Wild division algebras over Laurent series fields

A.B. Zheglov ^{1} ^{2}

Abstract

In this paper we study some special classes of division algebras over a Laurent series field with arbitrary residue field. We call the algebras from these classes as splittable and good splittable division algebras. It is shown that theses classes contain the group of tame division algebras. For the class of good division algebras a decomposition theorem is given. This theorem is a generalization of the decomposition theorems for tame division algebras given by Jacob and Wadsworth in [6]. For both clases we introduce a notion of a δ -map and develop a technique of δ -maps for division algebras from these classes. Using this technique we reprove several old well known results of Saltman and get the positive answer on the period-index conjecture: the exponent of A is equal to its index for any division algebra A over a C_{2} -field F , when F = F_{1}((t_{2})) , where F_{1} is a C_{1} -field (see [10], 3.4.5.). The paper includes also some other results about splittable division algebras, which, we hope, will be useful for the further investigation of wild division algebras.

1 Introduction

In this paper we study some class of division algebras over a Laurent series field with arbitrary residue field. Namely, we study division algebras which satisfy the following condition: there exists a section D̄ ↪ D of the residue homomorphism D → D̄, where D is a central division algebra over a complete discrete valued field F = k((t)) . We say that these division algebras are splittable. If chark = 0 , all such division algebras are tame and therefore belong to the group of tame division algebras, which was carefully studied in the papers [6] and [10] even in a much more general situation of a henselian field F of arbitrary characteristic. So, we consider mostly wild division algebras.

An extensive analysis of the wild division algebras of degree p over a field F with complete discrete rank 1 valuation with char(F̄) = p was given by Saltman in [11] (Tignol in [13] analyzed more general case of the defectless division algebras of degree p over a fild F with Henselian valuation). Here we study splittable division algebras of arbitrary index. This class (which is not a subgroup in Br(F) ) contains a class of good splittable division algebras (see the definition in section 2), which posess several beautiful properties. In particular, we prove a decomposition theorem for such algebras. This theorem is a generalization of the decomposition theorems for tame division algebras given by Jacob and Wadsworth in [6].

For arbitrary splittable division algebras we give only several assorted results, and the study of this class is far from to be complete. Nevertheless, we investigate here technical tools, which are important for the study of such algebras, and prove a relation between the level and a higher order level for some splittable division algebras (see section 6). We hope this technique will be applied to the study of the cyclisity question for certain division algebras od degree p^{k}.

1Supported by the Graduiertenkolleg ”Geometrie und Nichtlineare Analisys” of the DFG

2e-mail address: azheglov@mathematik.hu-berlin.de

As an application we get several results, which are partly well known (see proposition 6) and party not. In particular, we get the positive answer on the following conjecture: the exponent of A is equal to its index for any division algebra A over a C_{2} -field F = F_{1}((t_{2})) , where F_{1} is a C_{1} -field.

Here is a brief overview of this paper.

In section 2 we give a definition of splittable and good splittable division algebras and prove that all tame division algebras over F = k((t)) are good splittable.

Section 3 contains the most important technical tools for the study of splittable division algebras. We define a notion of δ -maps and investigate a theory of δ -maps for such algebras. In this section we define also the notion of a local height, which is a possible generalization of Saltman’s level.

In section 4 we prove the period-index conjecture metioned above. This section contains also a small history of the question known to the author. We note that the proof does not use all the results from section 3.

In section 5 we study good splittable division algebras and prove the decomposition theorem.

In section 6 we reprove some results of Saltman about semiramified division algebras of index p over F using the technique from section 3. Then we define a notion of a higher order level and prove several general properties of splittable division algebras satisfying the following condition: Z(D̄)/F̄ is a simple extension. At the end of section we put several open questions.

We use the notation of [6]. We always denote by D a division algebra finite dimensional over its center F = k((t)) = Z(D) . Recall that any Henselian valuation on F has a unique extension to a valuation on D. We denote the valuation on F by v and its unique extension on D by w.

Given a valuation w on D, we denote by Γ_{D} its value group, by V_{D} its valuation ring, by M_{D} its maximal ideal and by D̄ = V_{D}/M_{D} its residue division ring.

By [12], p.21 one has the fundamental inequality

[D : F] ≥ |Γ_{D} : Γ_{E}| · [D̄ : F̄].

D is called defectless over F if equality holds and defective otherwise. It is known that D is defectless if it has a discrete valuation of rank 1.

Jacob and Wadsworth in [6] introduced the basic homomorphism

θ_{D} : Γ_{D}/Γ_{F} → Gal(Z(D̄)/F̄)

induced by conjugation by elements of D. They showed that θ_{D} is surjective and Z(D̄) is the compositum of an abelian Galois and a purely inseparable extension of F̄.

We say D is tame division algebra if char(F̄) = 0 or char(F̄) = q ≠ 0 , D is defectless over F , Z(D̄) is separable over F̄, and q |̸|ker(θ_{D})|. We say D is wild division algebra if it is non tame.

We call a division algebra D inertially split if Z(D̄) is separable over F̄, the map θ_{D} is an isomorphism, and D is defectless over F .

Acknowledgements

I am grateful to Professor A. N. Parshin, Professor E.-W. Zink, and M. Grabitz for useful discussions and attention to my work. I am very grateful to Professor A.Wadsworth for carefully reading my paper and for showing me a mistake in the very first version of this paper and to Professor V.I.Yanchevskii for valuable discussions during his visit in Berlin. Finally, I thank my wife Olga for her support and encouragement.

2 Cohen’s theorem

Recall one definition from [14].

Definition 1 A division algebra D is said to be splittable if there is a homomorphism D̄ ↪ O_{D} ⊂ D that is a section of the map O_{D} → D̄.

There is a natural question if there exists a generalization of Cohen’s theorem, i.e. is any central division algebra splittable or not. It is not true if a division algebra is not finite dimensional over its centre, as Dubrovin’s example in [14] shows. It is not true also for some finite dimensional division algebras, as the example to theorem 2.7. in [11] shows. But it is true for tame division algebras over complete discrete valued fields. This easily follows from results of Jacob and Wadsworth [6] (compare with [14], Th.1).

Theorem 1 Let (F, v) be a valued field which is complete with respect to a discrete rank 1 valuation v. Suppose charF = charF̄. Let D be a tame division algebra with Z(D) = F and [D : F] < ∞.

Then there exists a section D̄ ↪ D of the residue homomorphism D → D̄.

Proof. Since F is a complete field, F is a Henselian field and v extends uniquely to a valuation w on D. Since D is tame, Z(D̄)/\overline{Z(D)} is a cyclic Galois extension. There exists an inertial lift Z of Z(D̄) over F , Z is Galois over F , and by classical Cohen’s theorem there exists a section Z̃(D̄) ↪ Z.

Consider the centraliser C = C_{D}(Z) of Z in D. Then we have C̄ = D̄.

Indeed, by Double Centraliser Theorem we have [D : F] = [C : F][Z : F] and [Z : F] = |Gal(Z(D̄)/F̄)|. By [6], prop.1.7 a homomorphism θ_{D} : Γ_{D}/Γ_{F} → Gal(Z(D̄)/F̄) is surjective, so for any parameter z we have θ_{D}(w(z)) = σ, where < σ >= Gal(Z(D̄)/F̄) . It is clear that z ∉ C. Now let u_{1}, …, u_{[C:F]} be a F -basis of C. It is easy to see that the elements u_{j}, zu_{j}, …, z^{n−1}u_{j}, j = 1, …, [C : F] , where n = ord(σ) , the order of σ, are linearly independent, so form a basis for D over F . Since

w(F ⟨zu_{j}, …, z^{n−1}u_{j}, j = 1, …, [C : F]⟩) ∩ Γ_{C} = 0,

where F ⟨zu_{j}, …, z^{n−1}u_{j}, j = 1, …, [C : F]⟩ denote a vector space in D over F generated by elements u_{j}z^{i}, this implies that for any element x ∈ D with w(x) = 0 we can find elements r_{1}, … r_{[C:F]} ∈ F such that x = r_{1}u_{1} +…+r_{[C:F]}u_{[C:F]} mod M_{D}. Hence C̄ = D̄.

Note that C is an unramified division algebra. Indeed, by [6], th.2.8, th.2.9 C contains a copy of the inertial lift of a maximal separable subfield in C̄, say C̃. Then the centralizer C_{C}(C̃) must be a totally ramified division algebra, i.e. it is trivial and C̃ is a maximal subfield. So, C must be unramified.

Fix an embedding i : F̄ ↪ F . It can be extended to the embedding i′ : Z̄ ↪ Z, i′|F̄ = i by Hensel lemma. Now consider the algebra A = C̄ ⊗Z̄ Z(C) . It is easy to see that A is an unramified division algebra with Ā = C̄ = D̄. Therefore by [3], Th.31, A ≅ C; so there exists a section D̄ ↪ C.

The theorem is proved. □

Later we will see that much more can be said about good splittable algebras:

Definition 2 A division algebra D is called good splittable if there exists a section s : D̄ ↪ D compatible with an embedding i : \overline{Z(D)} ↪ Z(D) , i.e. s(\overline{Z(D)}) = i(\overline{Z(D)}) ⊂ Z(D) .

It’s easy to see that all tame division algebras are good splittable, because by Hensel lemma any embedding \overline{Z(D)} ↪ Z(D) can be uniquely extended to any separable extension of Z(D) .

It is interesting to know what kind of splittable division algebras are good splittable. By theorem 3.9. in [11] even a splittable division algebra D of degree p = charD is

not a good splittable algebra if the level of D (the notion of level we will recall in section 3, see remark to lemma 7) is divisible by p. Nevertheless, it is an open question whether it is true or not, for example, for division algebras with D̄ = Z(D̄) such that D̄/F̄ is a simple extension and the local height (see the definition in the same remark) is not divisible by p. We will discuss this question in section 6.

3 Delta-maps of splittable algebras

In this section we develop some ideas from [14], where some properties of δ -maps for special kind of local skew fields were studied. Technical properties of δ -maps play the main role in all our results. Here we will give a list of these properties.

Let D be a finite dimensional division algebra over a complete valued field F = k((t)) . Let w be a unique extension of the valuation v to D. We will denote by z any parameter of D, i.e. any element with ⟨w(z)⟩ = Γ_{D}. Consider the ring ℤ⟨α, δ⟩ of noncommutative polinomials in two variables. Define the map

σ : ℤ⟨α, σ⟩ → ℤ⟨α, δ, δ_{i}; i ≥ 1⟩,
σ(α^{a_{1}}δ^{b_{1}} … α^{a_{n}}δ^{b_{n}}) = α^{a_{1}}δ_{b_{1}} … δ_{b_{n−1}}α^{a_{n}−1}δ^{b_{n}},

where a_{1}, b_{n} ≥ 0 , a_{i}, b_{j} ≥ 1 , i > 1 , j < n for every word in ℤ⟨α, δ⟩.

Let S_{i}^{k} ∈ ℤ⟨α, δ⟩, i ≥ k, i ≥ 1 be polynomials given by the following formula:

S_{i}^{k} = ∑_{τ∈S_{i}/G} τ(\underbrace{α … α}_{i−k} \underbrace{δ … δ}_{k}),

where S_{i} is a permutation group and G is an isotropy subgroup.

Lemma 1 ([14], lemma 2) The polynomials S_{i}^{k} satisfy the following property:

S_{i}^{i} = δ^{i}, S_{i}^{0} = α^{i}, S_{i+1}^{k+1} = αS_{i}^{k+1} + δS_{i}^{k}

For any splittable division algebra can be defined a notion of δ -maps:

Proposition 1 ([14], prop. 1,2) Let D be a splittable division algebra. Fix some parameter z and some embedding u : D̄ ↪ D. Then D is isomorphic to a division algebra D̄((z)) , which is defined to be the vector space of series with multiplication defined by the formula

zaz^{−1} = α(a) + δ_{1}(a)z + δ_{2}(a)z^{2} + …, a ∈ D̄,

where α : D̄ → D̄ is an automorphism and δ_{i} : D̄ → D̄ are linear maps such that the map δ_{i} satisfy the identity

δ_{i}(ab) = ∑_{k=0}^{i} σ(δ^{i−k}α)(a)σ(S_{i}^{k}α)(b), a, b ∈ D̄

Remark Note that the values σ(S_{i}^{k}α) and σ(δ^{i−k}α) belong to the subring ℤ⟨α, δ_{i}, i ≥ 1⟩, so the formula is well defined.

Note that δ -maps depend on the choice of a parameter and an embedding. The automorphism α, as it easy to see, depend only on the choice of a parameter. In the proposition we identify D̄ with u(D̄) .

Corollary 1 ([14], corol. 1) Suppose α = Id. Then

δ_{i}(ab) = δ_{i}(a)b + ∑_{k=1}^{i} δ_{i−k}(a) ∑_{(j_{1},...,j_{l})} C_{i−k+1}^{l}δ_{j_{1}} … δ_{j_{l}}(b),

where δ_{0} = α and the second sum is taken over all the vectors (j_{1}, …, j_{l}) such that 0 < l ≤ min{i − k + 1, k}, j_{m} ≥ 1 , ∑ j_{m} = k.

Further we will need even more general definition.

Definition 3 In the situation of proposition 1 let us define maps _{m}^{(z,u)}δ_{i} : D̄ → D̄, m ∈ ℤ, i ∈ ℕ as follows.

z^{m}az^{−m} = u(^{(z)}α^{m}(ā)) + u(_{m}^{(z,u)}δ_{1}(ā))z + u(_{m}^{(z,u)}δ_{2}(ā))z^{2} + …, a ∈ u(D̄).

If m = 0 , put _{m}^{(z,u)}δ_{i} = 0 .

Note that ^{(z)}α|_{Z(D̄)} does not depend on the choice of z.

Note that if ^{(z)}α = id, then _{m}^{(z,u)}δ_{i} = 0 for m = p^{k}, where k is sufficiently large, k depends on i. Moreover, _{m}^{(z,u)}δ_{i} = _{m+p^{k}}^{(z,u)}δ_{i} for k sufficiently large. We will use also the following notation:

_{m}^{(z,u)}δ̃_{i} = _{−m}^{(z,u)}δ_{i}, _{1}^{(z,u)}δ_{i} = ^{(z,u)}δ_{i}

Sometimes, we will write _{m}δ_{i} instead of _{m}^{(z,u)}δ_{i} and _{m}^{(z,u)}δ_{i}(a) instead of u(_{m}^{(z,u)}δ_{i}(ā)) whenever the context is clear.

Immediately from the definition follows

Lemma 2 In the situation of definition 3 we have (i) for |m| > 1

_{m}^{(z,u)}δ_{i}(a) = ^{(z)}α^{sign(m)}(_{sign(m)(|m|−1)}^{(z,u)}δ_{i}(a)) + _{sign(m)}^{(z,u)}δ_{i}(^{(z)}α^{sign(m)(|m|−1)}(a))+

∑_{j=1}^{i−1} _{sign(m)}^{(z,u)}δ_{j}(_{sign(m)(|m|−1)}^{(z,u)}δ_{i−j}(a)),

where sign(m) = m/|m|, a ∈ D̄; (ii) for any m ≠ 0

(z)α^{−m}(_{m}^{(z,u)}δ_{i}) + _{−m}^{(z,u)}δ_{i}(^{(z)}α^{m}) + ∑_{j=1}^{i−1} _{−m}^{(z,u)}δ_{j}(_{m}^{(z,u)}δ_{i−j}) = 0

Proposition 2 For fixed z, u from proposition 1 we have (i) The maps _{m}^{(z,u)}δ_{i} satisfy the following identities:

mδ_{i}(ab) = _{m}δ_{i}(a)α^{i+m}(b) + α^{m}(a)_{m}δ_{i}(b) + ∑_{k=1}^{i−1}_{m}δ_{i−k}(a)_{i−k+m}δ_{k}(b)

(ii) Suppose α = id. Then the maps _{m}^{(z,u)}δ_{i} satisfy the following identities:

mδ_{i}(ab) = _{m}δ_{i}(a)b + a_{m}δ_{i}(b) + ∑_{k=1}^{i−1}_{m}δ_{i−k}(a) ∑_{(j_{1},...,j_{l})} C_{i−k+m}^{l}δ_{j_{1}} … δ_{j_{l}}(b)

where the second sum is taken over all the vectors (j_{1}, …, j_{l}) such that 0 < l ≤ min{i − k + m, k}, j_{m} ≥ 1 , ∑ j_{m} = k; C_{j}^{k} = 0 if j = 0 , and C_{j}^{k} = C_{j+p^{q}}^{k} for q >> 0 if j ≤ 0 .

Proof. For any a, b ∈ D̄ we have

α^{m}(ab)z^{m} +_{m}δ_{1}(ab)z^{m+1} +_{m}δ_{2}(ab)z^{m+2} + … = z^{m}(ab) =
(α^{m}(a)z^{m} +_{m}δ_{1}(a)z^{m+1} +_{m}δ_{2}(a)z^{m+2} + …)b  (1)

If we represent the right-hand side of (1) as a series with coeffitients shifted to the left and then compare the corresponding coeffitients on the left-hand side and right-hand side, we get some formulas for _{m}δ_{i}(ab) . We have to prove that these formulas are the same as in our proposition.

Let

z^{i+m−k}b = α^{i+m−k}(b)z^{i+m−k} + … + x′_{k}z^{i+m} + …

and

(α^{m}(a)z^{m} +_{m}δ_{1}(a)z^{m+1} +_{m}δ_{2}(a)z^{m+2} + …)b = α^{m}(ab)z^{m} + y_{m+1}z^{m+1} + y_{m+2}z^{m+2} + …

Then we have

y_{i+m} = α^{m}(a)x′_{i} + ∑_{k=0}^{i−1}_{m}δ_{i−k}(a)x′_{k}

In the proof of [14], prop.2 we have shown that

z^{i+1−k}b = α^{i+1−k}(b)z^{i+1−k} + … + σ(S_{i}^{k}α)(b)z^{i+1} + …

Hence x′_{k} = σ(S_{i+m−1}^{k}α)(b) for k < i. It is easy to see that x′_{i} = _{m}δ_{i}(b) , x′_{0} = α^{i+m}(b) and σ(S_{i+m−1}^{k}α) = _{i+m−k}δ_{k}, which proves (i).

For α = id, by corollary 1,

σ(S_{i+m−1}^{k}α)(b) = ∑_{(j_{1},...,j_{l})} C_{i−k+m}^{l}δ_{j_{1}} … δ_{j_{l}}(b),

where l, j_{1}, …, j_{l} were defined in our proposition. This proves (ii). The proposition is proved. □

Lemma 3 ([14], lemma 3 )

In the situation of proposition 1 suppose _{i}^{(z,u)}δ_{j} is the first map such that _{i}^{(z,u)}δ_{j}(a) ≠ 0 for given a ∈ D̄, i ∈ ℤ\{0}, i.e. _{i}^{(z,u)}δ_{1}(a) = … = _{i}^{(z,u)}δ_{j−1}(a) = 0 , _{i}^{(z,u)}δ_{j}(a) ≠ 0 (so we have a map i ↦ j(i) ). Then

(i) for z′ = z + u(b)z^{q+1} , b ∈ D̄ we have ^{(z′)}α^{i}(a) = ^{(z)}α^{i}(a) , _{i}^{(z′,u)}δ_{k}(a) = _{i}^{(z,u)}δ_{k}(a) for k < q and

_{i}^{(z′,u)}δ_{q}(a) = _{i}^{(z,u)}δ_{q}(a) + b′^{(z)}α^{q+i}(a) −^{(z)}α^{i}(a)b′,

where b′ = ∑_{k=0}^{i−1} ^{(z)}α^{k}(b) .

(ii) Suppose ^{(z)}α^{n}|_{Z(D̄)} = id, n ≥ 1 , a ∈ Z(D̄) and _{1}^{(z,u)}δ_{1}(^{(z)}α^{k}(a)) = … = _{1}^{(z,u)}δ_{j−1}(^{(z)}α^{k}(a)) = 0 for any k.

Then for z′ = z +u(b)z^{q+1} , b ∈ D̄ we have ^{(z′)}α^{i}(a) = ^{(z)}α^{i}(a) , _{i}^{(z′,u)}δ_{k}(a) = _{i}^{(z,u)}δ_{k}(a) for k < q + j and

_{i}^{(z′,u)}δ_{q+j}(a) = _{i}^{(z,u)}δ_{q+j}(a) + b′^{(z)}α^{q}(_{i}^{(z,u)}δ_{j}(a)) − _{i}^{(z,u)}δ_{j}(a)^{(z)}α^{j}(b′)+

b′ ∑_{k=1}^{q}^{(z)}α^{q−k}(^{(z,u)}δ_{j}(^{(z)}α^{k+i−1}(a))) − _{i}^{(z,u)}δ_{j}(a) ∑_{k=0}^{j−1}^{(z)}α^{k}(b),

where b′ = ∑_{k=0}^{i−1} ^{(z)}α^{k}(b) , if n|q or ^{(z)}α(a) = a.

In particular, if ^{(z)}α = id and (i, p) = 1 , then

_{i}^{(z′,u)}δ_{q+j}(a) = _{i}^{(z,u)}δ_{q+j}(a) + (q − j)_{i}^{(z,u)}δ_{j}(a)b

(iii) for z′ = u(b)z, b ∈ Z(D̄) , b ≠ 0 we have ^{(z′)}α(a) = ^{(z)}α(a) , ^{(z′,u)}δ_{k}(a) = (z,u)δ_{k}(a) for k < j and

(z′,u)δ_{j}(a) = ^{(z,u)}δ_{j}(a)^{(z)}α(b^{−1}) ⋯^{(z)}α^{j}(b^{−1})

if i = 1 .

Proof. (i) We have z′^{i}az′^{−i} = (1+b′z^{q}+…)z^{i}az^{−i}(1+b′z^{q}+…)^{−1} = (z^{i}az^{−i}+b′z^{q}z^{i}az^{−i}+…)(1−b′z^{q}+…) =

(z^{i}az^{−i} − z^{i}az^{−i}b′z^{q} + … + b′z^{q}z^{i}az^{−i} − …) =

(z^{i}az^{−i} − [^{(z)}α^{i}(a) + _{i}^{(z,u)}δ_{j}(a)z^{j} + …]b′z^{q} + b′z^{q}[^{(z)}α^{i}(a) + _{i}^{(z,u)}δ_{j}(a)z^{j} + …] + …) =

(z^{i}az^{−i} − [^{(z)}α^{i}(a)b′ + _{i}^{(z,u)}δ_{j}(a)^{(z)}α^{j}(b′)z^{j} + …]z^{q} + b′^{(z)}α^{q+i}(a)z^{q} + …) = (z^{i}az^{−i} + (−^{(z)}α^{i}(a)b′ + b′^{(z)}α^{q+i}(a))z^{q} + …) = ^{(z)}α^{i}(a) + … + _{i}^{(z,u)}δ_{q−1}(a)z^{′q−1}+

(_{i}^{(z,u)}δ_{q}(a) + b′^{(z)}α^{q+i}(a) −^{(z)}α^{i}(a)b′)z^{′q} + …

(ii) Put c = z^{′i}z^{−i} − 1 − b′z^{q+i}. So, w(c) > q + i. Note that c^{(z)}α^{k}(a) = ^{(z)}α^{k}(a)c, since n|q or ^{(z)}α(a) = a and a ∈ Z(D̄) . We have z^{′i}az^{′−i} = (1+b′z^{q}+c)z^{i}az^{−i}(1+b′z^{q}+c)^{−1} = (z^{i}az^{−i}+b′z^{q}z^{i}az^{−i}+cz^{i}az^{−i})(1+b′z^{q}+c)^{−1} = (^{(z)}α^{i}(a)+_{i}^{(z,u)}δ_{j}(a)z^{j}+…+_{i}^{(z,u)}δ_{q+j}(a)z^{q+j}+…+b′z^{q}(^{(z)}α^{i}(a)+_{i}^{(z,u)}δ_{j}(a)z^{j}+…))(1+b′z^{q}+c)^{−1} =

(^{(z)}α^{i}(a) + b′^{(z)}α^{q+i}(a)z^{q} +^{(z)}α^{i}(a)c + _{i}^{(z,u)}δ_{j}(a)z^{j} + … + _{i}^{(z,u)}δ_{q+j}(a)z^{q+j} + … +

b′ ∑_{k=1}^{q}(^{(z)}α^{q−k}(^{(z,u)}δ_{j}(^{(z)}α^{k+i−1}(a))))z^{q+j} + b′(^{(z)}α^{q}(_{i}^{(z,u)}δ_{j}(a)))z^{q+j} + …)(1 + b′z^{q} + c)^{−1} =

(z)α^{i}(a)+[_{i}^{(z,u)}δ_{j}(a)z^{j}+…+_{i}^{(z,u)}δ_{q+j}(a)z^{q+j}+…+b′ ∑_{k=1}^{q}(^{(z)}α^{q−k}(^{(z,u)}δ_{j}(^{(z)}α^{k+i−1}(a))))z^{q+j}+

b′(^{(z)}α^{q}(_{i}^{(z,u)}δ_{j}(a)))z^{q+j} + …)](1 − b′z^{q} − c + …) =

(z)α^{i}(a)+_{i}^{(z,u)}δ_{j}(a)z^{j} +…+_{i}^{(z,u)}δ_{q+j}(a)z^{q+j} +…+b′ ∑_{k=1}^{q}(^{(z)}α^{q−k}(^{(z,u)}δ_{j}(^{(z)}α^{k+i−1}(a))))z^{q+j}+

b′(^{(z)}α^{q}(_{i}^{(z,u)}δ_{j}(a))z^{q+j} + … − _{i}^{(z,u)}δ_{j}(a)^{(z)}α^{j}(b′)z^{q+j} + … =

(z)α^{i}(a) + … + _{i}^{(z,u)}δ_{q+j−1}(a)z^{′q+j−1} + (_{i}^{(z,u)}δ_{q+j}(a) + b′^{(z)}α^{q}(_{i}^{(z,u)}δ_{j}(a)) − _{i}^{(z,u)}δ_{j}(a)^{(z)}α^{j}(b′)

+b′ ∑_{k=1}^{q}^{(z)}α^{q−k}(^{(z,u)}δ_{j}(^{(z)}α^{k+i−1}(a))) − _{i}^{(z,u)}δ_{j}(a) ∑_{k=0}^{j−1}^{(z)}α^{k}(b))z^{′q+j} + …,

since z^{′j} = z^{j} + ∑_{k=0}^{j−1} ^{(z)}α^{k}(b)z^{q+j} + . . ..

(iii) We have

z′az^{′−1} = bzaz^{−1}b^{−1} = ^{(z)}α(a) + b^{(z,u)}δ_{j}(a)^{(z)}α^{j}(b^{−1})z^{j} + … =
(z)α(a) +^{(z,u)}δ_{j}(a)^{(z)}α(b^{−1}) …^{(z)}α^{j}(b^{−1})z^{′j} + …,

since ^{(z′)}α|_{Z(D̄)} = ^{(z)}α|_{Z(D̄)} . □

Corollary 2 In the situation of lemma 3 we have

j = w(xu(a)x^{−1} − u(a)),

where x ∈ D is any element with w(x) = i, if a ∈ Z(D̄) , α(a) = a and (i, p) = 1 , where p = charD.

If i = 1 , we will denote j by j(u, a) or by i(u, a) .

Proof. Since for some parameter z we have x = b(1+x_{1}z+…)z^{i}, where b, x_{k} ∈ u(D̄) , the proof is easily follows from the proof of (ii) in lemma 3. □

In the sequel we will need the following definition.

Definition 4 Let (α, β) be endomorphisms of a division algebra D. A map δ : D → D′, where D ⊂ D′ are algebras, is called a (α, β) -derivation if it is linear and satisfy the following identity

δ(ab) = δ(a)α(b) + β(a)δ(b)

where a, b ∈ D. We will say that (α, 1) -derivation is an α -derivation.

Lemma 4 (cf. [14], lemma 4) Let δ be an (α, β) -derivation of an arbitrary division algebra D such that α, β preserve Z(D) and α|_{Z(D)} ≠ β|_{Z(D)} .

Then δ is an inner derivation, i.e. there exists d ∈ D such that

δ(a) = dα(a) − β(a)d

for all a ∈ D.

Proof. Put d = δ(a)(a^{α} − a^{β})^{−1} , where a ∈ Z(D) is any element such that α(a) ≠ β(a) . Put δ_{in}(x) = dα(x) − β(x)d. We claim that δ = δ_{in}. Indeed, consider the map δ̄ = δ − δ_{in}. It is an (α, β) -derivation. Take arbitrary b ∈ D. Then δ̄(ab) = δ̄(ba) . But we have

δ̄(ab) = δ̄(a)α(b) + β(a)δ̄(b) = β(a)δ̄(b),

and

δ̄(ba) = δ̄(b)α(a) + β(b)δ̄(a) = α(a)δ̄(b)

Therefore, δ̄(b) = 0 for any b. □

Proposition 3 (cf. [14], lemma 10) Let D be a splittable division algebra. Let n = Gal(Z(D̄)/\overline{Z(D)}) . There exists a parameter z′ such that

_{m}^{(z′,u)}δ_{j} = 0

if n |̸j.

Proof. Since for n = 1 there is nothing to prove, we will assume that n > 1 . Let z be some fixed parameter. By [6], prop. 1.7 ^{(z)}α|_{Z(D̄)} has order n.

By proposition 2, ^{(z,u)}δ_{1} is a (^{(z)}α^{2},^{(z)}α) -derivation. Since n > 1 , ^{(z)}α^{2}|_{Z(D̄)} ≠ (z)α|_{Z(D̄)} . Therefore, by lemma 4, ^{(z,u)}δ_{1} is an inner derivation and ^{(z,u)}δ_{1}(a) = d^{(z)}α^{2}(a)− (z)α(a)d, a ∈ D̄. Put z_{1} = z − u(d)z^{2} . By lemma 3, (i) we have for any a ∈ D̄ (z_{1},u)δ_{1}(a) = 0 and ^{(z)}α(a) = ^{(z_{1})}α(a) . So, ^{(z_{1},u)}δ_{1} = 0 and ^{(z)}α = ^{(z_{1})}α.

By proposition 2, ^{(z_{1},u)}δ_{2} is a (^{(z_{1})}α^{3},^{(z_{1})}α) -derivation. If n ≠ 2 then it is inner and we can apply lemma 3. By induction we get that there exists a parameter z_{n−1} such that (z_{n−1},u)δ_{j} = 0 for j < n and ^{(z)}α = ^{(z_{n−1})}α. It is easy to see that then _{m}^{(z_{n−1},u)}δ_{j} = 0 for j < n and all m ∈ ℤ. Note that ^{(z_{n−1},u)}δ_{n} is a (^{(z_{n−1})}α^{n+1},^{(z_{n−1})}α) = (^{(z_{n−1})}α,^{(z_{n−1})}α) - derivation, i.e. ^{(z_{n−1},u)}δ_{n}^{(z_{n−1})}α^{−1} is a derivation.

Note that ^{(z_{n−1},u)}δ_{n+1} is a (^{(z_{n−1})}α^{2},^{(z_{n−1})}α) -derivation. This follows by proposition 2, since _{m}^{(z_{n−1},u)}δ_{j} = 0 for j < n and all m ∈ ℤ. So, by lemma 4, ^{(z_{n−1},u)}δ_{n+1} is an inner derivation. Using lemma 3, (i) with z_{n+1} = z_{n−1} + bz_{n−1}^{n+2} for an appropriate b, we have ^{(z_{n+1},u)}δ_{j} = 0 for j < n + 2 , n |̸j and ^{(z)}α = ^{(z_{n+1})}α. Moreover, _{m}^{(z_{n+1},u)}δ_{j} = 0 for j < n + 2 , n |̸j and all m ∈ ℤ. This easily follows from lemma 2.

By induction we can assume that there exists a parameter z_{k} such that _{m}^{(z_{k},u)}δ_{j} = 0 for j < k + 1 , n |̸j and all m ∈ ℤ, and ^{(z)}α = ^{(z_{k})}α.

So, by proposition 2, if n |̸k+1 , then ^{(z_{k},u)}δ_{k+1} is an inner (^{(z_{k})}α^{k+2},^{(z_{k})}α) -derivation. And if n|k + 1 , we can apply the same arguments and conclude that ^{(z_{k},u)}δ_{k+2} is a (^{(z_{k})}α^{k+2},^{(z_{k})}α) -derivation. Therefore, by lemma 3 there exists a parameter z_{k+1} = z_{k} + bz_{k}^{k+2} ( z_{k} + bz_{k}^{k+3} if n|k + 1 ) such that _{m}^{(z_{k+1},u)}δ_{j} = 0 for j < k + 2 , n |̸j and all m ∈ ℤ, and ^{(z)}α = ^{(z_{k+1})}α (or _{m}^{(z_{k+1},u)}δ_{j} = 0 for j < k + 3 , n |̸j and all m ∈ ℤ, and (z)α = ^{(z_{k+1})}α if n|k + 1 ).

Since z_{l+1} = (1 + b_{l}z_{l}^{k_{l}})z_{l} for every l, the sequence {z_{l}}_{l=1}^{∞} converges in D, which completes the proof of the proposition. □

Lemma 5 Let D be a splittable division algebra as in proposition 1, of characteristic p > 0 . Let t ∈ Z(D̄) be an element such that α(t) = t.

Let j = i(u, t) be the minimal positive integer such that ^{(z,u)}δ_{j}|_{𝔽_{p}(t)} ≠ 0 (see corollary 2), and we assume j < ∞. Then the maps _{n}^{(z,u)}δ_{m}, kj ≤ m < (k+1)j, k ∈ {1, …, p−1} satisfy the following properties: i) there exist elements c_{n,m,k} ∈ D̄ such that

_{n}^{(z,u)}δ_{m}|_{𝔽_{p}(t)} = c_{n,m,1}δ + … + c_{n,m,k}δ^{k},

where δ : 𝔽_{p}(t) → 𝔽_{p}(t) is a derivation such that δ(t) = 1 , and

c_{n,kj,k} = (k!)^{−1}_{n}^{(z,u)}δ_{j}(t)_{n+j}^{(z,u)}δ_{j}(t) … _{n+(k−1)j}^{(z,u)}δ_{j}(t).

ii) Let ζ = ord(^{(z)}α|_{Z(D̄)}) . Then ζ|j and c_{n,kj,k} ≠ 0 if (n, j) = 1 and ^{(z)}α(^{(z,u)}δ_{j}(t)) ≠ ^{(z,u)}δ_{j}(t) ; c_{n,kj,k} ≠ 0 if ^{(z)}α(^{(z,u)}δ_{j}(t)) = ^{(z,u)}δ_{j}(t) and n, (n + j), …, (n + (k − 1)j) ≠ 0 mod p.

If ^{(z)}α = id, then c_{n,kj,k} ≠ 0 iff n, (n + j), …, (n + (k − 1)j) ≠ 0 mod p.

Proof. i) The proof is by induction on k. Let a, b ∈ 𝔽_{p}(t) . For k = 1 , by proposition 2, (ii) we have

nδ_{m}(ab) = _{n}δ_{m}(a)b + a_{n}δ_{m}(b)

because all the maps δ_{q}, q < j are equal to zero on 𝔽_{p}(t) . Hence, _{n}δ_{m} is a derivation on 𝔽_{p}(t) , _{n}δ_{m}|_{𝔽_{p}(t)} = c_{n,m,1}δ and c_{n,j,1} = _{n}δ_{j}(t) .

For arbitrary k, by proposition 2, (i) and by the induction hypothesis we have

nδ_{m}(t^{q}) = q_{n}δ_{m}(t)t^{q−1} +_{n}δ_{j}(t)(∑_{l=0}^{q−2}(c_{n+j,m−j,1}δ + … + c_{n+j,m−j,k−1}δ^{k−1})(t^{q−1−l})t^{l})+

… +_{n}δ_{m−j}(t)(∑_{l=0}^{q−2}(c_{m−j+n,m−s,1}δ)(t^{q−1−l})t^{l}).  (2)

Therefore, _{n}δ_{m}(t^{p}) = 0 , because k ≤ p − 1 and ∑_{l=0}^{p−2} δ^{i}(t^{p−1−l})t^{l} = 0 for i ≤ p − 2 . Hence, _{n}δ_{m}|_{𝔽_{p}(t)} = c_{n,m,1}δ + … + c_{n,m,p−1}δ^{p−1} and we only have to show that c_{n,m,q} = 0 for q > k.

Using (2) we can calculate c_{n,m,j}. We have

c_{n,m,1} = _{n}δ_{m}(t);

c_{n,m,2} = \frac{1}{2!}(_{n}δ_{m}(t^{2}) − 2c_{n,m,1}t) = \frac{1}{2}(_{n}δ_{j}(t)(c_{n+j,m−j,1}δ(t)) + … +_{n}δ_{s}(t)(c_{s+n,m−s,1}δ(t)))

…

c_{n,m,q} = \frac{1}{q!}(_{n}δ_{j}(t)(∑_{l=0}^{q−2} c_{n+j,m−j,q−1}δ^{q−1}(t^{q−1−l})t^{l}) + …

+_{n}δ_{m−(q−1)j}(t)(∑_{l=0}^{q−2} c_{m+n−(q−1)j,(q−1)j,q−1}δ^{q−1}(t^{q−1−l})t^{l}))

= \frac{1}{q}(_{n}δ_{j}(t)c_{n+j,m−j,q−1} + … +_{n}δ_{m−(q−1)j}(t)c_{m+n−(q−1)j,(q−1)j,q−1})  (3)

Hence, c_{n,m,k+1} = … = c_{n,m,p−1} = 0 and

c_{n,kj,k} = q^{−1}_{n}δ_{j}(t)c_{n+j,kj−j,k−1} = (k!)^{−1}_{n}^{(z,u)}δ_{j}(t)_{n+j}^{(z,u)}δ_{j}(t) … _{n+(k−1)j}^{(z,u)}δ_{j}(t).

ii) Let us prove first that ζ divide i. For, if i is not divisible by ζ, we have, by proposition 2,

(z,u)δ_{j}(tx) = ^{(z,u)}δ_{j}(t)^{(z)}α^{j+1}(x) +^{(z)}α(t)^{(z,u)}δ_{j}(x) = ^{(z,u)}δ_{j}(xt) =
(z,u)δ_{j}(x)^{(z)}α^{j+1}(t) +^{(z)}α(x)^{(z,u)}δ_{j}(t),

where x ∈ Z(D̄) , α(x) ≠ x. But then ^{(z)}α^{j+1}(x) = ^{(z)}α(x) , a contradiction. If ^{(z)}α = id, the same arguments show that ^{(z,u)}δ_{j}(t) ∈ Z(D̄) . If x ∈ D̄ is an arbitrary element, this formulae shows ^{(z)}α^{j} is an inner automorphism

ad(^{(z,u)}δ_{j}(t)^{−1}) . Therefore, ^{(z)}α^{j}(^{(z,u)}δ_{j}(t)) = ^{(z,u)}δ_{j}(t) .

Assume ^{(z)}α(^{(z,u)}δ_{j}(t)) ≠ ^{(z,u)}δ_{j}(t) . It’s clear then that

_{n+qj}^{(z,u)}δ_{j}(t) = ∑_{l=0}^{n+qj−1}^{(z)}α^{l}(^{(z,u)}δ_{j}(t)) ≠ 0

if (n, j) = 1 . So, c_{n,kj,k} ≠ 0 by (i) in this case.

If ^{(z)}α(^{(z,u)}δ_{j}(t)) = ^{(z,u)}δ_{j}(t) , then _{n+qj}^{(z,u)}δ_{j}(t) = (n + qj)^{(z,u)}δ_{j}(t) ≠ 0 iff p does not divide (n + qj) . So, by (i) c_{n,kj,k} ≠ 0 in this case iff n, (n + j), …, (n + (k − 1)j) ≠ 0 mod p.

The lemma is proved. □

Lemma 6 Let D be a splittable division algebra as in lemma 5. Let s ∈ Z(D̄) be an element such that α(s) = s. Let i = i(u, s) be the minimal positive integer such that (z,u)δ_{i}(s) ≠ 0 (see corollary 2).

If p|i, then for any positive integral k there exists a map ^{(z,u)}δ_{j(k)} such that (z,u)δ_{j(k)}(s^{p^{k}}) ≠ 0 .

Proof. We claim that ^{(z,u)}δ_{p^{q}i} is the first map such that ^{(z,u)}δ_{p^{q}i}|_{𝔽_{p}(s^{p^{q}})} ≠ 0 . The proof is by induction on q. For q = 0 , there is nothing to prove. For arbitrary q, put t = s^{p^{q−1}} . By proposition 2 we have

δ_{p^{q}i}(t^{p}) = δ_{p^{q−1}i}(t) ∑_{r=0}^{p−2}_{1+p^{q−1}i}δ_{p^{q−1}i(p−1)}(t^{p−1−r})t^{r} + ∑_{l=p^{q−1}i+1}^{p^{q}i−1} δ_{l}(t) ∑_{r=0}^{p−2}_{1+l}δ_{p^{q}i−l}(t^{p−1−r})t^{r}

By induction and lemma 5, _{1+l}δ_{p^{q}i−l}|_{𝔽_{p}(t)} = c_{1+l,p^{q}i−l,1}δ + … + c_{1+l,p^{q}i−1,p−2}δ^{p−2} for l > p^{q−1}i. Therefore, ∑_{r=0}^{p−2} _{1+l}δ_{p^{q}i−l}(t^{p−1−r})t^{r} = 0 . By lemma 5, (ii), _{1+p^{q−1}i}δ_{p^{q−1}i(p−1)}|_{𝔽_{p}(t)} = c_{1+p^{q−1}i,p^{q−1}i(p−1),1}δ + … + c_{1+p^{q−1}i,p^{q−1}i(p−1),p−1}δ^{p−1} with c_{1+p^{q−1}i,p^{q−1}i(p−1),p−1} ≠ 0 . Hence, δ_{p^{q}i}(t^{p}) = −c_{1+p^{q−1}i,p^{q−1}i(p−1),p−1}δ_{p^{q−1}i}(t) ≠ 0 .

The same arguments show that ^{(z,u)}δ_{j}(t^{p}) = 0 for j < p^{q}i. So, ^{(z,u)}δ_{p^{q}i} is the first non-zero map on 𝔽_{p}(s^{p^{q}}) . □ Lemma 7 Let D be a splittable division algebra. Let z be a fixed parameter and ^{(z)}α = id, let u be some fixed embedding u : D̄ ↪ D.

Let ^{(z,u)}δ_{i}, i ∈ ℕ ∪ ∞ be the first non-zero map on D̄. Assume (i, p) = 1 , where p = charD. Let ^{(z,u)}δ_{j}, j > i, j ∈ ℕ ∪ ∞ be the first map such that ^{(z,u)}δ_{j} ≠ 0 if j is not divisible by i and ^{(z,u)}δ_{j} ≠ c_{j/i}^{(z,u)}δ_{i}^{j/i} for some c_{j/i} ∈ D̄ otherwise. Then

a) for k < p = charD (arbitrary k if charD = 0 ) we have ^{(z,u)}δ_{ki} = c_{k}^{(z,u)}δ_{i}^{k}, where

c_{k} = \frac{(i + 1) … (i(k − 1) + 1)}{k!},  (4)

if ki < j.

b) if condition (4) is satisfied for any k with ki < j, then _{−i}^{(z,u)}δ_{q} = 0 for i < q < j and _{−i}^{(z,u)}δ_{j} is a derivation.

Remark. We will call the number i(u, z) = min_{a∈D̄}{w(zu(a)z^{−1} − u(a))} defined in this lemma a local height. The number i = i(z, u) in lemma coinside with the level of D defined in [11] if D has index p = charD and D is splittable. As it follows from lemmas 3, 10 (see below), i(z, u) does not depend on z, u in this case. Corollary 2 completes then the proof that it coinside with the level defined by Saltman in the case D is splittable. This number will play an important role in this work. It was one of the important parameters in [14]. Recall the definition of level: h(D) = min{w(ab − ba) − w(a) − w(b)}.

Proof. If we compare coefficients in formulae for δ_{ki}(ab) from proposition 2 with coefficients in formulae for δ_{i}^{k}(ab) multiplied by c_{k}, we must have

c_{k}k = ((k − 1)i + 1)c_{k−1},

where from follows a).

¿From the other hand side, if _{−i}δ_{q}, q > i is the first nonzero map after _{−i}δ_{i}, it must be a derivation by proposition 2, (i). Note that in characterictic zero case this can happens only if q ≥ j, because a map cδ_{i}^{k} can not be a derivation if k > 1 , which proves b) in this case.

Since the maps δ_{q} are uniquely defined, by lemma 2, by the maps δ̃_{l}, l ≤ q, and the maps δ̃_{q} are uniquely defined by the maps _{−i}δ_{l}, l ≤ q, and _{−i}δ_{q} are linear combinations of δ_{l}, l ≤ q with integer coefficients, we see that b) holds in arbitrary characteristic. □

Remark. So we see that the maps _{i}δ_{q} in this lemma satisfy the same identities as δ_{q/i}. This can be thought of as a possible reduction from level i to level 1 .

Definition 5 Let D be a splittable division algebra. Let u be some fixed embedding u : D̄ ↪ D. Let s ∈ Z(D̄) be an element such that α(s) = s. Let i = i(u, s) be the minimal positive integer such that ^{(z,u)}δ_{i}(s) ≠ 0 ( corollary 2 shows that i does not depend on z). Assume (i, p) = 1 , where p = charD. Define

d(u, s) = max_{z}{w(z^{−i}u(s)z^{i} − u(s) − u(_{−i}^{(z,u)}δ_{i}(s))z^{i})} ∈ ℕ ∪ ∞,

As we can see from lemma 7 b), d(u, s) can be interpreted under some conditions as the number j there. So, this definition was motivated by this lemma.

Lemma 8 In the definition above for p = charD > 0 and ^{(z)}α|_{Z(D̄)} = id we have i) d(u, s) = 2i mod p if d(u, s) < ∞; ii) If _{−i}^{(z,u)}δ_{i}(s) ≠ 0 , the map _{−i}^{(z,u)}δ_{d(u,s)+(p−1)i} is the first map such that _{−i}^{(z,u)}δ_{d(u,s)+(p−1)i}(s^{p}) ≠

0 for any parameter z. In particular, if d(u, s) = ∞, [u(s^{p}), z^{i}] = 0 .

Proof. (ii) Let _{−i}^{(z,u)}δ_{κ} be the first map such that _{−i}^{(z,u)}δ_{κ}(s^{p}) ≠ 0 . By corollary 2 κ does not depend on z. By the same reason, _{−i}^{(z,u)}δ_{i} is the first map such that _{−i}^{(z,u)}δ_{i}(s) ≠ 0 for any z.

Put w := d(u, s) + (p − 1)i and fix u, z. By proposition 2 we have

−iδ_{w}(s^{p}) = _{−i}δ_{d(u,s)}(s) ∑_{q=0}^{p−2}_{d(u,s)−i}δ_{(p−1)i}(s^{p−1−q})s^{q}+

∑_{k=d(u,s)+1}^{w−1}_{−i}δ_{k}(s) ∑_{q=0}^{p−2}_{k−i}δ_{w−k}(s^{p−1−q})s^{q}

By lemma 5, _{k−i}δ_{w−k}|_{𝔽_{p}(s)} = c_{k−i,w−k,1}δ + … + c_{k−i,w−k,p−2}δ^{p−2} for w − k < (p − 1)i and d(u,s)−iδ_{(p−1)i}|_{𝔽_{p}(s)} = c_{d(u,s)−i,(p−1)i,1}δ+…+c_{d(u,s)−i,(p−1)i,p−1}δ^{p−1} with c_{d(u,s)−i,(p−1)i,p−1} ≠ 0 if d(u, s) − i = i mod p. Indeed, as we have shown in the proof of lemma 5, (ii), the order n of the automorphism ^{(z)}α on ^{(z,u)}δ_{i}(s) must divide i, so (n, p) = 1 . Now we have two possibilities: n |̸d(u, s) and n|d(u, s) .

In the first case we can repeat the arguments to the first assertion in lemma 5, (ii) to show that c_{d(u,s)−i,(p−1)i,p−1} ≠ 0 . In the second case we have _{d(u,s)−i+qi}δ_{i}(s) = (d(u, s)−i+ qi)/i_{i}δ_{i}(s) ≠ 0 if d(u, s)−i+qi is not divided by p. So, by lemma 5, (i) c_{d(u,s)−i,(p−1)i,p−1} ≠ 0 iff d(u, s) − i = i mod p in this case.

Hence,

−iδ_{w}(s^{p}) = −_{−i}δ_{d(u,s)}(s)c_{d(u,s)−i,(p−1)i,p−1} ≠ 0

if d(u, s) − i = i mod p.

This also shows that _{−i}δ_{w} is the first map such that _{−i}δ_{w}|_{𝔽_{p}(s^{p})} ≠ 0 if d(u, s) − i = i mod p.

i) By Skolem-Noether theorem there exists a parameter z′ in D such that ^{(z′)}α = id. Put

d′(u, z′, s) = w(z^{′−j}u(s)z^{′j} − u(s) − u(_{−i}^{(z′,u)}δ_{i}(s))z^{′i}).

Since ^{(z′)}α = id, the map ^{(z′,u)}δ_{i} is the first map such that _{−i}^{(z′,u)}δ_{i}(s) ≠ 0 . If d′(u, z′, s) ≠ 2i mod p, we can find a parameter z′′ such that d′(u, z′′, s) > d′(u, z′, s) using lemma 3, (ii). Continuing this procedure, we find a parameter z such that d′(u, z, s) = 2i mod p or d′(u, z, s) = ∞.

Using arguments from ii) we get that the map _{−i}^{(z,u)}δ_{d′(u,z,s)+(p−1)i} is the first map such that _{−i}^{(z,u)}δ_{d′(u,z,s)+(p−1)i}(s^{p}) ≠ 0 for the parameter z. As it was noted in the beginning of the proof, the number κ = d′(u, z, s) + (p − 1)i does not depend on the parameter.

Since d′(u, z, s) ≤ d(u, s) , we get d′(u, z, s) = d(u, s) . For, otherwise we can repeat the arguments from (ii) and conclude that _{−i}^{(z,u)}δ_{d(u,s)+(p−1)i}(s^{p}) = 0 , a contradiction. The lemma is proved. □

It would be interesting to know more about a behaviour of _{m}^{(z,u)}δ_{j} with respect to the embedding u. We will give an answer in one special case, namely, when D̄ = Z(D̄) and Z(D̄)/\overline{Z(D)} is a simple extension.

Lemma 9 Let D be a division algebra such that charD = p > 0 , D̄ = Z(D̄) , Z(D̄) is not perfect and Z(D̄)/\overline{Z(D)} is a simple extension (so, D is splittable). Let ū be a primitive element of the extension Z(D̄)/\overline{Z(D)} such that ū ∉ (Z(D̄))^{p} and let u be any lift of ū in D.

Then there exists an embedding u : D̄ ↪ D such that u(ū) = u and any map _{m}^{(z,u)}δ_{j} is uniqely defined by the values _{m}^{(z,u)}δ_{j}(u^{q}) or, equivalently, by the values _{l}^{(z,u)}δ_{k}(u) , k ≤ j.

In particular, if _{m}^{(z,u)}δ_{k}(u) = 0 for k ≤ j, then _{m}^{(z,u)}δ_{j} = 0 .

Proof. Consider a field Z(D)(u) . It is a complete discrete valued field as a finite extension of Z(D) . By classical Cohen theorem, there exists an embedding \overline{Z(D)(u)} = D̄ ↪ Z(D)(u) ⊂ D. By [4], lemmas 11,12 the embedding is completely defined by a p -basis Γ of the field \overline{Z(D)(u)} . Namely, for any lift G of a given p -basis Γ there exists an embedding s such that G ⊂ s(\overline{Z(D)(u)}) .

Let’s show that there exists a p -basis Γ of the field D̄ such that ū ∈ Γ and Γ ∋ γ ∈ \overline{Z(D)} if γ ≠ ū.

Consider a set of all non-void sets Γ′ of elements γ_{τ} ∈ D̄ satisfying the following property: A) ū ∈ Γ′, Γ′ ∋ γ ∈ \overline{Z(D)} if γ ≠ ū and [D̄^{p}(γ_{1}, …, γ_{r}) : D̄^{p}] = p^{r} for any r distinct elements of Γ′.

This set is not void, since it contains the set Γ′ = {ū}. By Zorn’s lemma, there exists a maximal set Γ satisfying A). Then D̄ = D̄^{p}(Γ) . Indeed, since \overline{Z(D)}^{p}(ū) ⊂ D̄^{p}(Γ) , it suffice to show that any element from \overline{Z(D)} lies in D̄^{p}(Γ) . Suppose a ∈ \overline{Z(D)} , a ∉ D̄^{p}(Γ) . Then the set Γ′ = {a ∪ Γ} satisfy A), a contradiction with maximality of Γ .

Now, we can take a lift of Γ in the following way. We take u as a lift of ū, and we take lifts of all other elements in Z(D) . This lift defines an embedding u : D̄ ↪ D.

Let us show that any map _{m}^{(z,u)}δ_{j} (for some fixed z) is uniqely defined by the values _{l}^{(z,u)}δ_{k}(u) , k ≤ j. We have u(D̄) = u(\overline{Z(D)})(u) and any element a ∈ u(D̄) can be represented as a polynomial in finite number of elements from Γ with coefficients from u(D̄)^{p^{k}} for any k > 0 .

Note that for any j there exists k > 0 such that for any b ∈ \overline{Z(D)}^{p^{k}} _{l}^{(z,u)}δ_{q}(b) = 0

for all q ≤ j and all l. Indeed, assume _{1}^{(z,u)}δ_{q}(b) ≠ 0 for some q ≤ j, b ∈ \overline{Z(D)}^{p^{k}} and

_{l}^{(z,u)}δ_{s}(c) = 0 for all l, all c ∈ \overline{Z(D)}^{p^{k}} and all s < q. Then, since ^{(z)}α|\overline{Z(D)} = id and by

proposition 2, _{l}^{(z,u)}δ_{s}(b^{p}) = 0 for all b ∈ \overline{Z(D)}^{p^{k}} , all l and all s ≤ q.

Now, since u(D̄)^{p^{k}} = u(\overline{Z(D)})^{p^{k}}(u^{p^{k}}) , any element a ∈ u(D̄) can be represented as a polynomial in finite number of elements from Γ with coefficients from u(\overline{Z(D)})^{p^{k}} . Since all elements except u in Γ belong to the center Z(D) , the value of _{m}^{(z,u)}δ_{j}(a) is uniqely determined by the values _{m}^{(z,u)}δ_{j}(u^{l}) that are uniqely defined, by proposition 2, by the values _{l}^{(z,u)}δ_{k}(u) , k ≤ j. □

Remark In the case Z(D̄) perfect field there is only one embedding u, which is compatible with the embedding \overline{Z(D)} ↪ Z(D) . So, the assertion of lemma is easy in this case.

Lemma 10 (cf. [14], lemma 8)

In the situation of lemma 9 suppose _{m}^{(z,u)}δ_{1} = … = _{m}^{(z,u)}δ_{j−1} = 0 , _{m}^{(z,u)}δ_{j} ≠ 0 . Let n be the order of ^{(z)}α. Then

(i) for u′ = u + bz^{q}, b ∈ u(D̄) , n|q we have _{m}^{(z,u′)}δ_{l} = _{m}^{(z,u)}δ_{l}, l < q and

_{m}^{(z,u′)}δ_{q}(ū) = _{m}^{(z,u)}δ_{q}(ū) +^{(z)}α^{m}(b̄) − \frac{∂}{∂ū}(^{(z)}α^{m}(ū))b̄,

where the derivative is taken in the field D̄ = D̄^{p}(Γ) .

(ii) Suppose ^{(z)}α = id. Then for u′ = u + bz^{q}, b ∈ u(D̄) we have _{m}^{(z,u′)}δ_{l} = _{m}^{(z,u)}δ_{l}, l < q + j and

_{m}^{(z,u′)}δ_{q+j}(ū) = _{m}^{(z,u)}δ_{q+j}(ū) + _{m}^{(z,u)}δ_{j}(b̄) − \frac{∂}{∂ū}(_{m}^{(z,u)}δ_{j}(ū))b̄,

where the derivative is taken in the field D̄ = D̄^{p}(Γ) .

(iii) Suppose ^{(z)}α = id. Let ū′ ∈ D̄ be any primitive element of the extension D̄/\overline{Z(D)} satisfying the conditions of lemma 9, and let u′ ∈ D be any lift of ū′. Then we have _{m}^{(z,u′)}δ_{l} = _{m}^{(z,u)}δ_{l}, l < j and

_{m}^{(z,u′)}δ_{j}(ū′) = _{m}^{(z,u)}δ_{j}(ū)\frac{∂}{∂ū}(ū′),

where the derivative is taken in the field D̄ = D̄^{p}(Γ) .

Proof. First of all, let’s note that there exists k ∈ ℕ such that for any a ∈ \overline{Z(D)}^{p^{k}} holds u(a) − u′(a) = 0 mod M_{D}^{q+1} , where u′ is any another embedding, q ∈ ℕ is any given number.

Indeed, assume for any c ∈ \overline{Z(D)}^{p^{s}} holds u(c) − u′(c) = 0 mod M_{D}^{l}, i.e. u(c) = u′(c) + c_{l}z^{l} + …, where c_{l} ∈ u′(D̄) . Then u(c^{p}) = (u(c))^{p} = (u′(c))^{p} + pu′(c)^{p−1}c_{l}z^{l} + …, so u(c^{p}) − u′(c^{p}) = 0 mod M_{D}^{l+1} .

¿From this immediately follows that u(a) − u′(a) = 0 mod M_{D}^{q} for any a ∈ D̄ if u′ is defined by the element u′ = u + bz^{q}, because u(ū) − u′(ū) = bz^{q}. Moreover, if we represent a as some polynomial P (γ_{1}, …, γ_{r}, ū) with coefficients from \overline{Z(D)}^{p^{k}} , then it is clear that

[u(a) − u′(a)]z^{−q} = −\frac{∂}{∂ū}(P (γ_{1}, …, γ_{r}, ū))b̄ mod M_{D}

if n|q, since u(γ_{l}) = u′(γ_{l}) for any l and z^{q}uz^{−q} = u mod M_{D}. It is also clear that the derivative can be taken even in the field D̄^{p}(Γ) . So, we have (i)

z^{m}u′z^{−m} = z^{m}(u + bz^{q})z^{−m} = u(^{(z)}α^{m}(ū)) + u(_{m}^{(z,u)}δ_{j}(ū))z^{j} + … + (u(^{(z)}α^{m}(b̄))

+u(_{m}^{(z,u)}δ_{j}(b̄))z^{j} + …)z^{q} = u(^{(z)}α^{m}(ū)) + … + (u(^{(z,u)}δ_{q}(ū)) + u(^{(z)}α^{m}(b̄)))z^{q} + … =

u′(^{(z)}α^{m}(ū)) + … + (u′(_{m}^{(z,u)}δ_{q}(ū)) + u′(^{(z)}α^{m}(b̄)) − u′(\frac{∂}{∂ū}^{(z)}α^{m}(ū)b̄))z^{q} + …,

(ii) We have

z^{m}u′z^{−m} = z^{m}(u+bz^{q})z^{−m} = u(ū)+u(_{m}^{(z,u)}δ_{j}(ū))z^{j} +…+(u(b̄)+u(_{m}^{(z,u)}δ_{j}(b̄))z^{j} +…)z^{q} =

u(ū) + u(_{m}^{(z,u)}δ_{j}(ū))z^{j} + … + (u(_{m}^{(z,u)}δ_{q}(ū)) + u(b̄))z^{q} + u(_{m}^{(z,u)}δ_{q+1}(ū))z^{q+1} + …
+u(_{m}^{(z,u)}δ_{q+j−1}(ū))z^{q+j−1} + (u(_{m}^{(z,u)}δ_{q+j}(ū)) + u(_{m}^{(z,u)}δ_{j}(b̄)))z^{q+j} + … =

u′(ū) + u′(_{m}^{(z,u)}δ_{j}(ū))z^{j} + … + u′(_{m}^{(z,u)}δ_{q+j−1}(ū))z^{q+j−1} + (u′(_{m}^{(z,u)}δ_{q+j}(ū)) + u′(_{m}^{(z,u)}δ_{j}(b̄))−

u′(\frac{∂}{∂ū}(_{m}^{(z,u)}δ_{j}(ū))b̄))z^{q+j} + …

(iii) Assume u′ = u(ū′) + a_{1}z + …, where a_{i} ∈ u(D̄) . Since, by proposition 2, the map _{m}^{(z,u)}δ_{j} is a derivation, we have

z^{m}u′z^{−m} = [u(ū′) + u(_{m}^{(z,u)}δ_{j}(ū′))z^{j} + …] + [a_{1} + u(_{m}^{(z,u)}δ_{j}(a_{1})z^{j} + …]z + … =

u′ +u(_{m}^{(z,u)}δ_{j}(ū′))z^{j} +… = u′ +u(_{m}^{(z,u)}δ_{j}(ū)\frac{∂}{∂ū}(ū′))z^{j} +… = u′ +u′(_{m}^{(z,u)}δ_{j}(ū)\frac{∂}{∂ū}(ū′))z^{j} +…

□

4 The period-index problem

In this section we will prove the following theorem.

Theorem 2 The following conjecture: the exponent of A is equal to its index for any division algebra A over a C_{2} -field F has the positive answer for F = F_{1}((t)) , where F_{1} is a C_{1} -field.

Recall that a field F is called a C_{i} -field if any homogeneous form f(x_{1}, …, x_{n}) of degree d in n > d^{i} variables with coefficients in F has a non-trivial zero. Some basic properties of C_{i} -fields see, for example, in [10].

This conjecture was proposed by M. Artin and was solved for some another examples of the field F by many authors. As it is known for me, the positive answer for all division algebras of index indA = 2^{a}3^{b} was given in [10], for division algebras over the field F = k((X))((Y)) , where k is a perfect field of characteristic p ≠ 0 such that dim_{𝔽_{p}} k/℘(k) = 1 , was given by Tignol in the Appendix in [2] (we include this case though F may not be a C_{2} -field), for division algebras of index prime to the characterictic of F , where F is a function field of a surface, was given in [7]. I propose, the positive answer was also known for division algebras over F = F_{1}((t)) of characteristic 0. We will give the prove of the theorem above in any characteristic.

Proof. 1) Recall that any extension of a C_{1} -field is simple. Indeed, suppose E = F̄(u_{1}, …, u_{r}) . Consider the field K = F̄(u_{1}^{p}, …, u_{r}^{p}) . By Tsen’s theorem, K and E are C_{1} -fields. So, the form x_{1}^{p} + x_{2}^{p}u_{1} + … + x_{p}^{p}u_{1}^{p−1} + x_{p+1}^{p}u_{2} has a non-trivial zero in E. But x_{i}^{p} ∈ K and elements 1, u_{1}, …, u_{1}^{p−1}, u_{2} are linearly independent over K, a contradiction.

2) Assume the theorem is known in the prime exponent case. We deduce the theorem by ascending induction on e = expA. If e is not a prime number, then write e = lm. By assumption A^{⊗m} can be split by a field extension F ⊂ F′ of degree l. This implies that A_{F′} has exponent dividing m. Note that F′ is also a Laurent series field. By the induction hypothesis applied to the pair (F′, A_{F′}) , there exists a field extension F′ ⊂ L of degree dividing m splitting A_{F′} . Therefore A is split by the extension F ⊂ L of degree dividing lm and we conclude the theorem.

3) So, let expA = l be a prime number. By the basic properties of the exponent and the index (see, e.g. [10]) we have then indA = l^{k} for some natural k.

Suppose (l, p = charF) = 1 .

It is known that the conjecture is true for all division algebras of index indA = 2^{a}3^{b}, so we can assume l ≠ 2, 3 . We can assume F contains the group μ_{l} of l -roots of unity, because [F (μ_{l}) : F] < l and we can reduce the problem to the algebra A ⊗_{F} F (μ_{l}) . Then by the Merkuriev-Suslin theorem A is similar to the tensor product of symbol-algebras of index l.

To conclude the statement of the corollary it is sufficient to prove that every two symbol algebras A_{1}, A_{2} contain F -isomorphic maximal subfields.

Since every division algebra over a C_{1} -field is trivial and every field extension is simple, every symbol-algebra of index l over F is splittable. Since (l, p) = 1 , it is good splittable and its residue field is a cyclic Galois extension of F̄. So, if z_{i} is a parameter from proposition 3 for algebra A_{i}, then z_{i} acts on Ā_{i} as a Galois automorphism and z_{i}^{l} ∈ F . We have v(z_{i}^{l}) = 1 ( v is the valuation on F ).

Let us show that A_{1} contains a l -root of any element u in F with v(u) ≠ 0 . So, A_{1} will contain a subfield isomorphic to F (z_{2}) . Since for any element 1 + b, v(b) > 0 there exists a l -root (1 + b)^{1/l} ∈ F , it is sufficient to prove that A_{1} contains any l -root of elements ct, c ∈ u(F̄) , where u is some fixed embedding u : Ā_{1} ↪ A_{1} .

Assume z_{1}^{l} = c_{1}t, c_{1} ∈ u(F̄) . Note that for any element b ∈ u(Ā_{1}) we have (bz_{1})^{l} = u(N_{Ā_{1}/F̄}(b))z_{1}^{l}. But the norm map N_{Ā_{1}/F̄} is surjective, since F̄ is a C_{1} -field (see, e.g. [10], 3.4.2), so there exists b such that (bz)^{l} = ct.

4) Suppose now expA = p. Then indA = p^{k}.

By Albert’s theorem (in [1]) there exists a field F′ = F (u_{1}^{1/p}, …, u_{k}^{1/p}) which splits A. Using the same arguments as in 1) one can show that every such a field has maximum two generators, say F′ = F (u_{1}^{1/p}, u_{2}^{1/p}) . Therefore, indA ≤ p^{2} . If indA = p, there is nothing to prove, so we assume indA = p^{2} and F′ is a maximal subfield in A.

5) Suppose F_{1} is a perfect field.

By Albert’s theorem, A ≅ A_{1} ⊗_{F} A_{2} , where A_{1}, A_{2} are cyclic algebras of degree p, A_{1} = (L_{1}/F, σ_{1}, u_{1}) , A_{2} = (L_{2}/F, σ_{2}, u_{2}) . Since F_{1} is perfect, Ā_{1}/F̄, Ā_{2}/F̄ are Galois extensions. So, A_{1}, A_{2} are good splittable. Let us show that A_{1}, A_{2} have common splitting field of degree p over F . This leads to a contradiction.

By proposition 3 there exist parameters z_{1} ∈ A_{1} , z_{2} ∈ A_{2} such that they act on Ā_{1} , Ā_{2} as Galois automorphisms. Note that then z_{1}^{p}, z_{2}^{p} ∈ F . Let us show that F (z_{1}) splits A_{2} .

Consider the centralizer D = C_{A}(F (z_{1})) . Consider the element t_{1} = z_{2}z_{1}^{−1} . We have t_{1}^{p} ∈ F , w(t_{1}) = 0 , where w denote the unique extension of the valuation v on F . Since D̄/\overline{Z(D)} is a Galois extension, there exists an element b_{1} ∈ F such that w(t_{1} − b_{1}) > 0 . Since (t_{1} − b_{1})^{p} ∈ F , there exists natural k_{1} such that w((t_{1} − b_{1})z_{1}^{−k_{1}}) = 0 . Denote t_{2} = (t_{1} − b_{1})z_{1}^{−k_{1}} . We have again t_{2}^{p} ∈ F . Repeating this arguments and using the completeness of D ⊂ A we get z_{2} = t_{1}z_{1} = (t_{2}z_{1}^{k_{1}} + b_{1})z_{1} = … = b_{1}z_{1} + b_{2}z_{1}^{k_{1}+1} + …, so, z_{2} ∈ F (z_{1}) = Z(D) .

6) Suppose F_{1} is not perfect.

Since F′ is generated by two elements over F , it contains all p -roots of F . Then, every two elements u, z ∈ F such that z^{1/p} ∉ F (u^{1/p}) , where z^{1/p}, u^{1/p} ∈ F′, also generate F′ over F . This follows from the same arguments as in 1), 4).

Now take u ∈ F_{1}\F_{1}^{p}, z = u + t. It’s clear that p -roots of these elements generate F′ over F . Moreover, the fields F (u^{1/p}), F (z^{1/p}) are ”unramified” over F , i.e. [\overline{F (u^{1/p})} : F̄] = p = [F (u^{1/p}) : F] , [\overline{F (z^{1/p})} : F̄] = p. Denote u_{1} = u^{1/p}, u_{2} = z^{1/p} in F′. Then by Albert’s theorem, A ≅ A_{1} ⊗_{F} A_{2} , where A_{1}, A_{2} are cyclic algebras of degree p, A_{1} = (L_{1}/F, σ_{1}, u) , A_{2} = (L_{2}/F, σ_{2}, z) .

Concider the centralizer D = C_{A}(F (u_{1})) . Suppose D̄/\overline{Z(D)} is a separable extension. Then there exist a lift u : D̄ ↪ D of arbitrary embedding u′ : \overline{F (u_{1})} ↪ F (u_{1}) . Consider the embedding u′ = u_{1} defined in lemma 9. Since F (u_{1})/F is a purely inseparable extension, u′ is a good embedding, so u is a good embedding of D̄ = Ā in D ⊂ A. So, we get A is a good splittable algebra, and u(Ā) contain a purely inseparable over F element. But this is a contradiction with lemma 6. So, Ā/F̄ can not contain a separable subextension, because in this case D̄/\overline{Z(D)} must be a separable extension.

Now we can use, for shorteness, lemmas A.4., A.6. of Tignol in Appendix to the paper [2]. These lemmas show that a tensor product A_{1} ⊗ A_{2} of any two symbols A_{1}, A_{2} is similar either to a single symbol in Br(F) (in which case we are done) or to a product of two symbols of level zero. Recall that, by Saltman’s results in [11], every division algebra of level zero is tame, which means in our case that the residue division algebra is a separable extension over F̄. A notion of level was already discussed above in remark to lemma 7.

So, assume A ∼ D_{1} ⊗ D_{2} , where D_{1}, D_{2} are tame division algebras of degree p over F . We can assume A and D_{1} ⊗ D_{2} are division algebras, so A ≅ D_{1} ⊗ D_{2} . Since D_{1}, D_{2} are tame, we conclude Ā must contain a separable element, a contradiction.

The theorem is proved. □

5 Good splittable algebras

In this section we prove a decomposition theorem for good splittable division algebras. This theorem shows how the studying of good splittable division algebras can be reduced to the studying of division algebras with simple described structure. So, good splittable algebras are the most easy and good algebras to study.

Lemma 11 Let D be a good splittable division algebra, F = Z(D) , and let Z(D̄) = F̄(s) be a purely inseparable over F̄ field of degree p = charD > 0 . Let u : D̄ ↪ D be a good embedding.

Then there exists a parameter z such that _{−i}^{(z,u)}δ_{j} = 0 for j > i, where i = i(z, u) is a local height, and u(^{(z,u)}δ_{i}(s)) = x, where x ∈ Z(D) . Moreover, (i, p) = 1 .

Proof. Since Z(D̄)/F̄ is a purely inseparable extension, ^{(z)}α|_{Z(D̄)} = id for any parameter z. By Skolem-Noether theorem there exists a parameter z in D such that (z)α = id. Suppose ^{(z,u)}δ_{i}(s) = 0 , where i = i(z, u) . Then ^{(z,u)}δ_{i}|_{Z(D̄)} = 0 , since u is a good embedding and Z(D̄)/F̄ is a simple extension. So, ^{(z,u)}δ_{i} is an inner derivation by Scolem-Noether theorem, and by lemma 3, (i) there exists a parameter z′ such that (z′,u)δ_{i} = 0 , ^{(z′)}α = id.

So, we can assume ^{(z,u)}δ_{i}(s) ≠ 0 for some parameter z. Since s^{p} ∈ Z(D) , by lemma 6 we have (i, p) = 1 . Since ^{(z,u)}δ_{i} is a derivation, ^{(z,u)}δ_{i}(s) ∈ Z(D̄) (see the arguments in lemma 5, (ii)). Since (i, p) = 1 , there exists k such that p|(1 − ki) . So, by lemma 3, (iii), for the parameter z′ = (^{(z,u)}δ_{i}(s))^{k} we have ^{(z′)}α = id, ^{(z′,u)}δ_{i}(s) ∈ F̄, i.e. u(^{(z′,u)}δ_{i}(s)) ∈ Z(D) . Since s^{p} ∈ Z(D) , by lemma 8 we must have d(u, s) = ∞. In the proof of lemma 8, (i) was shown that d(u, s) = d′(u, z, s) for some parameter z, and the construction of this element uses lemma 3, (ii), so it preserves the initial values of ^{(z′)}α, (z′,u)δ_{i}. So, _{−i}^{(z,u)}δ_{j} = 0 for j > i and the lemma is proved. □

Proposition 4 Let D be a splittable division algebra. Then we have D ≅ D_{1} ⊗_{F} D_{2} , where D_{1}, D_{2} are splittable division algebras such that D_{1} is an inertially split algebra.

If D is a good splittable division algebra, then Z(D̄_{2})/F̄ is a purely inseparable extension and D_{2} is a good splittable algebra ( D_{1} or D_{2} may be trivial).

So, D ∼ A ⊗_{F} B ⊗_{F} D_{2} , where A is a cyclic division algebra and B is an unramified division algebra.

Proof. If charD = 0 , the proposition is obvious, so we assume charD > 0 .

By [9], p.261, D ≅ D_{1} ⊗_{F} … ⊗_{F} D_{k}, where [D : F] = p_{1}^{r_{1}} … p_{k}^{r_{k}} and [D_{i} : F] = p_{i}^{r_{i}} . Let p_{2} = p. Since D_{i} are defectless over F , D_{1}, D_{3}, … D_{k} are inertially split. Therefore, by theorem 1 the algebra B = D_{1} ⊗ D_{3} ⊗ … ⊗ D_{k} is good splittable.

Assume first that D is good splittable. By proposition 1.7. in [6], if s ∈ Z(D̄) is an element such that α(s) = s, then this element is a purely inseparable element over F̄. So, if D is a good splittable division algebra, then by lemma 6 D_{2} is either inertially split or Z(D̄_{2})/F̄ is a purely inseparable extension. For, otherwise there exists an element s ∈ Z(D̄_{2}) ⊂ Z(D̄) as above and by proposition 3 p|i(u, s) for any embedding u. If u is a good embedding, then s^{p^{k}} ∈ Z(D) for some k, a contradiction.

So, we assume below Z(D̄_{2})/F̄ is a purely inseparable extension. Now, we have (see, e.g. th.1 in [8]) D̄ ≅ D̄_{2} ⊗F̄ B̄ and so u(D̄) ≅ u(D̄_{2}) ⊗_{u(F̄)} u(B̄) , where u is a good embedding. So, E = u(Z(D̄_{2})) is a purely inseparable field over u(F̄) ⊂ Z(D) .

Consider the field E′ = u(K)⊗_{u(F̄)}F , where K is a maximal separable subfield in B̄. This is an inertial lift of K in D. Consider the centralizer C = C_{D}(E′) ≅ D_{2} ⊗_{F} E′. Let M be a maximal subfield in D̄_{2} . Note that u(D̄_{2}) ⊂ C, so L ⊂ C, where L = u(M)F is the composit of u(M) and F , and E ⊂ L. Note that [L : F] = indD_{2} = indC. The field L splits C by dimension arguments. So, it must split D_{2} , since ([E′ : F], p) = 1 , and D_{2} is a p -algebra. Therefore, L is isomorphic to a maximal subfield in D_{2} , so D_{2} contain a copy of purely inseparable ”unramified” subfield, whose residue field is isomorphic to Z(D̄_{2}) . Therefore, D_{2} is a god splittable algebra. For, the centralizer of this field is an unramified division algebra, so by theorem 1 is splittable. So, D_{2} is good splittable if the purely inseparable field is good splittable. But it is good splittable since it contains a subfield isomorphic to u(Z(D̄_{2})) by the construction. (Another way to see it is to use arguments from lemma 9 to show that there exists an appropriate p -basis).

Let D be a splittable algebra. Then the same arguments as in the previous paragraph show that L is isomorphic to a maximal subfield in D_{2} (it is not important that Z(D̄_{2})/F̄ may be not a purely inseparable extension). Now, the composit EF ⊂ L, EF ≠ L, since every element from E commute with u(D̄_{2}) , where u is some fixed embedding. So we must have \overline{C_{D_{2}}(EF) = D̄_{2}} and C_{D_{2}}(EF) is an unramified division algebra. Therefore, D_{2} is splittable division algebra.

Decomposition theorems [6], Thm. 5.6-5.15 complete the proof. □

This proposition shows that the study of splittable division algebras can be reduced to the study of splittable p -algebras. So, below in this section and in the next section we will deal with p -algebras only.

Proposition 5 Let D be a good splittable division algebra such that Z(D̄)/\overline{Z(D)} is a purely inseparable extension. Then D ≅ D_{1}⊗_{Z(D)}D_{2} , where D_{1} is an unramified division algebra and D_{2} is a good splittable division algebra such that D̄_{2} is a field, D̄_{2}/\overline{Z(D)} is a purely inseparable extension, [D̄_{2} : \overline{Z(D)}] = [Γ_{D_{2}} : Γ_{Z(D)}] .

Proof. The proof is by induction on the degree [Z(D̄) : \overline{Z(D)}] .

Assume [Z(D̄) : \overline{Z(D)}] = p. Let ^{(z,u)}δ_{i} be the map from lemma 11. Then ^{(z,u)}δ_{i}^{p} is a derivation trivial on the centre Z(D) , hence by Scolem-Noether theorem it is an inner derivation.

We claim that z^{p} ∈ Z(D) . We have

z^{−i}az^{i} = a +_{−i}δ_{i}(a)z^{i}, a ∈ u(D̄)

Therefore,

z^{−pi}az^{pi} = a +_{−i}δ_{i}^{p}(a)z^{pi}, a ∈ u(D̄)

and

z^{pi}az^{−pi} = a + δ′_{1}(a)z^{pi} + δ′_{1}^{2}(a)z^{2pi} + …,

where δ′_{1} = (−1)_{−i}δ_{i}^{p} = i^{p}δ_{i}^{p}. So,

z^{p}az^{−p} = a + \frac{1}{i}δ′_{1}(a)z^{pi} + c_{2} \frac{1}{i^{2}} δ′_{1}^{2}(a)z^{2pi} + …,

where c_{k} are given by (4) in lemma 7. So, z^{p} ∈ Z(D) iff δ_{i}^{p} = 0 . Suppose δ_{i}^{p} ≠ 0 . Consider an element Y ∈ Z(D) , w(Y) > 0 . Let

Y = a_{1}z^{p} + …, a_{1} ∈ u(D̄).

First note that

Y = a_{1}z^{p} + a_{2}z^{2p} + a_{3}z^{3p} + …, a_{i} ∈ u(D̄)

Indeed, Y must satisfy [Y, s] = 0 , where s is a generator of u(Z(D̄)) over u(F̄) . Since s ∈ u(Z(D̄)) and w([z^{k}, s]) = k + i if (k, p) = 1 and w([z^{k}, s]) = ∞ otherwise, we then have [z^{i_{k}}, s] = 0 for every k, where

Y = ∑_{k=1}^{∞} a_{k}z^{i_{k}}

Therefore, p|i_{k}.

Then, Y must satisfy Y a = aY for any a ∈ u(D̄) . Therefore, a_{1}, … a_{i} ∈ u(Z(D̄)) and we must have

aa_{i+1} − a_{i+1}a = a_{1}δ′_{1}(a)/i

and

aa_{2i+1} − a_{2i+1}a = a_{i}δ′_{1}(a) + a_{1}c_{2}δ′_{1}^{2}(a).

Since Δ(a) = aa_{2i+1} − a_{2i+1}a is an inner derivation, we get δ′_{1}^{2} = δ, where δ is a derivation, which is a contradiction if δ ≠ 0 and charD ≠ 2 . In the last case we can use the same arguments with a_{3i+1} . Therefore, δ′_{1}^{2} = δ = 0 and δ′_{1} = 0 , and z^{p} ∈ Z(D) .

Consider the algebra W = u(Z(D̄))((z)) . Since z^{p} ∈ Z(D) and u(F̄) ⊂ Z(D) , we have Z(W) = u(F̄)((z^{p})) = F . So, D ≅ W ⊗_{F} C_{D}(W) by Double Centralizer theorem. It is clear that C_{D}(W) is an unramified division algebra.

Now suppose the proposition is proved for [Z(D̄) : \overline{Z(D)}] = p^{k−1} . By Albert’s theorem (th.13 in [1]) D_{2} then is a cyclic algebra as a product of cyclic subalgebras D_{i}, where D̄_{i}/F̄ is a simple purely inseparable extension and D_{i} is a good splittable algebra.

Assume [Z(D̄) : \overline{Z(D)}] = p^{k}. For a good embedding there exists a lift K̃ of a subfield \overline{Z(D)} ⊂ K ⊂ Z(D̄) such that the extension K/\overline{Z(D)} has degree p, i.e. ¯ K̃ = K, ΓK̃ = Γ_{Z(D_{2})} , u(K) ⊂ K̃, K̃/Z(D) is a purely inseparable extension of degree p. By the induction hypothesis the centralizer C_{D}(K̃) ≅ A_{1} ⊗K̃ A_{2} , where A_{2} is a cyclic division algebra and Ā_{2} is a field. Note that Ā_{2} = Z(D̄) .

By theorem 6 in [1] we can assume A_{2} = (L/K̃, σ, a) , where a generate K̃ over Z(D) . So, A_{2} contains a maximal purely inseparable Kummer subfield E = K̃(y) with y^{p^{k−1}} = a, so E = Z(D)(y) . By theorem 3 in [1] L = L_{0} × K̃, where L_{0} is cyclic of degree p^{k−1} over Z(D) and yx_{0} = σ(x_{0})y, where x_{0} ∈ L_{0} .

Consider the centralizer B = C_{D}(L_{0}) . We claim B ≅ B_{1} ⊗_{L_{0}} B_{2} , where B_{2} is a cyclic division algebra of degree p and B_{2} contains K̃.

Note that B contains Z(D)(a) = K̃ and A_{1} . If K̃L_{0} = L is ”unramified” over L_{0} , then we apply the arguments for the first step of our induction to the algebra B. By construction, B_{2} then will contain L, so K̃. Suppose L is totally ramified over L_{0} and let z be a parameter of L, i.e. an element with the least possible positive mean of valuation on L. Since L is purely inseparable over L_{0} , z^{p} is a parameter of L_{0} .

We have W := C_{B}(L) = C_{D}(L) ≅ A_{1}⊗K̃ L is an unramified division algebra. Consider an embedding u′ : L̄ = L̄_{0} ↪ L_{0} . As it was shown in the proof of theorem 1 there is a lift ũ′ of u′, ũ′ : W̄ ↪ W . Now consider the subalgebra W′ = ũ′(W̄)((z^{p})) . We have Z(W′) = u′(L̄)((z^{p})) = L_{0} , so W′ is an unramified subalgebra in B. By Double Centralizer theorem, B ≅ W′ ⊗_{L_{0}} C_{B}(W′) , where C_{B}(W′) is a division algebra of degree p and contains L_{0}(z) = L, so it contains K̃ and it is cyclic by Albert’s theorem (th.12 in [1]).

Now we can word by word repeat the arguments in the proof of theorem 12 in [1] to show that there exists a cyclic Galois extension L′ of L_{0} which is cyclic Galois over Z(D) , and y acts as a Galois automorphism on L′/Z(D) which generates Gal(L′/Z(D)) . So, there is the cyclic subalgebra D_{2} = (L′/Z(D), ad(y), y^{p^{k}}) in D. Note that A_{2} ⊂ D_{2} , and A_{2} is known to be a good splittable algebra with [Ā_{2} : \overline{Z(A_{2})}] = [Γ_{A_{2}} : Γ_{Z(A_{2})}] . Since Ā_{2} = D̄_{2} and Z(A_{2}) = K̃ is a purely inseparable extension of Z(D) , D_{2} is a good splittable algebra such that D̄_{2} a field and [D̄_{2} : \overline{Z(D)}] = [Γ_{D_{2}} : Γ_{Z(D)}] . By Double Centralizer theorem D ≅ D_{1} ⊗_{Z(D)} D_{2} , where D_{1} = C_{D}(D_{2}) must be an unramified division algebra, which completes the proof. □

Combining all results in this section, we get the following theorem.

Theorem 3 Let D be a finite dimensional good splittable central division algebra over a field F = k((t)) .

If char(F) = p > 0 , then D ≅ D_{1} ⊗_{F} D_{2} ⊗_{F} A_{1} ⊗_{F} … ⊗_{F} A_{m}, where A_{i} are cyclic division algebras such that [Ā_{i} : \overline{Z(D)}] = [Γ_{A_{i}} : Γ_{Z(D)}] and Ā_{i}/\overline{Z(D)} are simple purely inseparable field extensions, D_{1} is an inertially split division algebra, (ind(D_{1}), p) = 1 , D_{2} is an unramified division algebra ( D_{1}, D_{2}, A_{i} may be trivial).

If charF = 0 , then D is an inertially split division algebra.

6 Splittability and good splittability

In this section we collect some assorted results about a relation between splittable and good splittable division algebras and about splittable division algebras. We consider here only division algebras with the following property: Z(D̄)/\overline{Z(D)} is a simple extension.

Proposition 6 Let D be a central division algebra over F of charD = p > 0 such that Z(D̄) = D̄ and [Z(D̄) : F̄] = p.

Then D is a splittable algebra and the local height i = i(u, z) (in the situatuion when it is defined, i.e. when α = id) does not depend on u and z. It is a good splittable algebra if (i, p) = 1 . If p|i, then there exists a parameter z such that z^{p} ∈ Z(D) and any ”unramified” maximal subfield is cyclic Galois.

So, in both cases D is a cyclic division algebra of degree p.

Proof. Since D̄/F̄ is a simple extension, we have [D̄ : F̄] = [Γ_{D} : Γ_{F}] . Indeed, consider the fields E = F (s) and E′ = F (z) , where s is any element such that s̄ is a primitive element of the extension D̄/F̄ and z is any parameter of D. Then [D̄ : F̄] ≤ [E : F] ≤ [D : F]^{1/2} = ([D̄ : F̄][Γ_{D} : Γ_{F}])^{1/2} , so [D̄ : F̄] ≤ [Γ_{D} : Γ_{F}] . From another hand side, [Γ_{D} : Γ_{F}] ≤ [E′ : F] ≤ ([D̄ : F̄][Γ_{D} : Γ_{F}])^{1/2} , so [D̄ : F̄] = [Γ_{D} : Γ_{F}] . So, D is splittable division algebra of degree p.

If Z(D̄)/F̄ is a separable extension, then D is a good splittable algebra by theorem 1. So, we assume it is a purely inseparable extension, Z(D̄) = F̄(ū) . For any lift u of the element ū let u be an embedding constructed in lemma 9, i.e. ^{(z,u)}δ_{j} is defined by the values ^{(z,u)}δ_{j}(u^{k}) for any j. By corollary 2 the local height i(u, z) does not depend on z, and by lemma 10 i(u, z) does not depend on u. For arbitrary embedding u′, since (z,u′)δ_{i(u′,z)} is a derivation and D̄/F̄ is a simple extension, ^{(z,u′)}δ_{i(u′,z)} is completely defined by a value at ū. Therefore, i(u′, z) = w(zu′(ū)z^{−1} − u′(ū)) and i(u′, z) is completely defined by the lift u′(ū) . But arbitrary lift of ū defines an embedding, on which we have proved i does not depend. So, i(u, z) does not depend on z and u.

Now assume p|i.

Using lemma 3, we can assume without loss of generality that ^{(z,u)}δ_{j} = 0 if j is not divisible by p.

Indeed, if ^{(z,u)}δ_{j} ≠ 0 , then we apply lemma 3, (ii) to show that there exists a parameter z_{j} such that ^{(z_{j},u)}δ_{j}(u) = 0 and ^{(z_{j},u)}δ_{k} = ^{(z,u)}δ_{k} for k < j, ^{(z_{j})}α = id. Since ^{(z_{j},u)}δ_{j} is a derivation by proposition 2 and by induction (similar arguments was already used in the proof of proposition 3), and since it is defined by the values on u^{k}, so by the values on u, we have ^{(z,u)}δ_{j} = 0 . Since for j_{1} > j_{2} we have w(z_{j_{1}} − z_{j_{2}}) > j_{1} − i, the sequence {z_{j}} convereges to a parameter z′, which satisfies our condition.

So, there exists the subalgebra A = u(D̄)((z^{p})) . Let’s show that Z(D) ⊂ A. Note that every element a ∈ D can be written as a = a_{0} + a_{1}z + … + a_{p−1}z^{p−1} , where a_{i} ∈ A. Note that z^{k}Az^{−k} ⊂ A for every k. So, if a ∈ Z(D) , then za_{j}z^{−1} = a_{j} and ua_{j}z^{j}u^{−1} = a_{j}z^{j} for every j. For j > 0 we have a_{j}z^{j} = ∑_{k} a_{jk}z^{kp+j}, so by corollary 2 ua_{j}z^{j}u^{−1} ≠ a_{j}z^{j}. Therefore, a = a_{0} ∈ A.

Since A ≠ D, A must be commutative, so z^{p} ∈ Z(D) . Moreover, A/Z(D) is cyclic Galois. Since the arguments work for arbitrary lift u of the element ū, arbitrary ”unramified” maximal subfield in D must be Galois over F .

Now let (i, p) = 1 .

Using lemma 3, (iii) we can find a parameter z and a primitive element s ∈ D̄ such that ^{(z,u)}δ_{i}(s) = sc, where c ∈ F̄. Indeed, since (i, p) = 1 , there exists k such that 1 − ki is divisible by p. So, by lemma 3, (iii) for a parameter z′ = u(^{(z,u)}δ_{i}(ū)^{k})z we have (z′,u)δ_{i}(ū) ∈ F̄, so by lemma 10, (iii) ^{(z′,u)}δ_{i}(s) = 1 , where s = ū^{(z′,u)}δ_{i}(ū)^{−1} . Now, there exists k_{1} such that −ik_{1} − 1 is divisible by p, so for z′′ = s^{k_{1}}z′ we have ^{(z′′,u)}δ_{i}(s) = sc, where c = s^{−ik_{1}−1} ∈ F̄. It is easy to see that, since s = ūa, where a ∈ F̄, the map ^{(z,u)}δ_{j} is uniquely defined also by ^{(z,u)}δ_{j}(s^{k}) , so by _{m}^{(z,u)}δ_{l}(s) for l ≤ j. So, we assume without loss of generality that s = ū, z = z′′.

Using lemma 10, (ii) we can find a converge sequence {u_{j}}, u_{j} ∈ D, j ≥ i such that u_{j+1} = u_{j} + b_{j}z^{j+1−i}, u_{i} = u, b_{j} ∈ u_{j}(D̄) (here u_{j} is an embedding defined by u_{j}, see lemma 9) and _{m}^{(z,u_{j})}δ_{k}(ū)ū^{−1} ∈ F̄ for all k ≤ j and all m.

Indeed, suppose it is true for j ≥ i. Let _{m}^{(z,u_{j})}δ_{j+1}(ū) = a_{0} + … a_{p−1}ū^{p−1} , a_{k} ∈ F̄. Since _{m}^{(z,u_{j})}δ_{i} = _{m}^{(z,u)}δ_{i} = m^{(z,u)}δ_{i}, we have

_{m}^{(z,u_{j})}δ_{i}(a_{k}ū^{k}) − \frac{∂}{∂ū}(_{m}^{(z,u_{j})}δ_{i}(ū))a_{k}ū^{k} = (k − 1)mca_{k}ū^{k}.

So, u_{j+1} = u_{j} − u_{j}(∑_{k,k≠1}(k − 1)^{−1}m^{−1}c^{−1}a_{k}ū^{k})z^{j+1−i} will satisfy our condition.

We will denote by u now a limit of the sequence {u_{j}}. Using induction and proposition 2 one can easily show that _{m}^{(z,u)}δ_{j}(ū^{k})ū^{−k} ∈ F̄ for any integer k. So, there is the subalgebra A = u(F̄)((z)) in D. Using similar arguments as in the case p|i, one can show that A contains Z(D) . Since A ≠ D, it must be commutative, so u^{p} ∈ Z(D) . Then u is a good embedding, which completes the proof. □

Let D be a splittable division algebra and let Z(D̄)/\overline{Z(D)} be a purely inseparable extension. As it was shown in the proof of lemma 11, then there exists a parameter z in D such that ^{(z,u)}δ_{i}|_{Z(D̄)} ≠ 0 , where i = i(u, z) is a local height. Though D may be not a good splittable algebra, the arguments from there are valid for every splittable algebra. We will call such a parameter an appropriate parameter, and the number i(u) = max_{z} i(u, z) = i(u, z) for an appropriate parameter a semilocal height. Let’s prove the following simple lemma.

Lemma 12 Let D be a splittable central division p -algebra over F , where p = charD > 0 , and let Z(D̄) = F̄(s) be a simple extension over F̄. Then

i) there exists an embedding u such that _{l}^{(z,u)}δ_{j}|_{Z(D̄)} is defined by the values _{l}^{(z,u)}δ_{j}(s^{k}) for any j, l, z (as in lemma 9);

ii) [Z(D̄) : F̄] = [Γ_{D} : Γ_{F}] ;

iii) if α|_{Z(D̄)} ≠ id or i(u) is divisible by p, then there exists a subalgebra A = u(D̄)((z)) for some appropriate parameter z such that Z(D) ⊂ Z(A) . Moreover, Z(A) is a cyclic Galois extension over Z(D) .

Proof. i) For arbitrary embedding u consider the field E = u(Z(D̄))F ⊂ D and the centralizer W = C_{D}(E) . We have W̄ = D̄ and so Z(W̄) = Ē. Therefore, W must be an unramified division algebra, and by theorem 1 there exists a lift on W̄ of arbitrary embedding Ē ↪ E. Now we can take an embedding defined by the element s as in lemma 9. It’s lift will be desired embedding. We will denote this embedding also by s.

ii) By proposition 1.7. in [6] the basic homomorphism θ_{D} (see introduction) is surjective. So, it is sufficient to prove the assertion only for the centralizer C_{D}(K) , where K is a lift of a Galois part of the extension Z(D̄)/F̄. So, we will assume below Z(D̄)/F̄ is a purely inseparable extension.

Consider a maximal separable subfield M in D̄, and let M′ be a separable part of the extension M/F̄. By [6], th.2.8, th.2.9. there exists an inertial lift of M′ in D, say M̃ . Consider the centralizer B = C_{D}(M̃) . Then B̄ is a field. Our assertion will be proved if we show it for B, since [M̃ : F] = ind(D̄) and [D : F] = ind(D̄)^{2}[Z(D̄) : F̄][Γ_{D} : Γ_{F}] .

Since B̄/\overline{Z(B)} is a simple extension, we can repeat the arguments from the beginning of proposition 6.

iii) If α|_{Z(D̄)} ≠ id, consider the parameter z from proposition 3. Then, clearly, A = u(D̄)((z)) will be a subalgebra with the center K, which is an inertial lift of a Galois part of the extension Z(D̄)/F̄.

Assume α|_{Z(D̄)} = id and i(u) is divisible by p. Let z be an appropriate parameter. Using lemma 3, we can prove that ^{(z,u)}δ_{j} = 0 if j is not divisible by p.

Indeed, let ^{(z,u)}δ_{j} ≠ 0 be the first map with this property for (j, p) = 1 . If (z,u)δ_{j}|_{Z(D̄)} = 0 , then we apply lemma 3, (i) to show that there exists a parameter z_{j} such that ^{(z_{j},u)}δ_{j} = 0 and ^{(z_{j},u)}δ_{k} = ^{(z,u)}δ_{k} for k < j, ^{(z_{j})}α = id, since ^{(z,u)}δ_{j} is a derivation by proposition 2 and by induction (similar arguments was already used in the proof of proposition 3) and so it is an inner derivation by Scolem-Noether theorem.

If ^{(z,u)}δ_{j}|_{Z(D̄)} ≠ 0 , then we apply lemma 3, (ii) to show that there exists a parameter z_{j} such that ^{(z_{j},u)}δ_{j}(s) = 0 and ^{(z_{j},u)}δ_{k} = ^{(z,u)}δ_{k} for k < j, ^{(z_{j})}α = id. Since ^{(z_{j},u)}δ_{j} is a derivation and since its restriction on Z(D̄) is defined by the values on s^{k}, so by the values on s, we have ^{(z,u)}δ_{j}|_{Z(D̄)} = 0 , and we reduce the problem to the previous case. Since for j_{1} > j_{2} we have w(z_{j_{1}} − z_{j_{2}}) > j_{1} − i, the sequence {z_{j}} convereges to a parameter z′, which satisfies our condition.

Therefore, there exists a subalgebra A = u(D̄)((z′)) in D. Using the same arguments as in proposition 6 one can show that Z(D) ⊂ Z(A) Since z′ preserves A, it preserves the centre Z(A) ¿From the other hand side, it acts nontrivially on it. So, Z(A) is a cyclic Galois extension of degree p, and ad(z′) generates its Galois group. □

This lemma shows that the study of splittable p -algebras over F can be reduced to the study of splittable p -algebras with a purely inseparable extension Z(D̄)/F̄ and (i(u), p) = 1 .

Definition 6 Let D be a splittable division p -algebra with a purely inseparable extension Z(D̄)/F̄. For any element a ∈ D̄ define the number

d_{D}(a) = max_{u,z} w(z^{−i(u,a)}u(a)z^{i(u,a)} − u(a) − u(_{−i(u,a)}^{(z,u)}δ_{i(u,a)}(a))z^{i(u,a)}) ∈ ℕ ∪ ∞,

where parameters z are taken from the set of appropriate parameters and i(u, a) was defined in corollary 2.

It seems that the number d_{D}(a) will play the role of a higher order level in a splittable division algebra. We will see that it codes a part of information about a division algebra.

Lemma 13 Let D be a splittable division p -algebra, p > 2 , with a purely inseparable simple extension Z(D̄)/F̄, let u be some fixed embedding u : D̄ ↪ D.

Suppose Z(D̄) = F̄(a) and (i(u, a), p) = 1 . Suppose d(u, a) ≤ 2i(u, a) .

Let z be a parameter such that ^{(z,u)}δ_{i(u,a)}(_{−i(u,a)}^{(z,u)}δ_{i(u,a)}(a)) = 0 , ^{(z)}α = id and _{−i(u,a)}^{(z,u)}δ_{q}|_{𝔽_{p}(a)} = 0 for i(u, a) < q < d(u, a) . Put j(k) := i(u, a^{p^{k}}) .

Suppose for every k ≥ 1 a parameter z_{k} such that _{−j(k)}^{(z_{k},u)}δ_{r}|_{𝔽_{p}(a^{p^{k}})} = 0 for j(k) < r < d(u, a^{p^{k}}) satisfy a condition ^{(z_{k},u)}δ_{i(u,a)} = ^{(z,u)}δ_{i(u,a)} , ^{(z_{k})}α = ^{(z)}α.

Suppose for every k ≥ 1 we have d(u, a^{p^{k}}) − j(k) = d(u, a) − j(0) .

Then the maps _{w+(p−1−r)j(k)}^{(z,u)}δ_{ζ} , rj(k) < ζ ≤ (r−1)j(k)+d(u, a^{p^{k}}) , r ∈ {1, …, p−1}, k ≥ 0 satisfy the following properties:

_{w+(p−1−r)j(k)}^{(z,u)}δ_{ζ}|_{𝔽_{p}(a^{p^{k}})} = c_{w+(p−1−r)j(k),ζ,1}δ + … + c_{w+(p−1−r)j(k),ζ,r}δ^{r},

where the derivation δ was defined in lemma 5, c_{w+(p−1−r)j(k),ζ,r} ∈ Z(D̄) , c_{w+(p−1−r)j(k),ζ,r} ≠ 0 only if ζ = (r − 1)j(k) + d(u, a^{p^{k}}) .

Moreover, c_{w+(p−1−r)j(k),(r−1)j(k)+d(u,a^{p^{k}}),r} ≠ 0 if w = i(u, a) mod p;

c_{w+(p−1−r)j(k),(r−1)j(k)+d(u,a^{p^{k}}),r} = r!c_{w+(p−r)j(k),(r−2)j(k)+d(u,a^{p^{k}}),r−1}_{w+(p−1−r)j(k)}δ_{j(k)}(a^{p^{k}}),

and _{w+(p−2)j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) = _{−j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) .

Proof. The proof is similar to the proof of lemma 5, (i). It is by induction on r simultaneously for all k ≥ 0 .

For r = 1 , using lemma 2 and induction, one can easily show that ^{(z_{k},u)}δ_{q}(a^{p^{k}}) = −(j(k))^{−1}_{−j(k)}^{(z_{k},u)}δ_{q}(a^{p^{k}}) for j(k) ≤ q < d(u, a^{p^{k}}) (we assume here z_{0} = z). By lemma 8, (i) we have d(u, a) − i(u, a) = i(u, a) mod p. So, by lemma 8, (ii) and by induction we have j(k) = j(0) mod p.

So, _{w+(p−2)j(k)}^{(z_{k},u)}δ_{q}|_{𝔽_{p}(a^{p^{k}})} = 0 if j(k) < q < d(u, a^{p^{k}}) and _{w+(p−2)j(k)}^{(z_{k},u)}δ_{q}|_{𝔽_{p}(a^{p^{k}})} ≠ 0 only if q = d(u, a^{p^{k}}) .

Since _{−j(k)}^{(z_{k},u)}δ_{j(k)}|_{𝔽_{p}(a^{p^{k}})} is a derivation and since, by proposition 2, (i), the map _{−j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}|_{𝔽_{p}(a^{p^{k}})} must be a derivation, we have _{w+(p−2)j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) ∈ Z(D̄) . For, as it was shown in the proof of lemma 5, (ii) for any derivation δ we have δ(b) ∈ Z(D̄) for any b ∈ Z(D̄) . Since _{w+(p−2)j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) = q_{1}_{−j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) + q_{2}_{m}^{(z_{k},u)}δ_{j(0)}(_{−j(k)}^{(z_{k},u)}δ_{j(k)}(a^{p^{k}})) for some integer q_{1}, q_{2}, m, we have proved our assertion. So, c_{w+(p−2)j(k),d(u,a^{p^{k}}),1} ∈ Z(D̄) .

If w = j(0) mod p, then _{w+(p−2)j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) = _{−j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) , since w + (p − 1)j(0) = 0 mod p and charD > 2 . So, we have c_{w+(p−2)j(k),d(u,a^{p^{k}}),1} ≠ 0 .

Put now t = a^{p^{k}} . For arbitrary r by proposition 2, (i) we have

_{w+(p−1−r)j(k)}^{(z_{k},u)}δ_{ζ}(t^{q}) = q_{w+(p−1−r)j(k)}δ_{ζ}(t)t^{q−1}+

w+(p−1−r)j(k)δ_{j(k)}(t) ∑_{l=0}^{q−2}_{w+(p−r)j(k)}δ_{ζ−j(k)}(t^{q−1−l})t^{l}+

w+(p−1−r)j(k)δ_{d(u,t)}(t) ∑_{l=0}^{q−2}_{w+(p−1−r)j(k)+d(u,t)}δ_{ζ−d(u,t)}(t^{q−1−l})t^{l}+

∑_{i=d(u,t)+1}^{ζ−1}_{w+(p−1−r)j(k)}δ_{i}(t) ∑_{l=0}^{q−2}_{w+(p−1−r)j(k)+i}δ_{ζ−i}(t^{q−1−l})t^{l}.

Using the same arguments as in the proof of lemma 5,(i) we see that _{w+(p−1−r)j(k)}δ_{ζ}(t^{p}) = 0 and _{w+(p−1−r)j(k)}δ_{ζ}|_{𝔽_{p}(t)} = c_{w+(p−1−r)j(k),ζ,1}δ + … + c_{w+(p−1−r)j(k),ζ,p−1}δ^{p−1} . To show that c_{w+(p−1−r)j(k),ζ,i} = 0 for i > r it suffice, by formulae (3) in lemma 5, to show that all the maps in the formula above are represented in the form c_{1}δ + … + c_{r−1}δ^{r−1} . Let us show it in details.

Since ζ − d(u, t) − 1 < (r − 1)j(k) , by lemma 5, (ii) _{m}δ_{ζ−i}|_{𝔽_{p}(t)} = c_{m,ζ−i,1}δ + … + c_{m,ζ−i,r−2}δ^{r−2} for any i > d(u, t) .

If w = j(0) mod p, then w + (p − 1 − r)j(k) + d(u, t) + (r − 2)j(k) = 0 mod p. Since ζ − d(u, t) ≤ (r − 1)j(k) , by lemma 5, (ii) we have _{w+(p−1−r)j(k)+d(u,t)}δ_{ζ−d(u,t)}|_{𝔽_{p}(t)} = c_{w+(p−1−r)j(k)+d(u,t),ζ−d(u,t),1}δ + … + c_{w+(p−1−r)j(k)+d(u,t),ζ−d(u,t),r−2}δ^{r−2} .

If w ≠ j(0) mod p, then by the same reason we have _{w+(p−1−r)j(k)+d(u,t)}δ_{ζ−d(u,t)}|_{𝔽_{p}(t)} = c_{w+(p−1−r)j(k)+d(u,t),ζ−d(u,t),1}δ + … + c_{w+(p−1−r)j(k)+d(u,t),ζ−d(u,t),r−1}δ^{r−1} and by lemma 5, (i) c_{w+(p−1−r)j(k)+d(u,t),ζ−d(u,t),r−1} ∈ Z(D̄) as a product of elements from Z(D̄) .

At last, by the induction hypothesis _{w+(p−r)j(k)}δ_{ζ−j(k)}|_{𝔽_{p}(t)} = c_{w+(p−r)j(k),ζ−j(k),1}δ+…+ c_{w+(p−r)j(k),ζ−j(k),r−1}δ^{r−1} and c_{w+(p−r)j(k),ζ−j(k),r−1} ≠ 0 only if ζ − j(k) = (r − 2)j(k) + d(u, t) , and c_{w+(p−r)j(k),ζ−j(k),r−1} ∈ Z(D̄) . Since _{w+(p−1−r)j(k)}δ_{j(k)}(t) ∈ Z(D̄) , by formulae (3) we get c_{w+(p−1−r)j(k),ζ,r} ∈ Z(D̄) and if w = j(0) mod p, then c_{w+(p−1−r)j(k),ζ,r} ≠ 0 iff ζ = (r − 1)j(k) + d(u, t) ,

c_{w+(p−1−r)j(k),(r−1)j(k)+d(u,t),r} = r!c_{w+(p−r)j(k),(r−2)j(k)+d(u,t),r−1}_{w+(p−1−r)j(k)}δ_{j(k)}(t) ≠ 0.

The lemma is proved. □

Lemma 14 Let D be a division algebra as in lemma 13. Suppose d(u, a) ≤ 2i(u, a) and charD > 2 .

Then for every k there exists a parameter z_{k} such that _{−j(k)}^{(z_{k},u)}δ_{r}|_{𝔽_{p}(a^{p^{k}})} = 0 for j(k) < r < d(u, a^{p^{k}}) and ^{(z_{k})}α = ^{(z)}α, ^{(z_{k},u)}δ_{j(l)} = ^{(z,u)}δ_{j(l)} for all l ≤ k (we use here the notation defined in lemma 13).

Moreover, for every k ≥ 1 we have d(u, a^{p^{k}}) − j(k) = d(u, a) − j(0) and

_{−j(k)}^{(z_{k},u)}δ_{d(u,a^{p^{k}})}(a^{p^{k}}) = −_{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,a^{p^{k−1}})}(a^{p^{k}})c_{d(u,t)−j(k−1),j(k)−j(k−1),p−1},

where c_{d(u,t)−j(k−1),j(k)−j(k−1),p−1} is defined in lemma 13.

Proof. The proof is by induction on k. By lemma 8 d(u, a) = 2j(0) mod p and j(1) = d(u, a) + (p − 1)j(0) . So, by the induction hypothesis we can assume for arbitrary k that d(u, a^{p^{k−1}}) = 2j(0) mod p and j(k − 1) = j(0) mod p, and j(k) = d(u, a^{p^{k−1}}) + (p − 1)j(k − 1) .

For the convinience we can start with a parameter z = z_{0} , which satisfy the conditions of lemma 13. Indeed, taking an appropriate parameter z and changing it by a parameter u(c)z for an appropriate c ∈ Z(D̄) (as in the proof of proposition 6), we can assume that _{−j(0)}^{(z,u)}δ_{j(0)}(a) ∈ Z(D̄)^{p}. Now, using arguments from the proof of lemma 8, (i), we can find such a parameter z_{0} .

The idea of the proof is the following. We prove first that _{−j(k−1)}^{(z_{k−1},u)}δ_{j(k)+d(u,a)−j(0)}(a^{p^{k}}) ≠ 0 . Then we prove that there exists a parameter z_{k} such that _{−j(k)}^{(z_{k},u)}δ_{ζ}(a^{p^{k}}) = 0 for j(k) < ζ < j(k)+d(u, a)−j(0) and _{−j(k)}^{(z_{k},u)}δ_{j(k)+d(u,a)−j(0)}(a^{p^{k}}) ≠ 0 . It will be shown that z_{k} satisfy the conditions of lemma.

So, assume j(k) ≤ ζ ≤ j(k) + d(u, a) − j(0) = j(k) + d(u, a^{p^{k−1}}) − j(k − 1) . Put t = a^{p^{k−1}} . By proposition 2, (i) we have

_{−j(k−1)}^{(z_{k−1},u)}δ_{ζ}(t^{p}) =

_{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,t)}(t) ∑_{l=0}^{p−2} _{d(u,t)−j(k−1)}^{(z_{k−1},u)}δ_{ζ−d(u,t)}(t^{p−1−l})t^{l} + … +

_{−j(k−1)}^{(z_{k−1},u)}δ_{ζ−(p−1)j(k−1)}(t) ∑_{l=0}^{p−2} _{ζ−pj(k−1)}^{(z_{k−1},u)}δ_{(p−1)j(k−1)}(t^{p−1−l})t^{l}+

∑_{i=ζ−(p−1)j(k−1)+1}^{ζ−1} _{−j(k−1)}^{(z_{k−1},u)}δ_{i}(t) ∑_{l=0}^{p−2} _{i−j(k−1)}^{(z_{k−1},u)}δ_{ζ−i}(t^{q−1−l})t^{l}.

By lemma 5, (i) in the last sum _{i−j(k−1)}^{(z_{k−1},u)}δ_{ζ−i}|_{𝔽_{p}(t)} = c_{i−j(k−1),ζ−i,1}δ+…+c_{i−j(k−1),ζ−i,p−2}δ^{p−2} , since ζ − i < (p − 1)j(k − 1) . So, this sum is equal to zero.

By lemma 5, (ii) we have _{ζ−pj(k−1)}^{(z_{k−1},u)}δ_{(p−1)j(k−1)}|_{𝔽_{p}(t)} = c_{ζ−pj(k−1),(p−1)j(k−1),1}δ + … + c_{ζ−pj(k−1),(p−1)j(k−1),p−1}δ^{p−1} and c_{ζ−pj(k−1),(p−1)j(k−1),p−1} ≠ 0 iff ζ = j(k−1) = j(0) mod p.

By lemma 5, (i) we have _{m}^{(z_{k−1},u)}δ_{q}|_{𝔽_{p}(t)} = c_{m,q,1}δ+…+c_{m,q,p−1}δ^{p−1} for (p−1)j(k−1) < q < (p − 1)j(k − 1) + d(u, a) − j(0) , and by lemma 13 c_{m,q,p−1} = 0 . By lemma 13 we have _{d(u,t)−j(k−1)}^{(z_{k−1},u)}δ_{ζ−d(u,t)}|_{𝔽_{p}(t)} = c_{d(u,t)−j(k−1),ζ−d(u,t),1}δ + … + c_{d(u,t)−j(k−1),ζ−d(u,t),p−1}δ^{p−1} with c_{d(u,t)−j(k−1),ζ−d(u,t),p−1} ≠ 0 if ζ − d(u, t) = j(0) .

So, we have the following picture: _{−j(k−1)}^{(z_{k−1},u)}δ_{ζ}(t^{p}) ≠ 0 only if ζ = j(0) mod p or if ζ = j(k) + d(u, a) − j(0) . In the last case

_{−j(k−1)}^{(z_{k−1},u)}δ_{ζ}(t^{p}) = −_{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,t)}(t^{p})c_{d(u,t)−j(k−1),j(k)−j(k−1),p−1},

where c_{d(u,t)−j(k−1),j(k)−j(k−1),p−1} can be calculated using lemma 13.

Let’s show that there exists a parameter z_{k} such that _{−j(k−1)}^{(z_{k},u)}δ_{ζ}(t^{p}) = 0 for j(k) < ζ < j(k)+d(u, a)−j(0) . By lemma 3, (ii) there exists a change of parameters z_{k−1} ↦ z′ = z_{k−1} + bz_{k−1}^{p+1} such that _{−j(k−1)}^{(z′,u)}δ_{j(k)+p}(t^{p}) = 0 . It suffice to prove that any such a change of parameters as in lemma 3, (ii) with p|q changes only the values of maps _{−j(k−1)}δ_{ζ} with ζ = j(0) mod p. For, if it is true, we can make several changes and kill all nonzero maps _{−j(k−1)}^{(z_{k−1},u)}δ_{ζ} with j(k) < ζ < j(k) + d(u, a) − j(0) , since they are derivations and therefore are completely defined by their values at t^{p}.

To prove it, we can use the calculations in the proof of lemma 3, (ii). Since d(u, a) − j(0) ≤ j(0) , it is easy to see that for a change z ↦ z′ = z + bz^{kp+1} , p > 2 we have there

z^{′−j(k−1)}t^{p}z^{′j(k−1)} = t^{p} + _{−j(k−1)}^{(z,u)}δ_{j(k)}(t^{p})z^{j(k)} + … + _{−j(k−1)}^{(z,u)}δ_{j(k)+j(0)}(t^{p})z^{j(k)+j(0)} + ….

Since z′ = z + bz^{kp+1} , any power z^{l} can be expressed as a series in z′, all powers of which are equal to l modulo p. So, this change will change only maps with right indexes equal to j(k) modulo p. Since _{−j(k−1)}^{(z_{k−1},u)}δ_{ζ}(t^{p}) ≠ 0 only if ζ = j(0) mod p for ζ < j(k) + d(u, a) − j(0) , our assertion is proved.

So, there exists a parameter z_{k} we have: _{−j(k−1)}^{(z_{k},u)}δ_{ζ}(t^{p}) ≠ 0 only if ζ = j(k) + d(u, a) − j(0) or ζ = j(k) . Since z_{k} was constructed as a sequence of changes as in lemma 3, (ii), we have ^{(z_{k})}α = ^{(z_{k−1})}α and ^{(z_{k},u)}δ_{j(q)} = ^{(z_{k−1},u)}δ_{j(q)} for any q ≤ k.

At last, let’s prove that _{−j(k)}^{(z_{k},u)}δ_{ζ}(t^{p}) ≠ 0 only if ζ = j(k) + d(u, a) − j(0) or ζ = j(k) . But this follows immediately from the definition of these maps, since j(k) = j(k − 1) mod p, d(u, a) − j(0) ≤ j(0) and charD > 2 . In particular, _{−j(k)}^{(z_{k},u)}δ_{j(k)}(t^{p}) = _{−j(k−1)}^{(z_{k},u)}δ_{j(k)}(t^{p}) , _{−j(k)}^{(z_{k},u)}δ_{j(k)+d(u,a)−j(0)}(t^{p}) = _{−j(k−1)}^{(z_{k},u)}δ_{j(k)+d(u,a)−j(0)}(t^{p}) .

The lemma is proved. □

Now we can prove the following theorem.

Theorem 4 Let D be a division p -algebra of charD = p > 2 with the center Z(D) = F . Suppose Z(D̄) = D̄ and D̄/F̄ is a simple purely inseparable extension, D̄ = F̄(a) . Suppose that the semilocal height i(u) , which does not depend on the embedding u in this case, is not divisible by p.

Then d_{D}(a) > i(u) .

Proof. By lemma 12, (ii) [D̄ : F̄] = [Γ_{D} : Γ_{F}] . So, the field F (ã) , where ã is a lift of a, is a maximal ”unramified” subfield and therefore D is a splittable division algebra. Obviously, α = id.

Since ^{(z,u)}δ_{i(u,z)} is a derivation and D̄/F̄ is a simple extension, ^{(z,u)}δ_{i(u,z)} is completely defined by a value at a. So, by lemma 3 i(u, z) does not depend on z and i(u, z) = i(u) . Therefore, i(u) = w(zu(a)z^{−1} − u(a)) and i(u) is completely defined by the lift u(a) . From the other hand side, any lift ã of a defines, by lemma 9, an embedding ã, and by lemma 10 i(ã) does not depend on ã. So, i(u) does not depend on u.

The idea of the proof is following. We consider linear spaces which are the images of the maps ^{(z,u)}δ_{j(k)}|_{F̄(a^{p^{k}})} in D̄, where j(k) were defined in lemma 14 and z, u are fixed. We show that every such spase has zero intersection with each other if d_{D}(a) ≤ i(u) . Then we show that this contradicts with the fact that u(a) generate a finite dimensional space over F .

So, assume d_{D}(a) ≤ i(u) . To calculate the spaces ^{(z,u)}δ_{j(k)}(F̄(a^{p^{k}})) ∈ D̄ we use lemmas 8, 13 and 14. We fix a parameter z defined in lemma 13. By lemmas 9, 10, (iii) we can find a primitive element ū ∈ D̄ of the extension D̄/F̄ such that ^{(z,u)}δ_{j(0)}(ū) = 1 , where u is an embedding defined in lemma 9 for some lift u of the element ū. Using lemma 3, (ii) we can find an embedding u such that ^{(z,u)}δ_{d(u,ū)}(ū) ∉ ^{(z,u)}δ_{j(0)}(D̄) . We fix this embedding. From lemmas 3, 10 immediately follows that d(u, ū) = d_{D}(ū) = d_{D}(a) . So, we assume without loss of generality a = ū.

Put J(k) := ^{(z,u)}δ_{j(k)}(a^{p^{k}}) . Put A(k) := ^{(z,u)}δ_{j(k)}(F̄(a^{p^{k}})) , A′(k) := F̄(a^{p^{k+1}}) · a^{p^{k}(p−1)}J(k) .

We have A(k) = ⊕_{q=0}^{p−2}F̄(a^{p^{k+1}}) · a^{p^{k}q}J(k) and D̄ · J(k) = A(k) ⊕ A′(k) as 𝔽_{p} -linear spaces.

¿From lemma 8 follows that

(z,u)δ_{j(k)}(a^{p^{k}}) = ^{(z_{k},u)}δ_{j(k)}(a^{p^{k}}) = q_{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,a^{p^{k−1}})}(a^{p^{k−1}})c_{d(u,a^{p^{k−1}})−j(k−1),(p−1)j(k−1),p−1},

where q ∈ 𝔽_{p}^{∗}, z_{k} were defined in lemma 13, c_{d(u,a^{p^{k−1}})−j(k−1),(p−1)j(k−1),p−1} is calculated

in lemma 5, (i) and it is not equal to zero by lemma 5, (ii), and _{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,a^{p^{k−1}})}(a^{p^{k−1}})

is calculated in lemma 14. By lemma 14 we have _{−j(k−1)}^{(z_{k−1},u)}δ_{d(u,a^{p^{k−1}})}(a^{p^{k−1}}) = −j(k − 1)^{(z,u)}δ_{d(u,a^{p^{k−1}})}(a^{p^{k−1}}) . Combining all these calculation together and using induction, we get J(k) = q_{k}J(k − 1)^{p}J(1) = q̃_{k}J(1)^{p^{k−1}+p^{k−2}+...+1} for k ≥ 1 , where q_{k} ∈ 𝔽_{p}.

Therefore, there is the following filtration

F̄ ⊂ … ⊂ F̄(a^{p^{k+1}})J(k + 1) ⊂ F̄(a^{p^{k}})J(k) ⊂ … ⊂ D̄,

and for every k ≥ 1 we have F̄(a^{p^{k}})·J(k) ⊂ A′(k−1) . So, A(k)∩A(k_{1}) = {0} if k ≠ k_{1} .

Now consider an element b ∈ F such that b̄ = a^{p^{l}} for some l > 0 . We assume l is a minimal possible integer. It exists, because D is a finite dimensional algebra over F . Let b = u(a^{p^{l}}) + b_{1}z + …, where b_{k} ∈ u(D̄) . Put I := min{w(zb_{k}z^{k−1} − b_{k}z^{k})} (we assume here that b_{0} = u(a^{p^{l}}) ). Note that I < ∞, since by lemma 14 j(l) < ∞, i.e. (z,u)δ_{j(l)}(a^{p^{l}}) ≠ 0 . Now we must have

zbz^{−1} = ∑_{k=0}^{∞} zb_{k}z^{k−1} = b + ∑_{r}^{(z,u)}δ_{j(r)}(b_{q_{r}})z^{I} + … = b,

where b_{q_{r}} ∈ F̄(a^{p^{r}}) and b_{q_{r}} ∉ F̄(a^{p^{r+1}}) . So, ∑_{r} ^{(z,u)}δ_{j(r)}(b_{q_{r}}) = 0 , but it is impossible, since A(k) ∩ A(k_{1}) = {0} if k ≠ k_{1} , a contradiction.

The theorem is proved. □

Remark. It would be interesting to know the answer on the following questions.

i) Suppose D is a division algebra as in the theorem 4. Does there exist a pair (z, u) such that all nonzero maps ^{(z,u)}δ_{q} satisfy the property i(u)|q? If it is true, there is a subalgebra D′ ⊂ D with [D : D′] < ∞ and D′ has level 1 (see remark before lemma 8). So, we can reduce studying of D to the algebra of level 1.

ii) Is it true that D is a good splittable algebra, i.e. cyclic? Probably, it is possible to apply our technique to give an answer to this question at least in the case of level 1.

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