have

A_{Irr_{K(\sqrt{π})|k}}(π^{d}w) ∼ (\frac{α , π^{d}w}{K(\sqrt{π})⟨…⟩}) ∼ (α, \sqrt{π}) ≁ 1.

5. In this case f(π^{d}w) ∼ α and f(0) ∼ 1. Then A is unramified and A_{x} ∼ (α, −π^{2d}w) ≁ 1.

6. f(0) ∼ 1, A_{x} ∼ (α, −π^{l}) ≁ 1

7. A_{x−π^{l}e} ∼ (α, π^{l}) ≁ 1.

8. f(π^{d}w) ∼ −αeπ, i.e. A is unramified. A_{x−π^{l}e} ∼ (−αeπ, πe) ≁ 1.

9. Let θ = π^{d−1/2}\sqrt{πw} ∈ K(\sqrt{πw}). Since d ∈ 1/2+Z, then d < m/3, h(\sqrt{w}) ∼ w^{2}+π^{k−2d}uw+π^{n−4d}δ = h^{0}(w), and f(θ) ∼ −h^{0}(w)we. Note that in view of h̄(t) = h̄^{0}(t^{2}) we have h̄^{0}(t) ∉ k̄[t]^{2}. So the condition

\{ \begin{matrix}w ≁ −δe, \\ h^{0}(w) ≁ δ \end{matrix}

can be satisfied in a suitable K. Under these conditions we have f(θ) ∼ 1 and

A_{Irr_{K(\sqrt{πw})|k}} ∼ (\frac{−πδe , w^{d−1/2}\sqrt{πw}}{K(\sqrt{πw})⟨…⟩} ) ∼ (\frac{\sqrt{πw} , −wδe}{K(\sqrt{πw})⟨…⟩}) ≁ 1.

Thus, case h̄(t) ∉ k̄[t]^{2} is completely considered. Let us make the form of polynomial g(t) more precise, provided h̄(t) ∈ k̄[t]^{2}. If x^{4} + \overline{π^{k−2d}u}x^{2} + \overline{π^{m−3d}v}x + \overline{π^{n−4d}δ} = (x^{2} + αx + β)^{2}, then α = 0, \overline{π^{m−3d}v} = 0, \overline{π^{k−2d}u} = 2β, and \overline{π^{n−4d}δ} = β^{2}, β ∈ k̄^{∗}. Then n = 2k, m > 3k/2, d = k/2 ∈ (1/2)Z, δ = γ^{2} + π^{s}τ, u = −2γ, τ, γ ∈ O_{k}^{∗}, s > 0. So g(x) has the following form.

g(x) = (x^{2} − π^{k}γ)^{2} + π^{m}vx + π^{2k+s}τ.

Let Δ = g(\sqrt{π^{k}γ}) = π^{2k+s}τ + π^{m}v\sqrt{π^{k}γ} ∈ k(\sqrt{π^{k}γ}). Note that one always can suppose v(Δ) = min{2k+s, m+k/2}. Indeed, let v(Δ) > min{2k+s, m+k/2}. This means that k ≡ 0 (2), γ = β^{2} ∈ (O_{k}^{∗})^{2}, 2k + s = m + k/2 and τ̄ + v̄β̄ = 0. Then τ̄ − v̄β̄ ≠ 0, and replacement β by −β gives us the required Δ. Moreover, if k ≡ 0 (2), then equality v(Δ) = 2m − k (or v(Δ)/2 − k = m − 3k/2) is impossible. Indeed, then 2m − k ≤ m + k/2, m ≤ 3k/2.

From now on we will consider only such g(x) and Δ.

Theorem 30 Let d < l. Then the following algebra A is unramified, nontrivial, and not isomorphic to the scalar one.

1. k ≡ 0 (2), γ = β^{2}, Δ ∼ π^{d}β ∼ α.

A = (π, x − π^{l}e).

2. k ≡ 0 (2), γ = β^{2}, Δ ∼ π^{d}β ∼ πρ, ρ ∈ O_{k}^{∗}.

A = (α, x − π^{l}e).

3. k ≡ 0 (2), γ = β^{2}, Δ ∼ π^{d}β ∼ 1.

Let θ = π^{d}β(1 + π^{r}w), r = min{v(Δ)/2 − k, m − 3k/2}, w ∈ O_{K}^{∗} such that ϕ(w) ∈ O_{K}^{∗} \ (O_{K}^{∗})^{2}, where

ϕ(x) = \{ \begin{matrix}4γ^{2}x^{2} + μ, if v(Δ)/2 − k < m − 3k/2 \\ 4γ^{2}x^{2} + vβx, if v(Δ)/2 − k > m − 3k/2, \end{matrix}

Δ = π^{v(Δ)}μ, and p = Irr_{K|k}(θ).

Then A = (α, p).

4. k ≡ 0 (2), γ = β^{2}, Δπ^{d}β ∼ α, d ≡ 1 (2).

A = (α, x − π^{d}β).

5. k ≡ 0 (2), γ = β^{2}, Δπ^{d}β ∼ α, d ≡ 0 (2), β ∼ 1.

A = (α, x − π^{d}β).

6. k ≡ 0 (2), γ = β^{2}, Δπ^{d}β ∼ α, d ≡ 0 (2), β ≁ 1.

A = (α, Irr_{k( ^{3} \sqrt{π})|k}(θ′)), where θ′ = π^{d}β(1 + ^{3} \sqrt{π}).

7. k ≡ 0 (2), γ = β^{2}, Δπ^{d}β ∼ πρ, v(Δ) ≡ 1 (2).

A = (πβμ, x − π^{d}β), Δ = π^{v(Δ)}μ.

8. k ≡ 0 (2), γ = β^{2}, Δπ^{d}β ∼ πρ, v(Δ) ≡ 0 (2).

A = (πβα, p), p is from item 3.

9. k ≡ 1 (2), m + k/2 < 2k + s, γ^{m}v ∼ 1.

A = (α, x − π^{l}e).

10. k ≡ 1 (2), m + k/2 < 2k + s, γ^{m}v ≁ 1.

A = (α, x^{2} − π^{k}γ).

11. k ≡ 1 (2), m + k/2 > 2k + s.

Let θ = \sqrt{π^{k}γ}(1 + (\sqrt{πγ})^{s}w) ∈ K(\sqrt{πγ}), w ∈ O_{K}^{∗} such that γ^{s}(4γ^{s+2}w^{2} + τ) ∈ O_{K}^{∗} \ (O_{K}^{∗})^{2}, q = Irr_{K(\sqrt{πγ})|k}(θ).

Then A = (αx, q(0)q(x)).

12. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), d ≡ 1 (2), −1 ∼ 1.

Let similarly as in item 3 θ = π^{d}\sqrt{γ}(1+π^{r}w) ∈ K(\sqrt{γ}), w ∈ O_{K(\sqrt{γ})}^{∗} such that ϕ(w) ≁ 1 in K(\sqrt{γ}), p = Irr_{K(\sqrt{γ})|k}(θ).

Then A = (π, p(x)).

13. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), d ≡ 1 (2), −1 ≁ 1.

Let p be as in item 12, μ ∈ O_{k}^{∗}, μ^{2} − γ ≁ 1.

Then A = (αp(x), x − π^{d}μ).

14. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), d ≡ 0 (2), −1 ≁ 1.

A = (αp(x), x − π^{d}μ), p and μ are as in 13.

15. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), d ≡ 0 (2), −1 ∼ 1.

A = (π, x − π^{l}e).

16. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), d ≡ 0 (2), μ ∼ \sqrt{γ} in k(\sqrt{γ}), Δ = π^{v(Δ)}μ.

A = (π, x^{2} − π^{k}γ).

17. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), d ≡ 0 (2), μ ≁ \sqrt{γ}.

A = (α(x^{2} − π^{k}γ), π(x − π^{d}η)), η ∈ O_{k}^{∗}, η^{2} − γ ≁ 1.

18. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), d ≡ 1 (2), −1 ∼ 1.

A = (π, p(x)), p is as in item 12.

19. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), d ≡ 1 (2), −1 ≁ 1.

A = (αp(x), x − π^{d}η), p is from item 12, η ∈ O_{k}^{∗}, η^{2} − γ ≁ 1.

Proof.

1. f(π^{d}β) ∼ Δπ^{d}β ∼ 1, so A_{x−π^{d}β} ∼ (π, π^{d}β) ≁ 1.

2. A_{x−π^{d}β} ∼ (α, π^{d}β) ≁ 1.

3. f(θ) ∼ g(θ)θ ∼ g(θ) ∼ 4π^{2k+2r}γ^{2}w^{2}(1+π^{r}w/2)^{2} +π^{v(Δ)}μ+π^{m+d+r}vβw ∼ 4γ^{2}w^{2} +π^{2[(v(Δ)/2−k)−r]}μ+ π^{(m−3k/2)−r}vβw ∼ ϕ(w) ∼ α. Thus A is unramified. Let θ′ = π^{d}β(1 + Π) ∈ k(Π), Π = ^{3} \sqrt{π}, p′ = Irr(θ′). Then f(θ′) ∼ g(θ′) ∼ 4γ^{2} + π^{2[(v(Δ)/2−k)−1/3]}μ + π^{(m−3k/2)−1/3} ∼ 1, and

A_{p′} ∼ (\frac{α , ∏_{σ∈G(K|k)}(π^{d}β(1 + Π) − π^{d}β(1 + π^{r}w^{σ}))}{k(Π)⟨p′⟩}) ∼ (α, Π) ≁ 1.

4. f(π^{d}β) ∼ α, so A is unramified. A_{x−π^{l}e} ∼ (α, π^{d}β) ≁ 1.

5. Let θ′, p′ be from item 3 of this proof. Then f(θ′) ∼ 1 and A_{p′} ≁ 1.

6. f(θ′) ∼ g(θ′)(π^{d}β − π^{l}e) ∼ π^{d}β ∼ α, A is unramified. Let θ, p be as in 3. Then f(θ) ∼ αϕ(w) ∼ 1 and A_{p} ≁ 1.

7. f(π^{d}β) ∼ πμβ, A is unramified. If η ∈ O_{K}^{∗} satisfies the conditions η ∼ 1, η̄ ≠ β̄, η − β ≁ 1, then f(π^{d}η) ∼ π^{d}η ∼ 1 and A_{Irr(π^{d}η)} ∼ (πβμ, η − β) ≁ 1.

8. f(θ) ∼ g(θ)π^{d}β ∼ πβϕ(w), so A is unramified.

A_{x−π^{l}e} ∼ (παβ, ∏_{σ∈G}(π^{l}e − π^{d}β(1 + π^{r}w^{σ}))) ∼ (παβ, −πβ) ≁ 1.

9. f(\sqrt{π^{k}γ}) ∼ Δ\sqrt{π^{k}γ} ∼ γ^{m}v ∼ 1 in k(\sqrt{πγ}). Therefore

A_{x^{2}−π^{k}γ} ∼ (\frac{α , \sqrt{πγ}}{k(\sqrt{πγ})⟨x^{2} − π^{k}γ⟩}) ≁ 1.

10. f(\sqrt{π^{k}γ}) ∼ α, A is unramified. A_{x−π^{l}e} ∼ (α, −π^{k}γ) ≁ 1.

11. We have g(θ) ∼ γ^{s}(4γ^{s+2}w^{2} + τ) ∼ α, then f(θ) ∼ αθ.

A is unramified. Indeed, A_{x} ∼ (αx, q(0)^{2}) ∼ 1 and

A_{q} ∼ (\frac{αθ , q(0)q}{K(\sqrt{πγ})(\sqrt{αθ})⟨q⟩}) ∼ 1.

A ≁ (π, α) in view of (π, α)_{x−π^{l}e} ≁ 1, but A_{x−π^{l}e} ∼ (απ^{l}e, q(0)^{2}) ∼ 1. Finally,

A_{∞} ∼ (\frac{α , q(0)}{k⟨\sqrt{1/x}⟩}) ≁ 1,

since v_{k}(q(0)) ≡ 1 (2).

12. f(θ) ∼ π\sqrt{γ}g(θ) ∼ π, A is unramified.

A_{x−π^{l}e} ∼ (π, ∏_{σ∈G(K|k)} π^{d}\sqrt{γ}(−π^{d}\sqrt{γ})) ∼ (π, −γ) ≁ 1.

13. p(π^{d}μ) ∼ ∏_{σ}(μ − \sqrt{γ})(μ + \sqrt{γ}) ∼ μ^{2} − γ ∼ α, so that A_{x−π^{d}μ} ∼ (α^{2}, x − π^{d}μ) ∼ 1.

f(θ) ∼ πg(θ)\sqrt{γ} ∼ π(μ − \sqrt{γ}) in k(\sqrt{γ}). Then

A_{p} ∼ (\frac{αp , π(μ − \sqrt{γ})}{K(\sqrt{γ})(\sqrt{f(θ)})⟨p⟩}) ∼ 1.

Thus, A is unramified.

A_{x−π^{l}e} ∼ (α ∏_{σ} π^{d}\sqrt{γ}(−π^{d}\sqrt{γ}), −π^{d}w) ∼ (π, −αγ) ≁ 1.

14. f(θ) ∼ g(θ) ∼ μ−\sqrt{γ}, hence A is unramified. Let θ′ = π^{d}\sqrt{γ}(1+Π), Π = ^{3} \sqrt{π}. Then f(θ′) = g(θ′) ∼ 1 and

A_{Irr_{k(Π,\sqrt{γ})}(θ′)} ∼ (\frac{∏_{σ∈G(K(\sqrt{γ})|k)}((\sqrt{γ} − (\sqrt{γ})^{σ}) + Πγ − (w\sqrt{γ})^{σ}π^{r}) , μ − \sqrt{γ}}{k(Π, \sqrt{γ})⟨…⟩} ) ∼

∼ (\frac{Π , μ − \sqrt{γ}}{k(Π, \sqrt{γ})⟨…⟩}) ≁ 1.

15. f(θ) ∼ g(θ)\sqrt{γ} ∼ 1, then

A_{p} ∼ (\frac{π , \sqrt{γ}}{K(\sqrt{γ})⟨p⟩}) ≁ 1.

16. f(π^{d}\sqrt{γ}) ∼ πμ\sqrt{γ} ∼ π, A is unramified. Let ζ ∈ O_{K}^{∗}, ζ^{2} − γ ≁ 1, ζ ∼ 1. Then f(π^{d}ζ) ∼ g(π^{d}ζ) ∼ 1 and

A_{Irr(π^{d}ζ)} ∼ (\frac{π , ζ^{2} − γ}{K⟨…⟩} ) ≁ 1.

17. A_{x−π^{d}η} ∼ (α^{2}, π(x − π^{d}η)) ∼ 1.

A_{x^{2}−π^{k}γ} ∼ (\frac{x^{2} − π^{k}γ , π(η − \sqrt{γ})}{k(\sqrt{γ})(\sqrt{f(π^{d}\sqrt{γ})})⟨x^{2} − π^{k}γ⟩}) ∼ 1,

because f(π^{d}\sqrt{γ}) ∼ πμ\sqrt{γ} ∼ π(η − \sqrt{γ}), i.e. A is unramified.

A_{∞} ∼ (\frac{α , π}{k⟨\sqrt{1/x}⟩}) ≁ 1.

It remains to prove that A ≁ (π, α). Choose ζ as in 16. Then A_{Irr(π^{d}ζ)} ∼ (α(ζ^{2} − γ), …) ∼ 1. On the other hand, (π, α)_{Irr(π^{d}ζ)} ≁ 1.

18. f(θ) ∼ π\sqrt{γ}g(θ) ∼ π, A is unramified.

A_{x−π^{l}e} ∼ (π, ∏_{σ∈G(K(\sqrt{γ})|k)} −π^{d}(\sqrt{γ})^{σ}(1 + π^{r}w^{σ})) ∼ (π, −γ) ≁ 1.

19. We have

A_{x−π^{d}η} ∼ (x − π^{d}η, α ∏_{σ∈G(K(\sqrt{γ})|k)}(η − (\sqrt{γ})^{σ}(1 + π^{r}w^{σ})) ∼ (x − π^{d}η, α(η^{2} − γ)) ∼ 1,

f(θ) ∼ π\sqrt{γ}g(θ) ∼ π(η − \sqrt{γ}) in k(\sqrt{γ}). So

A_{p} ∼ (\frac{p , π^{d}(\sqrt{γ} − η)}{k(\sqrt{γ})(\sqrt{π(η − \sqrt{γ})})⟨p⟩}) ∼ 1,

i.e. A is unramified. To prove the theorem it is enough to check that A is nontrivial.

A_{x−π^{l}e} ∼ (α ∏_{σ∈G(K(\sqrt{γ})|k)} −π^{d}(\sqrt{γ})^{σ}(1 + π^{r}w^{σ}), −π^{d}η) ∼ (−αγ, −πη) ≁ 1.

Theorem 31 Let d ≥ l and if d = l, π^{k}γ = (π^{d}β)^{2} ∈ k^{2}, then b̄ ≠ ē. Assume

ξ(x) = \{ \begin{matrix}−e, if d > l, \\ x − e, if d = l. \end{matrix}

Then the following algebra is unramified, nontrivial, and not isomorphic to the scalar one.

1. π^{k}γ = (π^{d}β)^{2} ∼ 1, v(Δ) ≡ 0 (2), l ≡ 0 (2), ξ(β) ≁ 1.

A = (π, x − π^{l}e).

2. π^{k}γ ∼ 1, v(Δ) ≡ 0 (2), l ≡ 0 (2), ξ(β) ∼ 1.

A = (α, p(x)), where p is defined in item 3 of the previous theorem.

3. π^{k}γ ∼ 1, v(Δ) ≡ 0 (2), l ≡ 1 (2).

A = (παξ(β), p(x)).

4. π^{k}γ ∼ 1, v(Δ) ≡ 1 (2), l ≡ 1 (2), μ ∼ ξ(β), where Δ = π^{v(Δ)}μ.

Let Π = ^{5} \sqrt{π}, t ∈ \{1, 2\}, t ≡ d (2), θ′ = π^{d}β(1 − Π^{t}αβ) ∈ k(Π), p′ = Irr(θ′).

Then A = (πξ(β), p′(x)).

5. π^{k}γ ∼ 1, v(Δ) ≡ 1 (2), l ≡ 1 (2), μ ≁ ξ(β).

A = (α, x − π^{d}β).

6. π^{k}γ ∼ 1, v(Δ) ≡ 1 (2), l ≡ 0 (2).

A = (πμξ(β), x − π^{d}β).

7. k ≡ 1 (2), v(Δ) ≡ 0 (2) in k(\sqrt{πγ}), −eγ^{l} ∼ 1.

Let θ = \sqrt{π^{k}γ}(1 + (\sqrt{πγ})^{r}w) ∈ K(\sqrt{πγ}), w ∈ O_{K}^{∗} such that g(θ) ∼ α in K(\sqrt{πγ}), p = Irr(θ).

Then A = (α, p(x)).

8. k ≡ 1 (2), v(Δ) ≡ 0 (2) in k(\sqrt{πγ}), −eγ^{l} ≁ 1.

Let θ′ = \sqrt{π^{k}γ}(1 + ^{6} \sqrt{πγ}) ∈ k( ^{6} \sqrt{πγ}), q = Irr(θ′).

Then A = (α, q(x)).

9. k ≡ 1 (2), v(Δ) ≡ 1 (2).

A = (ρπ^{(k−1)/2}x, −πγ(x^{2} − π^{k}γ)), ρ = −ev^{l+m+(k−1)/2}.

10. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), l ≡ 1 (2), l = d, ξ(\sqrt{γ}) ≁ 1 in k(\sqrt{γ}).

Let θ = π^{d}\sqrt{γ}(1 + π^{r}w), w ∈ O_{K(\sqrt{γ})}^{∗} such that g(θ) ∈ O_{K(\sqrt{γ})}^{∗} \ (O_{K(\sqrt{γ})}^{∗})^{2}, p = Irr(θ).

Then A = (π, p(x).

11. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), l ≡ 0 (2), l = d, ξ(\sqrt{γ}) ≁ 1 in k(\sqrt{γ}).

A = (π, x − π^{l}e).

12. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 0 (2), ξ(\sqrt{γ}) ∼ 1 in k(\sqrt{γ}) (the last condition always holds if l < d).

A = (αp, π^{l+d}(x − π^{d}β)), where β ∈ O_{k}^{∗}, β^{2} − γ ≁ 1, p is from item 10.

13. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), μ ≁ ξ(\sqrt{γ}) in k(\sqrt{γ}).

A = (α(x^{2} − π^{k}γ), π^{l+d+1}(x − π^{d}β)), where β ∈ O_{k}^{∗}, β^{2} − γ ≁ 1.

14. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), μ ∼ ξ(\sqrt{γ}) in k(\sqrt{γ}), l ≡ 0(2).

A = (π, x^{2} − π^{k}γ).

15. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), μ ∼ ξ(\sqrt{γ}) ∼ 1, l ≡ 1(2).

Let Π = ^{3} \sqrt{π}, θ′ = π^{d}\sqrt{γ}(1 + Πη), η ∈ O_{k(\sqrt{γ})}^{∗} \ (O_{k(\sqrt{γ})}^{∗})^{2}, q = Irr_{k(Π,\sqrt{γ})|k}(θ′).

Then A = (π, q(x)).

16. k ≡ 0 (2), γ ≁ 1, v(Δ) ≡ 1 (2), μ ∼ ξ(\sqrt{γ}) ≁ 1, l ≡ 1(2).

A = (π, x − π^{l}e).

Proof.

1. If θ, p are the same as in item 3 of the previous theorem, then g(θ) ∼ α and f(θ) ∼ απ^{l}ξ(β) ∼ 1, so A_{p} ∼ (π, ξ(β)) ≁ 1.

2. f(θ) ∼ α, A is unramified. Let θ′ = π^{d}β(1 + Π^{t}), Π = ^{5} \sqrt{π}, t ∈ \{1, 2\}, t ≢ d (2), p′ = Irr_{k(Π)|k}(θ′). Then f(θ′) ∼ ξ(β) ∼ 1 and

A_{p′} ∼ (\frac{α , ∏_{σ}[π^{d}β(1 + Π^{t}) − π^{d}β(1 + π^{r}w^{σ})]}{k(Π)⟨p′⟩}) ∼ (\frac{α , Π^{d+t}}{k(Π)⟨p′⟩} ) ≁ 1.

3. A is unramified because of f(θ) ∼ παξ(β). We have also

A_{x−π^{l}e} ∼ (παξ(β), ∏_{σ}(π^{l}e − π^{d}β(1 + π^{r}w^{σ}))) ∼ (παξ(β), −πξ(β)) ≁ 1.

4. f(π^{d}β) ∼ Δπ^{l}ξ(β) ∼ 1, f(θ′) ∼ πξ(β), so that A is unramified. If χ^{5} = 1, χ is a primitive root, then

A_{x−π^{d}β} ∼ (πξ(β), ∏_{i=0}^{5}(π^{d}β − π^{d}β(1 − Π^{t}χ^{it}αβ))) ∼ (Πξ(β), Π^{d+t}α)) ≁ 1.

5. f(π^{d}β) ∼ α, A is unramified. A_{x−π^{l}e} ∼ (α, π^{l}ξ(β)) ≁ 1.

6. f(π^{d}β) ∼ πμξ(β). Find η ∈ O_{K}^{∗} satisfying the following conditions

\{ \begin{matrix}η̄ ≠ β̄, \\ η − e ∼ 1, \\ η ≁ 1, if l < d, \\ η − β ≁ 1, if l = d. \end{matrix}

Then f(ηπ^{l}) ∼ (η − e)g(ηπ^{l}) ∼ 1 and A_{Irr(ηπ^{l})} ∼ (πμξ(β), η − π^{d−l}β) ≁ 1.

7,8. In these cases we have l < d, d ∈ 1/2 + Z.

f(θ) ∼ g(θ)(−π^{l}e) ∼ −eγ^{l}α in K(\sqrt{πγ}),

f(θ′) ∼ g(θ′)(−π^{l}e) ∼ −eγ^{l} in K(\sqrt{πγ}).

If −eγ^{l} ∼ 1, then A = (α, p) is nontrivial and p(θ′) ∼ ∏_{σ∈G(K|k)}(\sqrt{π^{k}γ}(1 + ^{6} \sqrt{πγ}) − \sqrt{π^{k}γ}(1 + (\sqrt{πγ})^{r}w^{σ}))(\sqrt{π^{k}γ}(1 + ^{6} \sqrt{πγ}) + \sqrt{π^{k}γ}(1 + (−1)^{r}\sqrt{(}πγ)^{r}w^{σ})) ∼ 2 ^{6} \sqrt{πγ}, i.e.

A_{q} ∼ (\frac{α , ^{6} \sqrt{πγ}}{k( ^{6} \sqrt{πγ})⟨q⟩}) ≁ 1.

Let now −eγ^{l} ≁ 1. Then A = (α, q) is unramified and q(θ) ∼ −2\sqrt{πγ}. So A_{p} ≁ 1.

9. A_{x} ∼ (ρπ^{(k−1)/2}x, (−πγ)^{2}) ∼ 1.

A_{x^{2}−π^{k}γ} ∼ (\frac{−(x^{2} − π^{k}γ) , ρ\sqrt{πγ}}{k(\sqrt{πγ})(\sqrt{f(\sqrt{π^{k}γ})})⟨x^{2} − π^{k}γ⟩}) .

Since v(Δ) ≡ 1 (2) in k(\sqrt{πγ}), then 2k +s < m+k/2 and Δ ∼ π^{m}v\sqrt{π^{k}γ} ∼ γ^{m+(k−1)/2}v\sqrt{πγ} in k(\sqrt{πγ}). So f(\sqrt{π^{k}γ}) ∼ ρ\sqrt{πγ} in k(\sqrt{πγ}) and A_{x^{2}−π^{k}γ} ∼ 1. Thus A is unramified. Furthermore,

A_{∞} ∼ (\frac{π , ργ^{(k−1)/2}}{k⟨\sqrt{1/x}⟩} ) .

If l ≡ 0 (2), then choose η ∈ O_{K}^{∗} satisfying the conditions

\{ \begin{matrix}η − e ∼ 1, \\ η ≁ ργ^{(k−1)/2}. \end{matrix}

Then f(π^{l}η) ∼ (ρπ^{(k−1)/2}η, −πγ(π^{2l}η^{2} − π^{k}γ)) ∼ (π, ρηγ^{(k−1)/2}), i.e. among the algebras A_{∞}, A_{Irr(π^{l}η)} there is exactly one nontrivial. Therefore A ≁ 1, (π, α).

Let now l ≡ 1 (2). One can find η ∈ O_{K}^{∗} such that

\{ \begin{matrix}γ − η ∼ 1, \\ η ∼ −e. \end{matrix}

Then we have f(\sqrt{π^{k}η}) ∼ −π^{l}e ∼ −ηe ∼ 1 in K(\sqrt{πη}) and

A_{Irr(\sqrt{π^{k}η})} ∼ (\frac{ρπ^{(k−1)/2}\sqrt{π^{k}η} , −πγπ^{k}(η − γ)}{K(\sqrt{πη})⟨…⟩}) ∼ (\frac{\sqrt{πγ} , γ}{K(\sqrt{πη})⟨…⟩}) ≁ 1.

On the other hand,

(πα)_{Irr(\sqrt{π^{k}η})} ∼ (\frac{π , α}{K(\sqrt{πη})⟨…⟩}) ∼ 1.

10,11. Since f(θ) ∼ π^{l}ξ(\sqrt{γ})δ, δ ∈ O_{K(\sqrt{γ})}^{∗} \ (O_{K(\sqrt{γ})}^{∗})^{2}, then in 10 A is unramified and in 11

A_{p} ∼ (\frac{π , π^{l}ξ(\sqrt{γ})}{K(\sqrt{γ})⟨p⟩}) ≁ 1.

Moreover, in the first case A_{x−π^{l}e} ∼ (π, e^{2} − γ) ≁ 1.

12. p(π^{d}β) ∼ β^{2} − γ ∼ α, therefore A_{x−π^{d}β} ∼ 1. f(θ) ∼ g(θ)ξ(\sqrt{γ})π^{l} ∼ δπ^{l}, δ ≁ 1 in k(\sqrt{γ}). Then

A_{p} ∼ (\frac{p , π^{l+d}(π^{d}\sqrt{γ} − π^{d}β)}{K(\sqrt{γ})(\sqrt{π^{l}δ})⟨p⟩}) ∼ (\frac{p , π^{l}(β − \sqrt{γ})}{K(\sqrt{γ})(\sqrt{π^{l}δ})⟨p⟩}) ∼ 1.

So, A is unramified. We have also

A_{∞} ∼ (\frac{α , π^{l+d}}{k⟨\sqrt{1/x}⟩}) , A_{x−π^{l}e} ∼ (α, π^{d}),

the latter equality because of p(π^{l}e) ∼ 1. If l ≡ 1 (2), then exactly one among these algebras is nontrivial, so A ≁ 1, (π, α).

Let l ≡ 0 (2). Suppose θ′ = π^{d}\sqrt{γ}(1 + Π), Π = ^{3} \sqrt{π}. Then f(θ′) ∼ ξ(\sqrt{γ}) ∼ 1 and

p(θ′) ∼ ∏_{σ}(π^{d}\sqrt{γ}(1 + Π) − (\sqrt{γ})^{σ}π^{d}(1 + π^{r}w^{σ})) ∼ Π.

We have

A_{Irr_{k(Π,\sqrt{γ})|k}(θ′)} ∼ (\frac{Π , β − \sqrt{γ}}{k(Π, \sqrt{γ})⟨…⟩}) ≁ 1.

Note, that (π, α) is trivial at this place.

13. A_{x−π^{d}β} ∼ (α^{2}, …) ∼ 1. f(π^{d}\sqrt{γ}) ∼ π^{l+1}μξ(\sqrt{γ}), then

A_{x^{2}−π^{k}γ} ∼ (\frac{x^{2} − π^{k}γ , π^{l+d+1}(π^{d}\sqrt{γ} − π^{d}β)}{k(\sqrt{γ})(\sqrt{f(π^{d}\sqrt{γ})})⟨x^{2} − π^{k}γ⟩}) ∼ 1,

i.e. A is unramified.

A_{∞} ∼ (α, π^{l+d+1}), A_{x−π^{l}e} ∼ (α \{ \begin{matrix}1, l < d \\ e^{2} − γ, l = d \end{matrix} \} , π^{d+1}) .

If l ≡ 1 (2), then we have just one nontrivial algebra, therefore A ≁ 1, (π, α). The same is true in case l ≡ 0 (2), ξ(\sqrt{γ}) ≁ 1, since then l = d and e^{2} − γ ≁ 1. Let l ≡ 0 (2), ξ(\sqrt{γ}) ∼ 1. For the θ′ from the previous item we have

A_{Irr_{k(Π,\sqrt{γ})|k}(θ′)} ∼ (\frac{(θ′)^{2} − π^{k}γ , π^{d+1}(θ′ − π^{d}β)}{k(Π, \sqrt{γ})⟨…⟩}) ∼ (\frac{2Π , π^{d+1}(Π(β − \sqrt{γ})}{k(Π, \sqrt{γ})⟨…⟩}) ≁ 1.

14,15,16. f(π^{d}\sqrt{γ}) ∼ π^{l+1}μξ(\sqrt{γ}) ∼ π^{l+1}. If μ ∼ ξ(\sqrt{γ}), l ≡ 0 (2) (case 14), then (π, x^{2} − π^{k}u) is unramified. If μ ∼ ξ(\sqrt{γ}), l ≡ 1 (2), then f(\sqrt{π^{k}γ}) ∼ 1 and

(π, x − π^{l}e)_{x^{2}−π^{k}γ} ∼ (\frac{π , π^{l}ξ(\sqrt{γ})}{k(\sqrt{γ})⟨x^{2} − π^{k}γ⟩}) ∼ (π, ξ(\sqrt{γ})) ≁ 1,

provided ξ(\sqrt{γ}) ≁ 1 (case 16). In case 15 we have

q(π^{d}\sqrt{γ}) ∼ ∏_{i=0}^{2}(π^{d}\sqrt{γ} − π^{d}\sqrt{γ}(1 + χ^{i}Πη))(π^{d}\sqrt{γ} + π^{d}\sqrt{γ}(1 + χ^{i}Πη^{σ})) ∼ ∏_{i=0}^{2} χ^{i}Πη ∼ πη,

where χ^{3} = 1, χ is a primitive root. So

A_{x^{2}−π^{k}γ} ∼ (\frac{π , πη}{k(\sqrt{γ})⟨x^{2} − π^{k}γ⟩}) ≁ 1.

To complete the proof one only need to check that A ≁ 1 in case 14 and A is nontrivial in 15. In these cases f(θ′) ∼ π^{l}ξ(\sqrt{γ}). Then f(θ′) ∼ π in 15 and A is unramified. If case 14 takes place and l < d, then f(θ′) ∼ −e ∼ 1 in k(\sqrt{γ}). Therefore

A_{q} ∼ (\frac{π , π^{k}γ(1 + Πη)^{2} − π^{k}γ}{k(Π, γ)⟨q⟩} ) ∼ (\frac{Π , η}{k(Π, γ)⟨q⟩}) ≁ 1.

Finally, let l = d. Assume ζ ∈ O_{K}^{∗} such that ζ̄ ≠ ē, ζ − e ∼ 1, and ζ^{2} − γ ≁ 1. Then f(π^{d}ζ) ∼ ζ − e ∼ 1 and A_{Irr_{K|k}(π^{d}ζ)} ∼ (π, ζ^{2} − γ) ≁ 1.

Theorem 32 Let d = l, π^{k}γ = (π^{d}β)^{2} ∈ k^{2}, and e = β + π^{n}δ, n > 0, δ ∈ O_{k}^{∗}.

Let firstly v(Δ) ≡ 0 (2) and θ = π^{d}β(1 + π^{r}w), w ∈ O_{K}^{∗} such that ϕ(w) ∈ O_{K}^{∗} \ (O_{K}^{∗})^{2}, p = Irr_{K|k}(θ), r = min{v(Δ)/2 − k, m − 3k/2} (see theorem 30, item 3). Then the following algebra A is unramified, nontrivial, and not isomorphic to the scalar one.

1. r ≤ n, r + d ≡ 1 (2). A = (αN_{K|k}(χ(w))π, p(x)), where

χ(x) = \{ \begin{matrix}βx, if r < n, \\ βx − δ, if r = n. \end{matrix}

2. r ≤ n, r + d ≡ 0 (2). A = (π, x − π^{d}e).

3. r > n, n + d ≡ 1 (2). A = (−πδα, p(x)).

4. r > n, n + d ≡ 0 (2), −δ ≁ 1. A = (π, x − π^{d}e).

5. r > n, n + d ≡ 0 (2), −δ ∼ 1. A = (α, p(x)).

Let now v(Δ) ≡ 1 (2), Δ = π^{v(Δ)}μ. Then the following algebra A is unramified, nontrivial, and not isomorphic to the scalar one.

6. n + d ≡ 1 (2), −μδ ≁ 1. A = (α, x − π^{d}β).

7. n + d ≡ 1 (2), −μδ ∼ 1. A = (α, x − π^{d}e).

8. n + d ≡ 0 (2), n ≤ r. A = (−μδπ, x − π^{d}β).

9. n + d ≡ 0 (2), n > r. A = (−πμα, x − π^{d}β(1 + π^{n}w)), where w ∈ O_{k}^{∗} such that δ − βw ≁ 1. (If there is no such w in k, then A = (−πμα, Irr_{K|k}(π^{d}β(1 + π^{n}w))), w ∈ O_{K}^{∗})

Proof. Let v(Δ) ≡ 0 (2). Then

f(θ) ∼ α(π^{d}β(1 + π^{r}w) − π^{d}(β + π^{n}δ)) ∼ \{ \begin{matrix}π^{d+r}αχ(w), if r ≤ n, \\ −π^{d+n}αδ, if r > n, \end{matrix}

χ(w) ∼ N_{K|k}(χ(w)) in K.

So, in cases 1,3,5 the algebra A is unramified.

In 4 f(θ) ∼ −αδ ∼ 1, then

A_{p} ∼ (\frac{π , π^{d}β(1 + π^{r}w) − π^{d}(β + π^{n}δ)}{K⟨p⟩}) ∼ (π, −π^{n+d}δ) ≁ 1.

In 2 f(θ) ∼ αχ(w). System

\{ \begin{matrix}ϕ(w) ≁ 1, \\ χ(w) ≁ 1 \end{matrix}

is always solvable in a sufficiently large extension K|k. For such w and p we have

A_{p} ∼ (π, π^{d}β(1 + π^{r}w) − π^{d}(β + π^{n}δ)) ∼ (\frac{π , χ(w)}{K⟨p⟩}) ≁ 1.

In 1,3

p(π^{d}e) ∼ π^{d} ∏_{σ∈G(K|k)}(π^{n}δ − π^{r}βw^{σ}) ∼ \{ \begin{matrix}−π^{d+r}N_{K|k}(χ(w)), r ≤ n, \\ π^{d+n}δ, r = n. \end{matrix}

Then in

1. A_{x−π^{d}e} ∼ (αN_{K|k}(χ(w))π, −πN_{K|k}(χ(w))) ≁ 1.

2. A_{x−π^{d}e} ∼ (−πδα, πδ) ≁ 1. Thus, for case v(Δ) ≡ 0 (2) it is enough to check that in 5 A ≁ 1. To do this, assume θ′ = π^{d}β(1 + Π^{t}) ∈ k(Π), Π = ^{5} \sqrt{π}, t ∈ \{1, 2\}, t ≢ d (2), p′ = Irr_{k(Π)|k}(θ′). Then f(θ′) ∼ g(θ′) ∼ 1 and

p(θ′) ∼ ∏_{σ∈G(K|k)}(π^{d}β(1 + Π^{t}) − π^{d}β(1 + π^{r}w^{σ})) ∼ π^{d}βΠ^{t}.

So,

A_{p′} ∼ (\frac{α , Π^{d+t}}{k(Π)⟨p′⟩} ) ≁ 1.

Consider now case v(Δ) ≡ 1 (2). We have f(π^{d}β) ∼ −π^{v(Δ)}μπ^{d+n}βδ ∼ −μδπ^{d+n+1}, therefore in 6 A is unramified and in 7 f(π^{d}β) ∼ 1, A_{x−π^{d}β} ∼ (α, π^{d+n}) ≁ 1. Furthermore, in 6 A is nontrivial. Indeed, then A_{x−π^{d}e} ∼ (α, π^{d+n}) ≁ 1.

In 8 f(π^{d}β) ∼ −μδπ^{d+n+1} and A is unramified. Let r = min{v(Δ)/2 − k, m − 3k/2} ∈ (1/2)Z, θ = π^{d}β(1 + (−αμδπ)^{r}w) ∈ K((−αμδπ)^{r}), w ∈ O_{K((−αμδπ)^{r})}^{∗}, p = Irr_{K((−αμδπ)^{r})|k}(θ). Then

g(θ) ∼ 4γ^{2}(−αμδ)^{2r}w^{2} + π^{v(Δ)−2k−2r}μ + π^{m−3k/2−2r}vβw(−αμδπ)^{r},

f(θ) ∼ g(θ)(−δ + βw(−αμδπ)^{r−n}) ∼ −g(θ)δ \{ \begin{matrix}1, r > n, \\ 1 + μβw, r = n. \end{matrix} \}

Moreover,

A_{p} ∼ (\frac{−μδπ , π^{d}β(−αμδπ)^{r}}{K((−αμδπ)^{r})(\sqrt{f(θ)})⟨p⟩}) .

If r ∈ Z, then

A_{p} ∼ (\frac{π , (−α)^{r}(μδ)^{d}βw}{K(\sqrt{f(θ)})⟨p⟩} ) .

One always can find K, w such that f(θ) ∼ 1 and w ≁ (−α)^{r}(μδ)^{d}β. Then we have A_{p} ≁ 1. Consider r ∈ 1/2 + Z. In this case

A_{p} ∼ (\frac{α , \sqrt{−αμδπ}}{K(\sqrt{−αμδπ})(\sqrt{f(θ)})⟨p⟩}) .

In addition, f(θ) ∼ −δ(4γ^{2}(−αμδ)^{2r}w^{2} + μ). So, A_{p} ≁ 1, provided f(θ) ∼ 1.

9. Since v(Δ) = min{2k + s, m + k/2}, then v(Δ) < m + n + k/2. Suppose θ_{0} = π^{d}β(1 + π^{n}w). g(θ_{0}) ∼ γ^{2}(2π^{n}w + π^{2n}w^{2})^{2} + π^{v(Δ)−2k}μ + π^{m−3k/2+n}vβw. In view of v(Δ) − 2k < m − 3k/2 + n and v(Δ)−2k < 2n (the latter inequality holds because of v(Δ)/2−k < r < n) we have g(θ_{0}) ∼ πμ. Therefore f(θ_{0}) ∼ πμ(βw − δ) and A is unramified. Finally, A_{x−π^{d}e} ∼ (−πμα, δ − βw) ≁ 1.

The theorem is proved.

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