Nondegenerate semiramified valued and graded division

algebras^{∗}^{†}

Karim Mounirh

Département des Mathématiques Faculté des Sciences Université Catholique de Louvain Belgium e.mail: akamounirh@hotmail.com

Abstract

In this paper, we define what we call (non)degenerate valued and graded division algebras [Definition 3.1] and use them to give examples of division p-algebras that are not tensor product of cyclic algebras [Corollary 3.17] and examples of indecomposable division algebras of prime exponent [Theorem 5.2, Corollary 5.3 and Remark 5.5]. We give also, many results concerning subfields of these division algebras.

∗Mathematics subject classification (MSC2000): 16K50, 16W50, 16W60 and 16W70.

†Key Words: (Graded) Brauer group, Valued division algebras, Henselization, Graded division algebras, Generic abelian crossed products.

Preliminaries

Let F be an associative ring with a unit and Γ a totally ordered abelian group. We say that F is a graded ring of type Γ if there are subgroups F_{γ} (γ ∈ Γ) of F such that F = ⊕_{γ∈Γ}F_{γ} and F_{γ}F_{δ} ⊆ F_{γ+δ}, for all γ, δ ∈ Γ. In this case, the set Γ_{F} = {γ ∈ Γ | F_{γ} ≠ 0} is called the support of F.

If F is a graded ring of type Γ and x ∈ F_{γ} for some γ ∈ Γ_{F}, we say that x is a homogeneous element of F. In particular, if x is a nonzero element of F_{γ}, we say that x has grade γ and we write gr(x) = γ. We denote by F^{h} [resp. F^{∗}] the set of homogeneous [resp. nonzero homogeneous] elements of F. A graded ring F which is commutative and for which all nonzero homogeneous elements are invertible is called a graded field. If in addition F has support Γ and there is a ring isomorphism σ : F ↦ F_{0}[Γ] such that σ(F_{0}) = F_{0} and σ(F_{γ}) = F_{0}γ for any γ ∈ Γ − {0}, then F is called a graded field of group-ring type.

Let F be a commutative graded ring of type Γ. A (left) algebra A over F is called a graded algebra over F (of type Γ) if A is a graded ring of type Γ and F_{γ} ⊆ A_{γ}, for all γ ∈ Γ. In particular, if F is a graded field, then graded algebras over F [resp. commutative graded algebras over F] for which homogeneous elements are invertible are called graded division algebras (GDA) over F [resp. graded field extensions of F]. If F is the center of a graded division algebra A, then A is called a graded central division algebra (GCDA) over F.

We recall that a graded field extension K of a graded field F is totally ramified over F if [K : F] = (Γ_{K} : Γ_{F}). It is inertial over F if [K : F] = [K_{0} : F_{0}] and K_{0} is separable over F_{0}. K is called tame over F if K_{0} is separable over F_{0} and Γ_{K}/Γ_{F} has no p-torsion, where p = char(F). Finally, K is purely wild over F if K_{0} is purely inseparable over F_{0} and Γ_{K}/Γ_{F} is a p-group. By [HW(1)99, Lemma 3.6], K/F is purely wild if and only if F rac(E)/F rac(F) is purely inseparable.

Let F be a graded field and F rac(F)_{alg} an algebraic closure of F rac(F). We recall that for any element λ of the divisible hull Δ_{F} (= Γ_{F} ⊗_{ZZ} Q ′ ) of Γ_{F}, there is a unique grading on the polynomial ring F [X] extending the grading of F and for which X is a homogeneous element with gr(X) = λ. We denote this grading by F [X]^{(λ)}. A polynomial f ∈ F [X] is called homogenizable if there is λ ∈ Δ_{F} such that f ∈ (F [X]^{(λ)})^{h}. Let x ∈ F rac(F)_{alg} and denote by f_{x,F rac(F)} its minimal polynomial over F rac(F). We say that x is algebraic over F if f_{x,F rac(F)} is a homogenizable polynomial of F [X]. If K is a graded field extension of F, we say that K is algebraic over F if every homogeneous element of K is algebraic over F. Let F_{alg} = F [{x ∈ F rac(F)_{alg} | x is algebraic over F }], then by [HW(1)99, Corollary 2.7(c)] F_{alg} is an algebraic graded field extension of F and all subrings of F rac(F)_{alg} that are algebraic graded field extensions of F are graded subfields of F_{alg}. We call F_{alg} ’the’ algebraic closure of F.

If F is a graded field and A is a graded division algebra over F, we denote by Cq(A) the algebra of central quotients of A. Obviously, to the graded structure of A corresponds a canonical valuation on Cq(A), defined by v(a) = gr(a) for any a ∈ A^{∗} (See [B98, §4] or [HW(2)99, §4]). We denote by HCq(A) the algebra Cq(A) ⊗_{F rac(F)} HF rac(F), where HF rac(F) is the Henselization of F rac(F) with respect to its canonical valuation (see [E72, §16]).

Conversely, let E be a field and v a valuation of E. Then, the filtration of E induced by v yields a canonical graded field GE. Namely, let E^{γ} = {x ∈ E | v(x) ≥ γ}, E^{>γ} = {x ∈ E | v(x) > γ}. Obviously, E^{>γ} is a subgroup of the additive group E^{γ}.

So, we can define the quotient group GE_{γ} = E^{γ}/E^{>γ}. For x ∈ E\{0}, we denote by x̃ the element x + E^{>v(x)} of GE_{v(x)}. One can easily see that the additive group GE = ⊕_{γ∈Γ}GE_{γ} with the multiplication law defined by x̃ỹ = x̃y is a graded field.

In the same way, if D is a valued division algebra over a field E (we refer to [JW90] for non-commutative valuative definitions used in this paper), then the filtration of D by the principal fractional ideals yields a graded division algebra GD (see [B98, §4] or [HW(2)99, §4]). We recall that if F is a graded field and A is a graded central division algebra over F, then A is graded isomorphic to GHCq(A) by means of the mapping x ↦ x̃, where x is an arbitrary homogeneous element of A. Indeed, we have A_{0} = (GHCq(A))_{0} and Γ_{A} = Γ_{GHCq(A)}. We write A ≅_{g} GHCq(A).

A valued central division algebra D over a field E is called defectless (over E) if [D : E] = [D̄ : Ē](Γ_{D} : Γ_{E}), where [D̄ : Ē] [resp. (Γ_{D} : Γ_{E})] is the residue degree [resp. ramification index] of D over E. If in addition, D̄ is a field and [D̄ : Ē] = (Γ_{D} : Γ_{E}), we say that D is semiramified. If D has an inertial and a totally ramified maximal subfields, we say that D is nicely semiramified (see [M05, Theorem 4]).

Let E be a Henselian valued field and D a defectless central division algebra over E. We know that the map

θ_{D} : Γ_{D}/Γ_{E} → Gal(Z(D̄)/Ē)
γ + Γ_{E} ↦ θ_{D}(γ + Γ_{E}) : ā ↦ \overline{dad^{−1}}

(d being an arbitrary element of D such that v(d) = γ) is a surjective group homomorphism [JW90, Proposition 1.7]. We say that D is tame (over E) if it is defectless over E, the center Z(D̄) of D̄ is separable over Ē and the characteristic of Ē does not divide the cardinality of the kernel of θ_{D}.

We recall that if F is a graded field and D is a graded central division algebra over F, then HCq(D) is a tame central division algebra over HF rac(F) (See [HW(2)99, Theorem 5.1]. It suffices also to apply [B95, Corollary 4.4]). Analogously, to the valued case, we say that D is semiramified [resp., nicely semiramified] if D_{0} is a field and [D_{0} : E_{0}] = (Γ_{D} : Γ_{E}) [resp., if D has an inertial and a totally ramified maximal graded subfields].

It is well known that graded central division algebras over a graded field F play the same role as central division algebras over a Henselian valued field. Indeed, their equivalent classes form a Graded Brauer group GBr(F) and there is a group isomorphism GBr(F) → T Br(HF rac(F)), where T Br(HF rac(F)) is the tame part of Br(HF rac(F)) [HW(2)99, Theorem 5.1]. Conversely, for any Henselian valued field E, there is a canonical group isomorphism T Br(E) → GBr(GE) [HW(2)99, Theorem 5.3].

Now, let M/E be a finite-dimensional abelian field extension with Galois group G; S = (σ_{i})_{1≤i≤r} be a basis of G (i.e. such that G = ⟨σ_{1}⟩ ⊕ ... ⊕ ⟨σ_{r}⟩); q_{i} = ord(σ_{i}) for 1 ≤ i ≤ r; M_{i} (resp. M_{ij}) be the subfield of M fixed by σ_{i} (resp. σ_{i} and σ_{j}); N_{i} (resp. N_{ij}) be the norm of M/M_{i} (resp. M/M_{ij}); and I = {(m_{1}, ..., m_{r}) ∈ IN^{r} | 0 ≤ m_{i} < q_{i} for 1 ≤ i ≤ r}. We recall that if M is a maximal subfield of a central simple E-algebra A, then there exist invertible elements x_{1}, ..., x_{r} of A such that the following conditions hold : (0.1) A = ⊕_{m̄∈I}Mxm̄, where xm̄ = x_{1}^{m_{1}}...x_{r}^{m_{r}} ; (0.2) x_{i}a = σ_{i}(a)x_{i} for a ∈ M and 1 ≤ i ≤ r. Let u_{ij} = x_{i}x_{j}x_{i}^{−1}x_{j}^{−1} and b_{i} = x_{i}^{q_{i}} for 1 ≤ i, j ≤ r, then the matrices U = (u_{ij})_{1≤i,j≤r} and B = (b_{i})_{1≤i≤r} satisfy the following conditions (for 1 ≤ i, j, k ≤ r) :

(0.3) u_{ij} and b_{i} are non-zero elements of M ; (0.4) u_{ji} = u_{ij}^{−1} ; (0.5) σ_{i}(u_{jk})σ_{j}(u_{ki})σ_{k}(u_{ij}) = u_{jk}u_{ki}u_{ij} ; (0.6) N_{ij}(u_{ij}) = 1 ; (0.7) σ_{j}(b_{i})b_{i}^{−1} = N_{i}(u_{ji}).

Conversely, for any pair of matrices (U, B) satisfying conditions 0.3 to 0.7, we can construct a central simple E-algebra (M, G, U, B) = ⊕_{m̄∈I}Mzm̄, where z_{i} are independent indeterminates that satisfy the following conditions : z_{i}a = σ_{i}(a)z_{i}, z_{i}z_{j} = u_{ij}z_{j}z_{i} and z_{i}^{q_{i}} = b_{i} (See [AS78, section 1]).

J.P. Tignol showed that conditions 0.3 to 0.7 mean that (U, B) is a 2-cocycle with respect to a G-complex described in [T81, §1, p. 421]. Hence he denoted (U, B) ∈ Z_{S}^{2}(G, M^{∗}). He wrote also U_{S}(G, M^{∗}) to denote the set of matrices U (with entries in M^{∗}) satisfying conditions (0.3) to (0.6). A matrix U ∈ U_{S}(G, M^{∗}) is said to be degenerate (with respect to the field extension M/E) if there exist m̄, n̄ ∈ I and a, b ∈ M^{∗} such that the subgroup ⟨σm̄, σn̄⟩ of G, generated by σm̄ and σn̄ is non-cyclic, and u_{m̄,n̄} = σm̄(a)a^{−1}σn̄(b)b^{−1}, where u_{m̄,n̄} = xm̄xn̄(xm̄)^{−1}(xn̄)^{−1} [AS78, Definition, p. 81].

Now, for a matrix U = (u_{ij})_{1≤i,j≤r} of U_{S}(G, M^{∗}), consider the amalgamated sum M.E⟨z_{1}, ..., z_{r}⟩ over E of M and the free E-algebra on the independent indeterminates z_{i} (1 ≤ i ≤ r); then let I be the ideal of M.E⟨z_{1}, ..., z_{r}⟩, generated by the elements z_{i}z_{j} − u_{ij}z_{j}z_{i} and z_{i}a − σ_{i}(a)z_{i}, for 1 ≤ i, j ≤ r and a ∈ M. We recall that the generic abelian crossed product (M, G, U) is the algebra of central quotients of M.E⟨z_{1}, ..., z_{r}⟩/I. Let B = (b_{i})_{1≤i≤r} be a matrix of M such that (U, B) ∈ Z_{S}^{2}(G, M^{∗}) and let Z_{i} = b_{i}^{−1}z_{i}^{q_{i}}, then the center of (M, G, U) is E(Z_{1}, ..., Z_{r}) [T81, §2, p. 428].

We have proved in [BM00, Theorem 1.1] that there is a graded field F and a semiramified graded division algebra D over F such that E(Z_{1}, ..., Z_{r}) = F rac(F) and (M, G, U) = Cq(D). The valuation of E(Z_{1}, ..., Z_{r}) defined by the graded structure of F will be called the canonical valuation of E(Z_{1}, ..., Z_{r}).

Let M/E be as above an abelian finite-dimensional field extension, F a graded field with F_{0} = E and with support Γ_{F} and MF the graded subfield of F_{alg} generated by M and F. Assume there is a totally ordered abelian group Δ containing Γ_{F} with ordering extending that of Γ_{F} and suppose ϵ : Δ → G is a group homomorphism with kernel Γ_{F}. Let δ_{i} ∈ Δ such that ϵ(δ_{i}) = σ_{i}, and consider a family (x_{q_{i}δ_{i}})_{1≤i≤r} of elements of F^{∗} with x_{0} = 1 and gr(x_{q_{i}δ_{i}}) = q_{i}δ_{i}. For a cocycle (U, B) ∈ Z_{S}^{2}(G, M^{∗}), consider the iterated twisted polynomial ring A := MF [X_{1}, ..., X_{r}; σ_{1}, ..., σ_{r}] defined by X_{i}X_{j} = u_{ij}X_{j}X_{i}, X_{i}a = σ_{i}(a)X_{i} for all 1 ≤ i, j ≤ r and a ∈ MF (see [AS, p. 78]), and let I be the left ideal of A, generated by the elements X_{i}^{q_{i}} − b_{i}x_{q_{i}δ_{i}}. Then, I is a two-sided ideal of A. Indeed, for any 1 ≤ i, j ≤ r, we have (X_{i}^{q_{i}} − b_{i}x_{q_{i}δ_{i}})X_{j} = N_{i}(u_{ij})X_{j}X_{i}^{q_{i}} − X_{j}σ_{j}^{−1}(b_{i})x_{q_{i}δ_{i}} = X_{j}(σ_{j}^{−1}(N_{i}(u_{ij}))X_{i}^{q_{i}} − σ_{j}^{−1}(b_{i})x_{q_{i}δ_{i}}) = X_{j}σ_{j}^{−1}(N_{i}(u_{ij}))(X_{i}^{q_{i}} − b_{i}x_{q_{i}δ_{i}}). So, we can consider the quotient ring (MF, ϵ, G, U, B) := A/I. One can easily see that (MF, ϵ, G, U, B) is a semiramified graded division algebra over F.

In the first section of the present paper we give some new results in Galois theory of graded fields. In particular, we show that if E is a field and M is a defectless normal finite-dimensional valued field extension of E, then GM is a normal graded field extension of GE (and that we can drop the condition ’defectless’ when E is Henselian) [Theorem 1.4]. We prove also that if E is a Henselian valued field, M is a finite-dimensional field extension of E, g = ∑_{i=0}^{n} ã_{i}X^{i} is a homogenizable irreducible polynomial of GE[X] (a_{i} ∈ E) with a simple root α in GM and f = ∑_{i=0}^{n} a_{i}X^{i}, then f is irreducible and has a simple root a in M such that ã = α [Proposition 1.5]. We show also that if E is a Henselian valued field with residue characteristic p > 0 and L is a purely wild [resp., a defectless simple purely wild] finite-dimensional graded field extension of GE, then there is a defectless field [simple field] extension K of E such that GK = L. Furthermore, K can be chosen to be a purely inseparable field extension of E when char(E) = p [Proposition 1.7]. Then, we use Proposition 1.5 to give a new proof of [HW(1)99, Theorem 5.2] which for an arbitrary Henselian valued field E establish a one-to-one correspondence between the set of isomorphism classes of finite-dimensional tame field extensions of E and the set of isomorphism classes of finite-dimensional tame graded field extensions of GE [Corollary 1.8]. Finally, in the last part of this section, we prove some results concerning cyclic and Kummer graded field extensions.

The second section deals with the structure of nicely semiramified valued division algebras. We recall that these algebras were defined in [JW90, Definition p.149] to be defectless finite-dimensional valued division algebras with inertial and totally ramified of radical type maximal subfields. As indicated in [JW90, p.128], nicely semiramified division algebras appeared in [Pl76] as examples of division algebras with nonzero SK_{1}. They appeared also in the similarity decomposition of any tame division algebra over a Henselian valued field [JW90, Lemma 5.14 and Lemma 6.2]. As seen above, we have proved in [M05, Theorem 4] that any finite-dimensional central division algebra over a Henselian valued field E with an inertial maximal subfield and a totally ramified maximal subfield (not necessarily of radical type) [resp. split by an inertial and a totally ramified field extensions of E] is nicely semiramified. We use in this section this new characterization of nicely semiramified division algebras to give many new properties describing them. We are precisely interested here in determining relations between cyclic division p-algebras and nicely semiramified division algebras over Henselian valued fields (see [Proposition 2.5 and Proposition 2.9]).

In the third section, we define what we call degenerate (valued and graded) division algebras [Definition 3.1] and we determine when a tame semiramified division algebra is degenerate [Proposition 3.6, Proposition 3.9, Proposition 3.11 and Proposition 3.12]. We prove also many equivalent statements for a semiramified graded division algebra D over a graded field F to be degenerate [Proposition 3.16]. In particular, we show that D is degenerate if and only if D_{0}F rac(F) is a degenerate maximal subfield of Cq(D) (see [BM00, Definition 0.12]). We use then nondegenerate division algebras to give examples of division p-algebras that are not tensor products of cyclic algebras [Corollary 3.17] (see also Remark 3.19(1)). As a consequence of Corollary 3.17, we get a new proof of [AS78, Theorem 3.2] (see [Corollary 3.18]).

We show in the forth section many results concerning subfields of nondegenerate tame semiramified division algebras of prime power degree over Henselian valued fields. In particular, we prove that a nondegenerate tame semiramified division algebra D over a Henselian valued field E with Γ_{D}/Γ_{E} non-cyclic cannot have a totally ramified (non-trivial) subfield [Proposition 4.2]. This enables us to prove that if E has residue characteristic p > 0, D is a nondegenerate tame semiramified division algebra of degree a power of p over E with Γ_{D}/Γ_{E} non-cyclic, and K is a normal subfield of D, then K is an inertial Galois subfield of D. Hence, there is a subgroup H of Gal(D̄/Ē) such that Gal(K/E) ≅ Gal(D̄/Ē)/H. In particular, if K is a Galois maximal subfield of D, then Gal(K/E) ≅ Gal(D̄/Ē) [Proposition 4.3]. This Proposition generalizes [S78, Theorem 3.1] and [McK05, Theorem 2.3]. We prove also that if char(Ē) is arbitrary, deg(D) is a power of a prime integer p and rk(Γ_{D}/Γ_{E}) ≥ 3, then all abelian subfield of D are inertial over E [Proposition 4.9]. Then, we show that if rk(Γ_{D}/Γ_{E}) is arbitrary but exp(Γ_{D}/Γ_{E}) = p, then any non-maximal subfield of D is inertial over E. Also, if K is a non-quaternion normal maximal subfield of D, then K is either cyclic with dimension ≤ p^{2} or inertial over E [Theorem 4.11]. More results for the case where char(Ē) does not divide deg(D) are proved at the end of this section.

Finally, in the last section we prove that nondegenerate tame semiramified division algebras of prime power degree over Henselian valued fields are indecomposable [Theorem 5.2]. This can be used to construct indecomposable division algebras of prime exponent (see [Remark 5.4]).

We will give in a next paper many results concerning prime to p-extensions of nondegenerate inertially split division algebras of degree a power of p over a Henselian valued field (where p is a prime positive integer).

1 Graded and valued field extensions

Let F be a graded field and E an algebraic graded field extension of F. We recall that E is called normal over F if every homogenizable irreducible polynomial of F [X] that has a root in E splits completely (i.e., into degree one factors) in E[X]. Let g be a homogenizable irreducible polynomial of F [X]. Assume g has a root x ∈ E\{0} and let f_{x,F rac(F)} be the minimal polynomial of x over F rac(F). By Claim A.1 comments below, we have g = af_{x,F rac(F)} for some a ∈ F^{∗}. So, by [HW(1)99, Proposition 2.2 and corollary 2.5] x ∈ E^{∗} . Therefore, E is normal over F if and only if for any x ∈ E^{∗}, f_{x,F rac(F)} splits into one degree polynomials in E[X].

Lemma 1. 1 Let E/F be an algebraic graded field extension. Then, E is normal over F if and only if F rac(E) is a normal field extension of F rac(F).

Proof. Suppose that E is normal over F and consider a F rac(F)-isomorphism σ from F rac(E) into F rac(E)_{alg}. Let x ∈ E^{∗} and let f_{x,F rac(F)} be its minimal polynomial over F rac(F). Obviously, we have f_{x,F rac(F)}(σ(x)) = 0. So, σ(x) ∈ E^{∗}. See that we have F rac(E) = EF rac(F), hence σ(F rac(E)) = F rac(E). Therefore, F rac(E) is a normal field extension of F rac(F). Conversely, suppose that F rac(E) is a normal field extension of F rac(F) and let x ∈ E^{∗}. Then, f_{x,F rac(F)} splits into one degree polynomials in F rac(E)[X]. Let y be a root of f_{x,F rac(F)} and let σ be a F rac(F)-automorphism of F rac(E) such that y = σ(x). By [HW(1)99, Corollary 2.5(d)] the restriction of σ to E is a graded F -automorphism of E. So, y ∈ E^{∗}.

Proposition 1. 2 Let E/F be a finite-dimensional graded field extension. Then the following statements are equivalent :

(1) E/F is tame and normal.

(2) E is a Galois graded field extension of F.

Proof. By [HW(1)99, Theorem 3.11(a), (b)] and Lemma 1.1.

Let E/F be a normal finite-dimensional graded field extension. As in the ungraded case, we call the group G consisting of graded F -isomorphisms of E into E_{alg} the Galois group of E/F.

Proposition 1. 3 Let E/F be a finite-dimensional normal graded field extension of Galois group G and let F ix_{G}(E) be the graded subfield of E, elementwise invariant under the action of G. Then, F ix_{G}(E) is purely wild over F and E is Galois over F ix_{G}(E). Moreover, if T is the tame closure of F in E, then E = T F ix_{G}(E) ≅_{g} T ⊗_{F} F ix_{G}(E).

Proof. Since E is normal over F, then by Lemma 1.1 F rac(E)/F rac(F) is a normal field extension. Moreover, we have F rac(E) ≅ E ⊗_{F} F rac(F), so Gal(E/F) = Gal(F rac(E)/F rac(F)) -up to a group isomorphism- (See that by [HW(1)99, Corollary 2.5(d)] the restriction to E of any σ ∈ Gal(F rac(E)/F rac(F)) is in Gal(E/F)). By identification of F rac(E) and E ⊗_{F} F rac(F), for any element x ∈ F rac(E) there are α ∈ E and β ∈ F rac(F) such that x = α ⊗ β (indeed, we can write x = ∑ a_{i} ⊗ b_{i} where a_{i} ∈ E and b_{i} ∈ F rac(F). Let 0 ≠ b ∈ F such that h_{i} := b_{i}b ∈ F for all i, then we have x = (∑ a_{i}h_{i}) ⊗ b^{−1}). So, for any σ ∈ Gal(F rac(E)/F rac(F)), σ(x) = x if and only if σ(α) = α. Accordingly, F ix_{G}(F rac(E)) = F ix_{G}(E) ⊗_{F} F rac(F) = F rac(F ix_{G}(E)). Therefore, by [Karp89, Proposition 7.7, p.283] and [HW(1)99, Theorem 3.11(b) and Lemma 3.6], E/F ix_{G}(E) is Galois and F ix_{G}(E)/F is purely wild. By [HW(1)99, (3.8)], F rac(T) is the separable closure of F rac(F) in F rac(E). Furthermore, we have T F ix_{G}(E) ⊗_{F} F rac(F) = F rac(T)F ix_{G}(F rac(E)). So, by [Karp89, Proposition 7.7, p.283] T F ix_{G}(E) ⊗_{F} F rac(F) = F rac(E). Therefore, T F ix_{G}(E) = E. Remark that F rac(T) is separable over F rac(F) and F rac(F ix_{G}(E)) is purely inseparable over F rac(F). Therefore, T and F ix_{G}(E) are linearly disjoint over F. Hence, E ≅_{g} T ⊗_{F} F ix_{G}(E).

Let E be a Henselian valued field, M a finite-dimensional field extension of E, (x_{i})_{i=1}^{n} a family of elements of M such that v(x_{i}) = v(x_{j}) for all 1 ≤ i, j ≤ n and such that f := ∏_{i=1}^{n}(X −x_{i}) is an element of E[X]. We aim to show here that ∏_{i=1}^{n}(X −x̃_{i}) is an element of GE[X]. Let for m̄ = (m_{1}, ..., m_{n}) ∈ IN^{n}, supp(m̄) = {i | such that m_{i} ≠ 0}, then let I_{k}^{n} = {m̄ = (m_{1}, ..., m_{n}) ∈ IN^{n} | card(supp(m̄)) = k and m_{i} = 0 or 1 for all 1 ≤ i ≤ n}. For any m̄ = (m_{1}, ..., m_{n}) ∈ I_{k}^{n}, denote xm̄ = ∏_{i∈supp(m̄)} x_{i}. Then, ∏_{i=1}^{n}(X − x_{i}) = ∑_{i=0}^{n} d_{k}X^{n−k} [resp. ∏_{i=1}^{n}(X − x̃_{i}) = ∑_{i=0}^{n} d′_{k}X^{n−k}] , where d_{0} = 1 [resp. d′_{0} = 1] and d_{k} = ∑_{m̄∈I_{k}^{n}}(−1)^{k}xm̄ [resp. d′_{k} = ∑_{m̄∈I_{k}^{n}}(−1)^{k}x̃m̄] (for 1 ≤ k ≤ n). See that v(xm̄) = kv(x) for any m̄ ∈ I_{k}^{n}. So, in GE_{kv(x)}(= E^{kv(x)}/E^{>kv(x)}) we have d̄_{k} = \overline{∑_{m̄∈I_{k}^{n}}(−1)^{k}xm̄} = ∑_{m̄∈I_{k}^{n}}(−1)^{k}x̃m̄ = d′_{k}. So, ∏_{i=1}^{n}(X − x̃_{i}) ∈ GE[X].

Theorem 1. 4 Let E be a field and M a defectless finite-dimensional valued field extension of E. If M is normal over E, then GM is normal over GE. If E is Henselian, then this is true even if M is not defectless over E.

Proof. Let v be the valuation of M and let HM be the henselization of M with respect to v [E72, §16]. Since M̄ ⊗Ē \overline{HE} = M̄ and Γ_{M} ∩ Γ_{HE} = Γ_{E}, then by [Mor89, Theorem 1] M ⊗_{E} HE is a field. So, HM = MHE = M ⊗_{E} HE. Moreover, because M ⊆ HM, (GM)_{0} = M̄ = (GHM)_{0} and Γ_{GM} = Γ_{GHM}, then GM = GHM. Similarly, GE = GHE. Therefore, we can assume that E is Henselian. Now, let x be an element of M and let f_{x,E} be the minimal polynomial of x over E. M being normal over E, then we can write f_{x,E} = ∏_{i=1}^{n}(X − x_{i}), where x_{i} = σ_{i}(x) for some σ_{i} ∈ Gal(M/E). Since E is Henselian, then v(x_{i}) = v(x). Therefore, by the above the polynomial ∏_{i=1}^{n}(X − x̃_{i}) is in GE[X]. Hence, the minimal polynomial f_{x̃,F rac(GE)} of x̃ over F rac(GE) splits into one degree polynomials in GM[X].

Proposition 1. 5 Let E be a Henselian valued field, M a finite-dimensional field extension of E, g = ∑_{i=0}^{n} ã_{i}X^{i} a homogenizable irreducible polynomial of GE[X], with a_{i} ∈ E, and f = ∑_{i=0}^{n} a_{i}X^{i}. If g has a simple root α in GM, then f is irreducible and has a simple root a in M such that ã = α.

Proof. Since X − α divides g and g is homogenizable, then α ∈ GM^{h} and gr(α) = gr(X). Therefore, α = ẽ for some e ∈ M. Let f_{α,F rac(GE)} be the minimal polynomial of α over F rac(GE). Then, by Claim A.1 comments below, g = βf_{α,F rac(GE)} for some β ∈ GE^{h}. Let Γ_{M} be the value group of M (for the extension of the valuation of E to M), Δ_{M} = Γ_{M} ⊗_{ZZ} Q ′ (the divisible hull of Γ_{M}) and μ = gr(g)(∈ Δ_{M}). Then, gr(ã_{i}) + igr(ẽ) = μ for all 1 ≤ i ≤ n when a_{i} ≠ 0. See that ã_{0} ≠ 0 (because g is irreducible). Hence e ≠ 0. We distinguish the following two cases:

i) If μ = 0, then setting Y = ẽ^{−1}X, we get g = ∑_{i=0}^{n} \overline{a_{i}e^{i}}Y^{i}, with Y = 1 as a simple root in M̄. Therefore, by Hensel Lemma [E72, Corollary 16.6(iv)], there is a b ∈ V_{M} (where V_{M} is the valuation ring of the extension of the valuation of E to M) such that b̄ = 1 and ∑_{i=0}^{n} a_{i}(eb)^{i} = 0. So, eb is a root of f and ẽb = ẽ = α. ii) If μ ≠ 0, then setting c = a_{k}e^{k}, where a_{k} is a nonzero coefficient of f, Y = ẽ^{−1}X and h = c̃^{−1}g, we get h = ∑_{i=0}^{n} \overline{c^{−1}a_{i}e^{i}}Y^{i}. So again by Hensel Lemma, there is b ∈ V_{M} such that c^{−1} ∑_{i=0}^{n} a_{i}(eb)^{i} = 0 and b̄ = 1. Hence, eb is a root of f with ẽb = α. Now to prove that f is irreducible, remark that because g = βf_{α,F rac(GE)}, then [GE[ã] : GE] = deg(g) = deg(f). Moreover, since f(a) = 0, then the minimal polynomial f_{a,E} of a over E divides f, hence [E[a] : E] ≤ deg(f). On the other hand, we have [E[a] : E] ≥ [G(E[a]) : GE] ≥ [GE[ã] : GE]. Therefore, [E[a] : E] = [G(E[a]) : GE] = [GE[ã] : GE] = deg(g) = deg(f). So, f is irreducible. (The reader can see that E[a] is necessaraly defectless over E). α being a simple root of g, then g has deg(g) distinct roots. Let N be a normal field extension of E which contains M. By Theorem 1.4, GN is a normal graded field extension of GE. Hence all the roots of g are in GN. Moreover, by the previous part of this proof, for any root δ of g there is a root d of f in N such that d̃ = δ. Therefore, f has deg(f) distinct roots. This shows that a is a simple root of f.

Remark 1.6 Let E be a Henselian valued field, g = ∑_{i=0}^{n} ã_{i}X^{i} a homogenizable irreducible polynomial of GE[X], with a_{i} ∈ E, and f = ∑_{i=0}^{n} a_{i}X^{i}. If f has a root a such that ã is a root of g, then using the same arguments as in Proposition 1.5, f is irreducible in E[X] and G(E[a]) = GE[ã].

Proposition 1. 7 Let E be a Henselian valued field with residue characteristic p > 0 and L a purely wild [resp., simple purely wild] finite-dimensional graded field extension of GE, then there is a defectless field extension [resp., simple field extension] K of E such that GK = L. If char(E) = p, then K can be chosen to be a purely inseparable field extension of E.

Proof. Let N be a normal field extension of E such that L ⊆ GN. Assume first that L = GE[ã] (for some a ∈ N) and let p^{n} = [L : GE]. Since F rac(L) is purely inseparable over F rac(GE), then the minimal polynomial of ã over F rac(GE) is g := X^{p^{n}} − ã^{p^{n}}. We have L ∩ F rac(GE) ⊆ GN ∩ F rac(GE) = GE. So, there is b ∈ E such that ã^{p^{n}} = b̃. Let f = X^{p^{n}} − b and let x be a root of f in N. Clearly, x̃ is a root of g. So, x̃ = ã (because x̃^{p^{n}} = ã^{p^{n}} in GN). By remark 1.6, we have G(E[x]) = GE[x̃] = GE[ã] = L. Now, let L be an arbitrary finite-dimensional purely wild graded field extension of GE. Then, we can write L = GE[ã_{1}, ..., ã_{r}]. Therefore, one can easily proceed by induction on r to end the proof.

As a consequence of Proposition 1.5, we have the following Corollary that gives explicitly the correspondence between (finite-dimensional) tame valued field extensions over a Henselian valued field and tame graded field extensions.

Corollary 1. 8 [HW(1)99, Theorem 5.2] Let E be a Henselian valued field. Then, the map K ↦ GK gives a one-to-one correspondence between the set of isomorphism classes of finite-dimensional tame field extensions of E and the set of isomorphism classes of finite-dimensional tame graded field extensions of GE. Moreover, K is a Galois tame (finite-dimensional) field extension of E if and only if GK is a Galois (finite-dimensional) graded field extension of GE and in such a case Gal(K/E) is isomorphic to Gal(GK/GE).

Proof. If K is a tame field extension of E, then obviously GK is a tame graded field extension of GE. Let K′ be a tame field extension of E such that K′ ≅ K. Since E is Henselian, then clearly GK ≅_{g} GK′. Conversely, if L is a tame finite-dimensional graded field extension of GE, then we can write L = GE[x̃_{1}, ..., x̃_{r}], where x_{i} ∈ E_{alg} with x̃_{i} separable over F rac(GE). Assume first that r = 1 and let M = E[x_{1}]. By Proposition 1.5, there is a_{1} ∈ M such that ã_{1} = x̃_{1} and [E[a_{1}] : E] = [L : GE]. Hence, G(E[a_{1}]) = L. By induction, assume there are a_{1}, ..., a_{r−1} ∈ E[x_{1}, ..., x_{r−1}] such that G(E[a_{1}, ..., a_{r−1}]) = GE[x̃_{1}, ..., x̃_{r−1}]. Then, again by Proposition 1.5, there is a_{r} ∈ E[x_{1}, ..., x_{r}] such that G(E[a_{1}, ..., a_{r}]) = L (this is obtained by applying Proposition 1.5 on the field extension M/E′, where E′ = E[a_{1}, ..., a_{r−1}]). Clearly, K := E[a_{1}, ..., a_{r}] is tame over E. If K′ is an other tame field extension of E such that GK′ ≅_{g} L, then by Proposition 1.5, there is a′_{1} ∈ K′ such that G(E[a′_{1}]) ≅_{g} G(E[a_{1}]) and both a′_{1} and a_{1} have the same minimal polynomial over E. So, E[a′_{1}] ≅ E[a_{1}]. By induction, we prove that K′ ≅ K.

Now, let K be a Galois tame finite-dimensional field extension of E. By Proposition 1.2 and Theorem 1.4, GK is a Galois graded field extension of GE. Since v is Henselian, then for any σ ∈ Gal(K/E), we can define a graded GE-automorphism σ̃ : GK → GK by extending the mapping x̃ ↦ σ̃(x). Let ϕ : G → Gal(GK/GE) be the group homomorphism defined by ϕ(σ) = σ̃, B be the valuation ring of the extension of v to K and let G^{v} the ramification group of B over E. See that ϕ(σ) = id_{GK} if and only if v(\frac{σ(x)}{x} − 1) > 0, for any x ∈ K^{∗}. Otherwise said, ϕ(σ) = id_{GK} if and only if σ ∈ G^{v}. But G^{v} = {id_{K}} (since K is tame over E). Therefore, ϕ is injective. Remark that Gal(K/E) and Gal(GK/GE) have the same cardinality (since Gal(GK/GE) ≅ Gal(F rac(GE)/F rac(GE))). So, ϕ is a group isomorphism. Let M be a tame finite-dimensional field extension of E such that GM is a Galois graded field extension of GE and consider a Galois tame finite-dimensional field extension N of E containing M. By the above GN is a Galois graded field extension of GE [resp., of GM] and Gal(N/E) ≅ Gal(GN/GE) [resp., Gal(N/M) ≅ Gal(GN/GM)]. Since GM is a Galois graded field extension of GE, then Gal(GN/GM) is a normal subgroup of Gal(GN/GE), therefore Gal(N/M) is a normal subgroup of Gal(N/E). Hence, M is a Galois field extension of E.

Let F be a graded field and K a finite-dimensional graded field extension of F. For an arbitrary abelian group A -namely for A = Γ_{K}/Γ_{F} - and a family a_{1}, a_{2}..., a_{r} of elements of A, we say that a_{1}, a_{2}..., a_{r} are independent if the subgroup ⟨a_{1}, a_{2}, .., a_{r}⟩ of A, generated by a_{1}, a_{2}..., a_{r}, equals ⊕_{i=1}^{r}⟨a_{i}⟩. We recall that K is called totally ramified of radical type (T RRT) over F if there are homogeneous elements t_{1}, ..., t_{r} ∈ F^{∗} and nonnegative integers n_{1}, ..., n_{r} such that the following conditions are satisfied:

(1) K = F [t_{1}^{1/n_{1}}, ..., t_{r}^{1/n_{r}}] and (2) gr(t_{i}^{1/n_{i}}) + Γ_{F} (1 ≤ i ≤ r) are independent elements of Γ_{K}/Γ_{F}, with order n_{i}, respectively.

One can see that in the same way as for T RRT valued field extensions (See [JW90, Lemma 4.1]), a totally ramified finite-dimensional graded field extension K of F is T RRT over F if and only if there is a subgroup G of K^{∗}/F^{∗} such that the mapping G → Γ_{K}/Γ_{F}, defined by xF^{∗} ↦ gr(x) + Γ_{F}, is a group isomorphism.

Lemma 1. 9 Let F be a graded field and K a totally ramified finite-dimensional graded field extension of F. Then, K is T RRT over F.

Proof. [M05, Lemma 1].

Corollary 1. 10 [Sch50, p.64, Theorem 3] Let E be a Henselian valued field and K a tame totally ramified finite-dimensional field extension of E, then K is totally ramified of radical type.

Proof. By Lemma 1.9 GK is totally ramified of radical type. Write GK = GE[α_{1}] ⊗_{GE} ... ⊗_{GE} GE[α_{r}] for some α_{1}, ..., α_{r} ∈ GK^{∗} where gr(α_{i}) + Γ_{GE} are independent elements of Γ_{GK}/Γ_{GE}. Assume first that r = 1. By Proposition 1.5 there is x_{1} ∈ K such that x̃_{1} = α_{1} and [E[x_{1}] : E] = [GE[α_{1}] : GE] = [GK : GE]. Hence, E[x_{1}] = K and v(E[x_{1}]) = ⟨gr(α_{1}) + Γ_{GE}⟩ = ⟨v(x_{1}) + Γ_{E}⟩. More generally, for an arbitrary positive integer r, by the above there are x_{i} ∈ K such that [E[x_{i}] : E] = [GE[α_{i}] : GE] and v(x_{i}) + Γ_{E} generates Γ_{E[x_{i}]}/Γ_{E}. Hence, by [Mor89, Theorem 1] K = E[x_{1}] ⊗_{E} ... ⊗_{E} E[x_{r}].

Let E/F be a Galois finite-dimensional graded field extension. In the same way as for ungraded fields, one may define the norm N_{E/F} of E/F (i.e. N_{E/F} = ∏_{σ∈Gal(E/F)} σ(x) for all x ∈ E). Then, the following lemma is a direct consequence of [L84, Theorem 6.1, p323].

Lemma 1. 11 Let E/F be a finite-dimensional cyclic graded field extension with dimension n and with Galois group generated by σ and x ∈ E^{∗}. Then, N_{E/F} (x) = 1 if and only if there exists y ∈ E^{∗} such that x = yσ(y)^{−1}

Proof. Assume that N_{E/F} (x) = 1. Then, N_{F rac(E)/F rac(F)}(x)(= N_{E/F} (x)) = 1. So, by [L84, Theorem 6.1, p. 323] there is z ∈ F rac(E)^{∗} such that x = zσ(z)^{−1}. We may assume z ∈ E^{∗}. Write z = z_{1} +...+z_{r}, where z_{i} (1 ≤ i ≤ r) are nonzero homogeneous elements of E with gr(z_{i}) < gr(z_{i+1}), for all 1 ≤ i < r. Since σ(z)x = z and x is homogeneous, then for all 1 ≤ i ≤ r, σ(z_{i})x = z_{i}. The converse is clear.

Remark that an alternative cohomological proof of Lemma 1.11 can be obtained by considering the G-module E^{∗}.

Proposition 1. 12 Let F be a graded field with characteristic not dividing a positive integer n, and assume F_{0} contains a primitive n^{th} root of unity ζ. Then, the following statements hold :

(1) If E is a cyclic graded field extension of F with dimension n, then there is x ∈ E^{∗} such that E = F [x], x^{n} ∈ F^{∗} and Gal(E/F) is generated by the graded F -isomorphism σ defined by σ(x) = ζx.

(2) Conversely, if a ∈ F^{∗} and x is a root of the polynomial X^{n} − a in F_{alg}, then F [x] is a cyclic graded field extension of F with [F [x] : F] = m dividing n and x^{m} ∈ F.

Proof. (1) Let σ a generator of Gal(E/F). We have N_{E/F} (ζ^{−1}) = 1, so by Lemma 1.11 there is x ∈ E^{∗} such that σ(x) = ζx. Accordingly, σ(x^{n}) = σ(x)^{n} = (ζx)^{n} = x^{n}. Hence, x^{n} ∈ F^{∗}. Since σ^{i}(x)(= ζ^{i}x) (1 ≤ i ≤ n) are pairwise distinct, then the minimal polynomial of x over F rac(F) is X^{n} − x^{n}. So, F [x] = E.

(2) By [L84, Theorem 6.2, p. 324] F rac(F)(x)/F rac(F) is cyclic with dimension m dividing n and with x^{m} ∈ F rac(F)^{∗}. Hence, F [x]/F is cyclic of dimension m and by [HW(1)99, Corollary 2.5(b)] x^{m} ∈ F [x]^{∗} ∩ F rac(F)^{∗} = F^{∗}.

Let F be a graded field with characteristic p > 0. Then, Galois graded p-extensions of F are inertial over F, so they are exactly graded fields of the form KF F, where K runs over Galois p-extensions of F_{0}. This because Galois graded p-extensions are necessarily inertial. Therefore, a graded field extension E/F of dimension a power of p is cyclic if E = F (x_{1}, ..., x_{n}), where x = (x_{1}, ..., x_{n}) ∈ W_{n}(E_{0}) and (x_{1}^{p}, ..., x_{n}^{p}) − (x_{1}, ..., x_{n}) ∈ W_{n}(F_{0}) (here W_{n}(E_{0}) is the ring of Witt vectors associated to the field E_{0}). In particular, cyclic extensions of degree p of F are F [x], where x is a root of a polynomial X^{p} − X − a for some a ∈ F_{0} with x ∉ F_{0}.

Let F be a graded field and K a finite-dimensional abelian graded field extension of F. We say that K is a Kummer graded field extension of F if F_{0} contains a primitive m^{th} root of unity, where m is the exponent of Gal(K/F). In such a case, we denote KUM(K/F) = {x ∈ K^{∗} | x^{m} ∈ F } and kum(K/F) = KUM(K/F)/F^{∗} (i.e., the quotient group). As for Kummer field extensions, one can see that kum(K/F) is isomorphic to Gal(K/F).

Let F be a graded field of characteristic not dividing a positive integer n. Readily, F_{0} contains a primitive n^{th} root of unity if and only if F rac(F) does (indeed, if ξ is a primitive n^{th} root of unity in F rac(F), then clearly F [ξ] is a graded field extension of F. So, [F [ξ] : F] = [F rac(F [ξ]) : F rac(F)] = 1. Hence ξ ∈ F_{0}). Therefore, for any finite-dimensional abelian graded field extension K of F, K is a Kummer graded field extension of F if and only if F rac(K) [resp. HF rac(K)] is a Kummer field extension of F rac(F) [resp. HF rac(F)]. Moreover, if K is a Kummer graded field extension of F, then kum(K/F) embeds canonically in kum(F rac(K)/F rac(F)) [resp. in kum(HF rac(K)/HF rac(F))]. Since both groups have the same cardinality, then -up to a group isomorphism- kum(F rac(K)/F rac(F)) = kum(K/F) [resp. kum(HF rac(K)/HF rac(F)) = kum(K/F)]. Obviously, this isomorphism can be deduced easily from the fact that Gal(K/F) ≅_{g} Gal(F rac(K)/F rac(F)) [resp., Gal(K/F) ≅_{g} Gal(HF rac(K)/HF rac(F))], but the above shows that every class of kum(F rac(K)/F rac(F)) [resp. kum(HF rac(K)/HF rac(F))] can be represented by an element of KUM(K/F).

Proposition 1. 13 Let F be a graded field and K a totally ramified graded field extension of F. Let m be the exponent of Γ_{K}/Γ_{F}. Then, the following statements are equivalent :

(1) K is a Galois graded field extension of F.

(2) K is a Kummer graded field extension of F.

(3) F_{0} contains a primitive m^{th} root of unity. If K satisfies these conditions, then Gal(K/F) ≅ kum(K/F) ≅ Γ_{K}/Γ_{F}.

Proof. By [HW(1)99, Proposition 3.3].

2 NICELY SEMIRAMIFIED DIVISION ALGEBRAS OVER HENSELIAN VALUED FIELDS22

2 Nicely semiramified division algebras over Henselian valued fields

Let F be a graded field and D a graded central division algebra over F. We recall that in the same way as for valued central division algebras, D is called nicely semiramified (NSR) (over F) if it has an inertial and a totally ramified (of radical type) maximal graded subfields. We recall also that D is NSR (over F) if and only if Cq(D) is NSR (over F rac(F)) ([B98, Proposition 6.4]). Moreover, Cq(D) is NSR (over F rac(F)) if and only if HCq(D) is NSR (over HF rac(F)). Indeed, assume that Cq(D) is NSR and let K [resp. L] be an inertial [resp. a T RRT] maximal subfield of Cq(D). Then, by [Mor89, Theorem 1], HK(= K ⊗_{F rac(F)} HF rac(F)) [resp. HL(= L ⊗_{F rac(F)} HF rac(F))] is an inertial [resp. a T RRT] maximal subfield of HCq(D). Conversely, if HCq(D) is NSR, then it has an inertial [resp. a T RRT] maximal subfield K′ [resp. L′]. So GK′ [resp. GL′] is an inertial [resp. a totally ramified] maximal graded subfield of D(≅_{g} GHCq(D)). Hence D is NSR. Therefore, by the above, Cq(D) is NSR.

The following lemma is the analogue of [JW90, Theorem 4.4]. It gives equivalent statements for a graded central division algebra over a graded field to be NSR. In condition 3(i) of this lemma, the graded field extensions L^{(i)} are said to be linearly disjoint if L^{(1)} ⊗_{F} .... ⊗_{F} L^{(k)} is a graded field.

Lemma 2. 1 Let F be a graded field and D a graded central division algebra of degree n over F. Then the following statements are equivalent :

(1) D is NSR.

(2) D is split by an inertial and a totally ramified graded field extensions of F.

(3) D ≅_{g} (L^{(1)}/F, σ_{1}, t_{1}) ⊗_{F} ... ⊗_{F} (L^{(k)}/F, σ_{k}, t_{k}), where L^{(i)}, σ_{i} and t_{i} satisfy the following conditions :

(i) L^{(i)} are linearly disjoint cyclic inertial graded field extensions of F with dimension [L^{(i)} : F] = n_{i} and with Galois group generated by σ_{i} (1 ≤ i ≤ k), ∏_{i=1}^{k} n_{i} = n,

(ii) t_{i} are nonzero homogeneous elements of F such that gr(t_{i}^{n/n_{i}}) + nΓ_{F} are independent elements of Γ_{F} /nΓ_{F}, with order n_{i} (1 ≤ i ≤ k), respectively.

In such a case, Γ_{D}/Γ_{F} = ⊕_{i=1}^{k}⟨gr(t_{i}) + Γ_{F} ⟩.

Proof. [M05, Lemma 2].

Lemma 2. 2 Let E be a Henselian valued field and D a defectless central division algebra over E. Then the following statements are equivalent :

(1) D is NSR over E.

(2) GD is NSR over GE.

Proof. [M05, Lemma 3].

Theorem 2. 3 Let E be a Henselian valued field and D a defectless central division algebra over E. Then the following statements are equivalent:

(1) D is NSR.

(2) D has an inertial and a totally ramified maximal subfields.

(3) D is split by an inertial and by a totally ramified field extensions of E.

Proof. [M05, Theorem 4].

Corollary 2. 4 Let E be a Henselian valued field and D a tame semiramified division algebra over E, then the following statements are equivalent.

(1) D is nicely semiramified.

(2) D has a totally ramified maximal subfield.

(3) D is split by a totally ramified field extension of E.

Proof. Since D is tame semiramified, then D̄ is an abelian field extension of Ē. So, D has an inertial maximal subfield (the inertial lift of D̄ over E in D). Therefore, our Lemma follows by Theorem 2.3.

Proposition 2. 5 Let E be a Henselian valued field of characteristic p > 0 and D a tame semiramified division p-algebra over E. Then, the following statements are equivalent.

(1) D is nicely semiramified.

(2) D is a tensor product of cyclic algebras.

(3) D has a purely inseparable maximal subfield.

Proof. (1) ⇒ (2) This follows by [JW90, Theorem 4.4].

(2) ⇒ (3) Assume that D = D_{1} ⊗_{E} ... ⊗_{E} D_{r}, where D_{i} are cyclic algebras. By [A61, Theorem 26, p. 107], each D_{i} contains a (simple) purely inseparable maximal subfield K_{i}. So, K := K_{1} ⊗_{E} ... ⊗_{E} K_{r} is a purely inseparable maximal subfield of D.

(3) ⇒ (1) Let K be a purely inseparable maximal subfield of D. Since D is defectless over E and D̄ is separable over Ē, then K is totally ramified over E. So, D is nicely semiramified (by Corollary 2.4).

Proposition 2. 6 Let E be a Henselian valued field and D an inertially split defectless central division algebra over E. Then the following statements hold :

(1) D is semiramified if and only if D̄ is a field.

(2) Let L be a field extension of E in D and let C be the centralizer of L in D. If D is semiramified, then C is also semiramified. Furthermore, if L is inertial over E and D is nicely semiramified, then C is nicely semiramified.

(3) Suppose D is semiramified and D = D_{1} ⊗_{E} D_{2} with D_{1} and D_{2} inertially split. Then, each D_{i} is semiramified, D̄ = D̄_{1} ⊗Ē D̄_{2}, and Γ_{D}/Γ_{E} = Γ_{D_{1}}/Γ_{E} × Γ_{D_{2}}/Γ_{E} (via the canonical inclusion Γ_{D_{i}}/Γ_{E} → Γ_{D}/Γ_{E}). Furthermore, if D_{1} and D_{2} are nicely semiramified, then D is nicely semiramified.

(4) If D is semiramified with a cyclic inertial maximal subfield, then D is nicely semiramified and Γ_{D}/Γ_{E} is cyclic.

Proof. (1) This follows easily from the equality (Γ_{D} : Γ_{E}) = [Z(D̄) : Ē] obtained from the isomorphism θ_{D}, since Z(D̄) is separable over Ē (see [JW90, Lemma 5.1]).

(2) Clearly, C is also inertially split. If D is semiramified, then by (1) C is also semiramified (because C̄ ⊆ D̄). Assume L is inertial over E and D is nicely semiramified, and let K be a totally ramified field extension of E that splits D, then by [Mor89, Theorem 1], K ⊗_{E} L is a totally ramified field extension of L that splits C, therefore by Corollary 2.4 C is nicely semiramified.

(3) By (1) D_{i} are semiramified. Let L_{i} be the inertial lift of the field D̄_{i} over E in D_{i} -see [JW90, Theorem 2.8]-, and let L = L_{1} ⊗_{E} L_{2}, which is a subfield of D. Clearly, L is inertial over E, since it is a compositum of two inertial extensions of E. Hence, L̄ = D̄_{1} ⊗Ē D̄_{2}, which by dimension count must be all of D̄. Now, Γ_{D_{1}}/Γ_{E} acts trivially on D̄_{2} via θ_{D}, as D_{1} centralizes D_{2}; likewise Γ_{D_{2}}/Γ_{E} acts trivially on D̄_{1}. Hence, (Γ_{D_{1}}/Γ_{E}) ∩ (Γ_{D_{2}}/Γ_{E}) must be trivial, since it acts trivially on all of D̄_{1} ⊗Ē D̄_{2} = D̄ and θ_{D} is injective. Hence, (Γ_{D_{1}}/Γ_{E}) × (Γ_{D_{2}}/Γ_{E}) ⊆ Γ_{D}/Γ_{E}, and since these groups have the same cardinality, they must be equal. Suppose that D_{1} and D_{2} are nicely semiramified, and let K_{i} be a totally ramified maximal subfield of D_{i}, then by [Mor89, Theorem 1] K_{1} ⊗_{E} K_{2} is a totally ramified maximal subfield of D. So, D is nicely semiramified.

(4) Let K be a cyclic inertial maximal subfield of D, then K̄(= D̄) is a cyclic field extension of Ē. So, Γ_{D}/Γ_{E} is cyclic (since θ_{D} is an isomorphism). Let γ + Γ_{E} be a generator of Γ_{D}/Γ_{E} and let x be an element of D with valuation γ. Then, E[x] is a totally ramified maximal subfield of D. Hence, D is nicely semiramified.

Corollary 2. 7 Let E be a Henselian valued field of residue characteristic p > 0 and D a tame central division algebra of degree a power of p over E. Then, D is semiramified if and only if D̄ is a field.

Proof. Let K be a tame maximal subfield of D. Since D has degree a power of p, then K is inertial over E. Therefore, our corollary follows by Proposition 2.6.

Remark 2.8 One can easily see that there are graded algebra analogues of Corollary 2.4, Proposition 2.5, Proposition 2.6 and Corollary 2.7. Indeed, let F be a graded field and D a graded central division algebra over F. Then D is semiramified if and only if HCq(D) is semiramified, and D is nicely semiramified if and only if HCq(D) is nicely semiramified.

Let A be a tame valued division algebra over a Henselian valued field E. We will say that A is t-indecomposable, if for any tame division algebras B and C over E such that A ≅ B ⊗_{E} C, necessarily B or C is trivial.

Proposition 2. 9 Let F be a graded field of characteristic p > 0, D a semiramified graded division p-algebra over F and K an immediate valued field extension of F rac(F) for its canonical valuation (i.e., such that K̄ = \overline{F rac(F)} and Γ_{K} = Γ_{F rac(F)}). Then, the following statements are equivalent :

(1) D is cyclic.

(2) D_{0} is a cyclic field extension of F_{0}.

(3) Γ_{D}/Γ_{F} is a cyclic group.

(4) D has a simple purely wild maximal graded subfield.

(5) D is indecomposable and nicely semiramified.

(6) D is nicely semiramified and exp(D) = deg(D).

(7) Cq(D)⊗_{F rac(F)}K is nicely semiramified and exp(Cq(D)⊗_{F rac(F)}K) = deg(Cq(D)⊗_{F rac(F)} K).

(8) Cq(D) ⊗_{F rac(F)} K is cyclic and indecomposable. Furthermore, if HF rac(F) ⊆ K, then the above statements are equivalent to the following condition :

(9) Cq(D) ⊗_{F rac(F)} K is cyclic and t-indecomposable.

Proof. (1) ⇒ (2) Assume that D has a cyclic maximal graded subfield K. By [HW(1)99, Theorem 3.11] K is tame (hence inertial) over F. So, by [HW(1)99, Remark 3.1] D_{0}(= K_{0}) is a cyclic field extension of F_{0}.

(2) ⇔ (3) By [B98, Proposition 6.1].

(3) ⇒ (4) Let x ∈ D^{∗} such that gr(x) + Γ_{F} generates Γ_{D}/Γ_{F}. Then, F [x] is a simple totally ramified -hence a purely wild- maximal graded subfield of D.

(4) ⇒ (6) Clearly D_{0}F is an inertial maximal graded subfield of D. Let L be a simple purely wild maximal graded subfield of D. As D_{0} is separable over F_{0}, then L is totally ramified over F. So, D is nicely semiramified, and Γ_{D}/Γ_{F} (= Γ_{L}/Γ_{F}) is cyclic (because L is simple over F). Therefore, by [B98, Proposition 6.5] exp(D) = exp(Γ_{D}/Γ_{F}) = (Γ_{D} : Γ_{F}) = deg(D).

(6) ⇒ (5) Clear.

(5) ⇒ (1) By Lemma 2.1.

(6) ⇒ (7) We have G(Cq(D) ⊗_{F rac(F)} K) ≅_{g} D. Indeed, let HK be the henselization of K with respect to the (considered) valuation of K, then by [HW(1)99, Corollary 5.7] the following diagram is commutative :

T Br(HF rac(F)) →^{Ext} T Br(HK)
≅↓  ≅↓
GBr(F)  →^{Ext} GBr(GHK)

(where the horizontal maps are the scalar extension homomorphisms and the vertical ones are the canonical group isomorphisms seen in the Preliminaries). Since K is immediate over F rac(F), then GK ≅_{g} GHK ≅_{g} F. Moreover, we have G(Cq(D) ⊗_{F rac(F)} K) = G(Cq(D) ⊗_{F rac(F)} HK). Hence, G(Cq(D) ⊗_{F rac(F)} K) ≅_{g} D. Therefore, Cq(D) ⊗_{F rac(F)} K is nicely semiramified and deg(Cq(D) ⊗_{F rac(F)} K) = deg(D) = exp(D) = exp(Cq(D) ⊗_{F rac(F)} K).

(7) ⇒ (8) Obviously, Cq(D) ⊗_{F rac(F)} K is indecomposable, so D is indecomposable. Moreover, as seen above, we have D ≅_{g} G(Cq(D)⊗_{F rac(F)} HK). So, D is nicely semiramified. Hence, by Lemma 2.1, D (so Cq(D)) is cyclic. Therefore, Cq(D)⊗_{F rac(F)} K is cyclic. Now, if HF rac(F) ⊆ K, then (8) ⇒ (9) is evident.

(9) ⇒ (1) Since Cq(D)⊗_{F rac(F)}K is t-indecomposable, GK ≅_{g} F and G(Cq(D)⊗_{F rac(F)} K) ≅_{g} D, then D is indecomposable (it suffices to use the canonical group isomorphism T Br(K) → GBr(GK)). Moreover, by [A61, Theorem 26, p. 107] Cq(D)⊗_{F rac(F)} K has a purely inseparable maximal subfield L. See that Cq(D) ⊗_{F rac(F)} K is defectless over K and \overline{Cq(D) ⊗_{F rac(F)} K}(≅ D_{0}) is separable over K̄(≅ F_{0}), so L is totally ramified over K. Therefore, Cq(D) ⊗_{F rac(F)} K (hence D) is nicely semiramified. So, by Lemma 2.1 D is cyclic.

Remark 2.10 More generally, if F is a graded field of characteristic p > 0 and D is a graded central division p-algebra over F, then in the same way as in [A61, Theorem 26, p. 107], one can show that D is cyclic if and only if D contains a simple purely wild maximal graded subfield (see Theorem C.6 comments below).

Let p be a prime positive integer and G a finite abelian p-group. Then, we can write G = ⟨σ_{1}⟩ ⊕ ... ⊕ ⟨σ_{r}⟩, where ⟨σ_{i}⟩ is the cyclic group generated by σ_{i}. We recall that the number r is called the rank of G. We write r = rk(G).

(2.11) Now, let E be a Henselian valued field, D a nicely semiramified division algebra of prime power degree over E and L a totally ramified of radical type maximal subfield of D. Write L = E[t_{1}^{1/n_{1}}, ..., t_{r}^{1/n_{r}}], where v(t_{i}^{1/n_{i}}) + Γ_{E} (1 ≤ i ≤ r) are independent elements of Γ_{L}/Γ_{E}, with order n_{i}, respectively. Then, the integer r depends only on D (because ⊕_{i=1}^{r}⟨v(t_{i}^{1/n_{i}}) + Γ_{E}⟩ = Γ_{L}/Γ_{E} = Γ_{D}/Γ_{E}). We call r the radical length of D and we write r = rl(D). By definition, we have rl(D) = rk(Γ_{D}/Γ_{E}).

In the same way, for a graded field F and a nicely semiramified graded division algebra D over F, we define rl(D) = rk(Γ_{D}/Γ_{F}). Obviously, we have rl(D) = rl(HCq(D)).

3 NONDEGENERATE SEMIRAMIFIED VALUED AND GRADED DIVISION ALGEBRAS 30

Corollary 2. 12 Let F be a graded field of characteristic p > 0 and D a semiramified graded division p-algebra over F. Then, D is nicely semiramified with radical length r if and only if D is a tensor product of r cyclic graded division algebras.

Proof. If D = D_{1} ⊗_{F} ... ⊗_{F} D_{r}, where D_{i} are cyclic graded division algebras, then by Proposition 2.5, HCq(D) is nicely semiramified. So, D is nicely semiramified. Moreover, by the graded version of Proposition 2.6(3) and Proposition 2.9, each D_{i} is semiramified (since D_{i} are split by inertial graded field extensions of F) ; Γ_{D}/Γ_{F} = Γ_{D_{1}}/Γ_{F} ⊕ ... ⊕ Γ_{D_{r}}/Γ_{F} and Γ_{D_{i}}/Γ_{F} are cyclic. So, rl(D) = r. The converse follows by Lemma 2.1.

3 Nondegenerate semiramified valued and graded division algebras

Let M/E be a field extension and D a central division algebra over E. We denote by D_{M} the central division algebra over M similar to D ⊗_{E} M with respect to Br(M). We recall that if M is a subfield of D, then up to an algebra isomorphism, we have D_{M} = C_{D}^{M}.

Definition 3.1 Let E be a Henselian valued field and D a central division algebra of prime power degree over E. We say that D is degenerate if there is an inertial field extension M of E in D such that D_{M} is nicely semiramified (over M) with radical length r ≥ 2.

Proposition 3. 2 Let E be a Henselian valued field and D an inertially split central division algebra of prime power degree over E. Then, D is degenerate if and only if there is a subfield K of D such that Γ_{K}/Γ_{E} is non-cyclic.

Proof. Assume D is degenerate and let M be an inertial subfield of D such that D_{M} is nicely semiramified with rl(D_{M}) ≥ 2. Let K be a totally ramified maximal subfield of C_{D}^{M}, then Γ_{C_{D}^{M}} /Γ_{M} = Γ_{K}/Γ_{E}. So Γ_{K}/Γ_{E} is non-cyclic. Conversely, suppose that there is a subfield K of D such that Γ_{K}/Γ_{E} is non-cyclic. Let L be a maximal subfield of D that contains K and let M be ’the’ inertial lift of L̄ over E in L, then L is a totally ramified maximal subfield of the inertially split division algebra C_{D}^{M}. So, C_{D}^{M} is nicely semiramified with rl(C_{D}^{M}) ≥ 2.

Corollary 3. 3 Let E be a Henselian valued field such that Ē is finite. Then, any inertially split division algebra D of prime power degree over E is nondegenerate cyclic nicely semiramified (over E) with exp(D) = deg(D).

Proof. Since Ē is finite, then D̄ is a cyclic field extension of Ē. Moreover, since D is inertially split over E, then Γ_{D}/Γ_{E} is isomorphic to Gal(D̄/Ē). So, D is semiramified and by Proposition 3.2, D is nondegenerate. Let v be the extension of the valuation of E to D and let x ∈ D such that v(x)+Γ_{E} generates Γ_{D}/Γ_{E}. Then, E[x] is a totally ramified maximal subfield of D. Hence, by Corollary 2.4 D is nicely semiramified. Therefore, D is cyclic (because it is nondegenerate and nicely semiramified). Moreover, by [JW90, Lemma 5.15], we have exp(D) = exp(Γ_{D}/Γ_{E}) = (Γ_{D} : Γ_{E}) = deg(D).

Remark : By [P82, Proposition 17.6] any local field E is Henselian. Therefore, Corollary 3.3 can be considered as a generalization of [P82, Corollary 17.8(b) and Corollary 17.10(b)]. Henselian valued fields with finite residue fields are called generalized local fields. The reader can see the thesis of F.H. Chang [Ch04] for more results concerning division algebras over these fields.

Corollary 3. 4 Let E be a Henselian valued field of characteristic p > 0 and D a nondegenerate tame semiramified division p-algebra over E. Then, the following statements are equivalent :

(1) D is cyclic.

(2) D is a tensor product of cyclic algebras.

(3) D is nicely semiramified.

(4) Γ_{D}/Γ_{E} is cyclic. In such a case, we have exp(D) = deg(D) (hence D is indecomposable) and all purely inseparable maximal subfields of D are simple (i.e. of the form E[x] for some x ∈ D^{∗})

Proof. (1) ⇒ (2) Clear.

(2) ⇒ (3) By Proposition 2.5.

(3) ⇒ (4) Obvious (since D is nondegenerate).

(4) ⇒ (1) Since D is tame semiramified, then D̄ is a Galois field extension of Ē with Gal(D̄/Ē) isomorphic to Γ_{D}/Γ_{E}. Therefore, the inertial lift of D̄ over E in D is a cyclic maximal subfield of D. Assume these conditions hold, then by [JW90, Lemma 5.15] exp(D) = exp(Γ_{D}/Γ_{E}) = (Γ_{D} : Γ_{E}) = deg(D). Moreover, in this case, if K is a purely inseparable maximal subfield of D, then K is totally ramified over E (because D is defectless over E and D̄ is separable over Ē). D being nondegenerate, then by Proposition 3.2 Γ_{K}/Γ_{E} is cyclic. Let x ∈ K^{∗} with v(x) + Γ_{E} generating Γ_{K}/Γ_{E}, then K = E[x].

Corollary 3. 5 Let E be a Henselian valued field of characteristic p > 0 and D a nondegenerate tame semiramified division p-algebra over E. Then, the following statements are equivalent :

(1) D is cyclic.

(2) GD is cyclic.

(3) D is nicely semiramified.

Proof. (1) ⇔ (2) This follows by Corollary 3.4 and Proposition 2.9.

(1) ⇔ (3) By Corollary 3.4.

Proposition 3. 6 Let E be a Henselian valued field and D a tame semiramified division algebra of prime power degree over E. Then, the following statements are equivalent.

(1) D is degenerate.

(2) There is an inertial subfield M of D such D_{M} is the tensor product of two (nontrivial) nicely semiramified division algebras.

(3) There is an inertial subfield M of D such D_{M} is the tensor product of two (nontrivial) cyclic nicely semiramified division algebras.

(4) There is an inertial subfield M of D such D_{M} is the tensor product of two (nontrivial) cyclic nicely semiramified division algebras D_{1} and D_{2} with Γ_{D_{i}}/Γ_{M} cyclic.

Proof. (2) ⇒ (1) Assume there is an inertial subfield M of D such D_{M} = D_{1}⊗_{M} D_{2}, where D_{i} are non-trivial nicely semiramified division algebras. Then, by Proposition 2.6(3), Γ_{D_{M}} /Γ_{M} = Γ_{D_{1}}/Γ_{M} ⊕ Γ_{D_{2}}/Γ_{M}. Let L_{i} be a totally ramified maximal subfield of D_{i} (i = 1, 2). Since Γ_{L_{i}}/Γ_{M} = Γ_{D_{i}}/Γ_{M}, then by [Mor89, Theorem 1] L_{1} ⊗_{M} L_{2} is a totally ramified maximal subfield of D_{M}. So, D_{M} is nicely semiramified with rl(D_{M}) ≥ 2.

(1) ⇒ (4) If M is an inertial subfield of D such that D_{M} is nicely semiramified with radical length r ≥ 2, then we can write C_{D}^{M} = (M_{1}/M, σ_{1}, a_{1})⊗_{M} ...⊗_{M} (M_{r}/M, σ_{r}, a_{r}), where M_{i}, σ_{i} and a_{i} satisfy the conditions of [JW90, Theorem 4.4(iii)]. We have C_{D}^{MM_{3}...M_{r}} = C_{C_{D}^{M}}^{M_{3}...M_{r}}, so C_{D}^{MM_{3}...M_{r}} ≅ (M_{1}M_{3}...M_{r}/M_{3}...M_{r}, σ_{1}, a_{1})⊗_{MM_{3}..M_{r}}(M_{2}M_{3}...M_{r}/ M_{3}...M_{r}, σ_{2}, a_{2}). Let D_{i} := (M_{i}M_{3}...M_{r}/M_{3}...M_{r}, σ_{i}, a_{i}) (1 ≤ i ≤ 2), then by [JW90, Theorem 4.4] D_{i} are nicely semiramified and Γ_{D_{i}}/Γ_{M} are cyclic.

(4) ⇒ (3) and (3) ⇒ (2) Obvious.

Remark 3.7 Let E be a Henselian valued field of characteristic p > 0 and D a tame semiramified division p-algebra. Then, Proposition 2.6 and Proposition 3.4, D is degenerate if and only if there is an inertial subfield M of D such that D_{M} is the tensor product of more than two (non-trivial) cyclic tame division algebras.

(3.8) Let E be a Henselian valued field and D a tame division algebra over E, then D is similar to a tensor product I ⊗_{E} N ⊗_{E} T, where I is an inertial central division algebra over E, N is a nicely semiramified division algebra over E and T is a totally ramified central division algebra over E (by [JW90, Lemma 5.15 and Lemma 6.2]). D is semiramified if and only if T is trivial and N̄ splits Ī. Indeed, if D is semiramified, then T will be inertially split, hence trivial. Moreover, in this case, by [JW90, Lemma 5.15] N̄ splits Ī. The converse follows also by [JW90, Lemma 5.15]. In what follows, we aim to determine necessary and sufficient conditions for such tame semiramified division algebra D to be nondegenerate.

Proposition 3. 9 Let E be a Henselian valued field, N a nicely semiramified division algebra over E, I an inertial central division algebra over E and D the central division algebra over E similar to I ⊗_{E} N with respect to Br(E). Assume that D is semiramified with prime power degree. Then, D is nondegenerate if and only if for any subfield K of N̄, N̄/K is cyclic or Ī_{K} ∉ Dec(N̄/K).

Proof. Let M be an inertial subfield of D. By [JW90, Lemma 5.15] we have D̄ ≅ N̄, so M is isomorphic to a subfield of N. Therefore, by Proposition 2.6(2) N_{M} is nicely semiramified. Moreover, since I is inertial, then by [JW90, Lemma 1.8] I_{M} is inertial (over M). Hence, by [JW90, Theorem 5.15] D_{M} is nicely semiramified if and only if ĪM̄ ∈ Dec(N̄/M̄) (see that N̄M̄ = N̄). So, D is nondegenerate if and only if for any subfield K of N̄, N̄/K is cyclic or Ī_{K} ∉ Dec(N̄/K).

(3.10) We recall that for a crossed product A over a field E and a Galois maximal subfield L of A, we say that L is degenerate in A if there is an intermediate field K of L/E such that :

(1) Gal(L/K) is the direct sum of two cyclic subgroups ⟨σ_{1}⟩ and ⟨σ_{2}⟩, and

(2) the centralizer C_{A}^{K} decomposes into a tensor product of two cyclic algebras, split by the subfields K_{1} and K_{2} of L fixed respectively by σ_{2} and σ_{1} (i.e. C_{A}^{K} = (K_{1}/K, σ_{1}, a_{1}) ⊗_{K} (K_{2}/K, σ_{2}, a_{2}) for some a_{1}, a_{2} ∈ K^{∗}) [BM00, Definition 0.12].

Proposition 3. 11 Let E be a Henselian valued field, N a nicely semiramified division algebra of prime power degree over E, I an inertial central division algebra over E such that N̄ is a maximal subfield of Ī, and D the central division algebra over E similar to I ⊗_{E} N with respect to Br(E). Then, D is semiramified. Furthermore, D is degenerate if and only if N̄ is degenerate in Ī.

Proof. Since N̄ is a maximal subfield of Ī, then by [JW90, Lemma 5.15] D is semiramified. Suppose that D is degenerate and let M be an inertial subfield of D such that D_{M} is nicely semiramified with rl(D_{M}) ≥ 2. As previously seen in the proof of Proposition 3.9, I_{M} is inertial ; N_{M} is nicely semiramified and D_{M} is similar to I_{M} ⊗_{M} N_{M}, so again by [JW90, Lemma 1.8 and Lemma 5.15] ĪM̄ ∈ Dec(N̄/M̄). Let r = rk(Gal(N̄/M̄))(= rk(Gal(\overline{D_{M}}/M̄)) = rl(D_{M})), and write N̄ = K_{1} ⊗M̄ ... ⊗M̄ K_{r}, where K_{i}/M̄ are cyclic field extensions (1 ≤ i ≤ r). Since ĪM̄ ∈ Dec(N̄/M̄), then ĪM̄ = (K_{1}/M̄, σ_{1}, a_{1}) ⊗M̄ ... ⊗M̄ (K_{r}/M̄, σ_{r}, a_{r}), where σ_{i} is a generator of the cyclic field extension K_{i}/M̄ and a_{i} ∈ M̄^{∗}. Let L = K_{3}...K_{r}, then Ī_{L} = (K_{1}L/L, σ_{1}, a_{1}) ⊗_{L} (K_{2}L/L, σ_{2}, a_{2}). So, N̄ is degenerate in Ī. Conversely, assume that N̄ is degenerate in Ī and let L be an intermediate field of N̄/Ē such that Ī_{L} = (L_{1}/L, τ_{1}, b_{1}) ⊗_{L} (L_{2}/L, τ_{2}, b_{2}), where L_{1} ⊗_{L} L_{2} = N̄, τ_{i} is a generator of the cyclic field extension L_{i}/L and b_{i} ∈ L^{∗}. Let M be the inertial lift of L over E in D, then ĪM̄ ∈ Dec(N̄/M̄) and rl(D_{M}) = rk(Gal(N̄/M̄)) = 2. So, D is degenerate.

Proposition 3. 12 Let E be a Henselian valued field, D a tame semiramified division algebra of prime power degree over E, and L ’the’ inertial lift of D̄ over E in D. Then, D is degenerate if and only if L is degenerate in D.

Proof. If D is degenerate, then by Proposition 3.6 there is an inertial subfield M of D such that C_{D}^{M} is a tensor product of two cyclic nicely semiramified division algebras D_{1} and D_{2} with Γ_{D_{i}}/Γ_{M} cyclic. Write D_{i} = (M_{i}/M, σ_{i}, a_{i}) (1 ≤ i ≤ 2), where M_{i} are inertial cyclic field extensions of E with Galois group generated by σ_{i}, and a_{i} ∈ M^{∗}. Clearly, this implies that M_{1} ⊗_{M} M_{2} is a degenerate maximal subfield of C_{D}^{M}. Moreover, by comparing the cardinalities, we have M̄_{1} ⊗M̄ M̄_{2} = D̄.

So, M_{1} ⊗_{M} M_{2} is isomorphic to L (because, M̄ = M̄_{1} ⊗M̄ M̄_{2} = D̄), whence L is degenerate in D. Conversely, assume that L is degenerate in D and let K be a subfield of L satisfying the conditions of (3.10). Write C_{D}^{K} = (K_{1}/K, σ_{1}, a_{1}) ⊗_{K} (K_{2}/K, σ_{2}, a_{2}), where K_{1} and K_{2} are the subfields of L fixed by σ_{2} and σ_{1} respectively and a_{1}, a_{2} ∈ K^{∗}, and let D_{i} = (K_{i}/K, σ_{i}, a_{i}) (i = 1, 2). By Proposition 2.6(2 and 3) C_{D}^{K} and D_{i} are semiramified. Moreover, K_{i} is a cyclic inertial maximal subfield of D_{i}, so again by Proposition 2.6(4) D_{i} is nicely semiramified and Γ_{D_{i}}/Γ_{K} is cyclic. Hence, by Proposition 2.6(3) C_{D}^{K} is nicely semiramified with radical length r = 2. Therefore, D is degenerate.

Proposition 3. 13 Let E be a Henselian valued field and D a tame semiramified division algebra of prime power degree over E. If K is a subfield of D with (Γ_{K} : Γ_{E}) maximal, then K is a maximal subfield of D. In particular, if K is totally ramified over E and D is nondegenerate, then D is cyclic.

Proof. By Proposition 2.6(2) C_{D}^{K} is semiramified. Since (Γ_{K} : Γ_{F}) is maximal, then (Γ_{C_{D}^{K}} : Γ_{K}) = 1. Hence, C_{D}^{K} = K. So, K is a maximal subfield of D. If K is totally ramified over E, then by Corollary 2.4 D is nicely semiramified. If, we suppose in addition that D is nondegenerate, then by Corollary 3.5 D is cyclic.

Remark 3.14 Let F be a graded field and D a graded central division algebra over F. As for the valuative case, we say that D is degenerate if there is an inertial graded subfield M of D such that D_{M} is nicely semiramified with rl(D_{M}) ≥ 2. One can then use the same arguments as in Proposition 3.2 to show that for an inertially split graded central division algebra D over F, D is degenerate if and only if there is a graded subfield K of D such that Γ_{K}/Γ_{F} is non-cyclic. One can also see that there are analogous graded versions for all the results we are given in this work.

Lemma 3. 15 Let E be a Henselian valued field and D a tame central division algebra of prime power degree over E. Then, D is degenerate if and only if GD is degenerate.

Proof. Suppose that GD is degenerate and let K be an inertial graded subfield of GD such that GD_{K} is nicely semiramified with rl(GD_{K}) ≥ 2. Let M be the inertial lift of K_{0} over E in D. Then, G(D_{M}) ≅ GD_{K}. So, by Lemma 2.2, D_{M} is nicely semiramified. Moreover, we have rl(D_{M}) = rl(GD_{K}) ≥ 2. Hence, D is degenerate. Conversely, suppose that D is degenerate and let L be an inertial subfield of D such that D_{L} is nicely semiramified with rl(D_{L}) ≥ 2. Then, clearly GD_{GL} is nicely semiramified and rl(GD_{GL}) = rl(D_{L}) ≥ 2.

Proposition 3. 16 Let F be a graded field, D a semiramified graded division algebra of prime power degree over F, (L, v) an immediate valued field extension of F rac(F) (for its canonical valuation) and HL the henselization of L with respect to v. Then, the following statements are equivalent.

(1) D is degenerate.

(2) Cq(D) ⊗_{F rac(F)} HL is degenerate.

(3) D_{0}HL is degenerate in Cq(D) ⊗_{F rac(F)} HL.

(4) D_{0}L is degenerate in Cq(D) ⊗_{F rac(F)} L.

(5) D_{0}F rac(F) is degenerate in Cq(D).

Proof. By [Mor89, Theorem 1] Cq(D) ⊗_{F rac(F)} HL is a division algebra (of center HL). Since L is an immediate valued field extension of F rac(F), then so is HL.

Therefore, GHL ≅ F and G(Cq(D) ⊗_{F rac(F)} HL) ≅_{g} D. So, (1) ⇔ (2) follows by Lemma 3.15.

(2) ⇔ (3) By Proposition 3.12.

(1) ⇒ (5) By the graded version of Proposition 3.6 there is an inertial graded subfield M of D such that C_{D}^{M} is a tensor product of two cyclic nicely semiramified graded division algebras (M_{i}/M, σ_{i}, a_{i}), where M_{i} are inertial cyclic graded field extensions of M with Galois group generated by σ_{i}, and a_{i} ∈ M^{∗}. So, up to an isomorphism, we have :

C_{Cq(D)}^{F rac(M)} = C_{D}^{M} ⊗_{M} F rac(M) = ((M_{1}/M, σ_{1}, a_{1}) ⊗_{M} (M_{2}/M, σ_{2}, a_{2})) ⊗_{M} F rac(M)
= ((M_{1}/M, σ_{1}, a_{1}) ⊗_{M} F rac(M)) ⊗_{F rac(M)} ((M_{2}/M, σ_{2}, a_{2}) ⊗_{M} F rac(M))
= (F rac(M_{1})/F rac(M), σ_{1}, a_{1}) ⊗_{F rac(M)} (F rac(M_{2})/F rac(M), σ_{2}, a_{2})

By the graded version of Proposition 2.6(3), we have M_{1}⊗_{M}M_{2} = D_{0}M, so F rac(M_{1})⊗_{F rac(M)} F rac(M_{2}) = D_{0}F rac(F) (see that D_{0}F rac(F) = D_{0}F rac(M)). Moreover, it is clear that F rac(M_{1}) [resp. F rac(M_{2})] is the subfield of D_{0}F rac(F) fixed by σ_{2} [resp. by σ_{1}]. This shows that D_{0}F rac(F) is degenerate in Cq(D).

(5) ⇒ (4) and (4) ⇒ (3) Obvious.

Corollary 3. 17 Let E be a field, K a non-cyclic abelian field extension of E with prime power dimension, G the Galois group of K over E, S a basis of G, U a nondegenerate matrix of U_{S}(G, K^{∗}), Z the center of the generic abelian crossed product (K, G, U) and L an immediate field extension of Z (for its canonical valuation), then (K, G, U) ⊗_{Z} L is semiramified but not nicely semiramified. In particular, if char(E) = p and G is a p-group, then (K, G, U) ⊗_{Z} L is not a tensor product of cyclic algebras. If p = 2 and exp((K, G, U) ⊗_{Z} L) = 2, then (K, G, U) ⊗_{Z} L has an involution of the first kind but is not a tensor product of 2-symbols.

Proof. We have seen in [BM00, Theorem 1.1] that there is a graded field F and a semiramified graded division algebra D over F such that (K, G, U) = Cq(D). Moreover, by [BM00, Proposition 0.13], U is nondegenerate (with respect to K/E or equivalently to D_{0}F rac(F)/F rac(F)) if and only if D_{0}F rac(F) is nondegenerate in Cq(D). So, by Proposition 3.16 D is nondegenerate. Therefore, D cannot be nicely semiramified (because Γ_{D}/Γ_{F} (≅ G) is non-cyclic). The rest follows by Corollary 2.5.

Corollary 3. 18 [AS, Theorem 3.2] Let E be a field of characteristic p > 0, K a finite dimensional non-cyclic abelian field extension of E with a Galois p-group G, S a basis of G and U a nondegenerate matrix of U_{S}(G, K^{∗}), then the generic abelian crossed product (K, G, U) is non-cyclic.

4 SUBFIELDS OF NONDEGENERATE TAME SEMIRAMIFIED DIVISION ALGEBRAS 41

Remark 3.19 (1) Let F be a graded field with support Γ_{F} ; E = F_{0} ; M a finitedimensional abelian field extension of E with Galois group G ; S a basis of G ; Δ a totally ordered abelian group that contains Γ_{F} with ordering extending that of Γ_{F} ; ϵ : Δ → G a group epimorphism with kernel Γ_{F} and (U, B) a cocycle of Z_{S}^{2}(G, M^{∗}). We have seen in the preliminaries how to construct the semiramified graded division algebra (MF, ϵ, G, U, B). Assume that U is nondegenerate in M/E and let L be an immediate field extension of F rac(F). One can easily see that Corollary 3.17 is still true if we replace the generic abelian crossed product (K, G, U) by Cq((MF, ϵ, G, U, B)). More generally, let F be a graded field and D a semiramified graded division algebra over F. Let x_{i} ∈ D^{∗} such that (gr(x_{i}) + Γ_{F})_{i=1}^{r} is a basis of Γ_{D}/Γ_{F} and let u_{ij} = x_{i}x_{j}x_{i}^{−1}x_{j}^{−1} and U = (u_{ij})_{1≤i,j≤r}. Then, by [BM00, Proposition 0.13] U is nondegenerate with respect to D_{0}/F_{0} if and only if D_{0}F rac(F) is nondegenerate in Cq(D). Hence, by Proposition 3.16, U is nondegenerate with respecto to D_{0}/F_{0} if and only if D is nondegenerate. It is easily seen that we can replace (K, G, U) in Corollary 3.17 by Cq(D) where D is a nondegenerate semiramified graded division algebra.

(2) Let E be a field, (D, v) a defectless valued finite-dimensional division algebra over E and HE the henselization of E with respect to the restriction of v to E. We say that D is tame over E if D ⊗_{E} HE is so over HE. In particular, if we suppose that E is the center of D, then D is tame over E if and only if GE is the center of GD. Indeed, one can easily see that GE = GHE and GD = G(D ⊗_{E} HE). In the same way, we can define nondegenerate valued central division algebras over a field E (and generalize many results of this paper) without assuming that E is Henselian.

(3) In [McK05] the author defined what she called strongly degenerate matrices. The reader can see that many results of this paper can be proved for non-strongly-degenerate tame semiramified division algebras. In particular, this can be done for Corollary 3.17 and all the results of next sections, but we dont know if there exist really a non-strongly-degenerate tame semiramified division algebra over a Henselian valued field that is degenerate or if the two definitions (strongly degenerate and degenerate) coincide.

4 Subfields of nondegenerate tame semiramified division algebras

Lemma 4. 1 Let F be a graded field and D a nondegenerate semiramified graded division algebra of prime power degree p^{n} over F with rk(Gal(D_{0}/F_{0})) ≥ 2. Let d ∈ D^{∗} such that d^{p} ∈ F. Then, d ∈ D_{0}F^{∗}. If in addition p = char(F), then d ∈ F^{∗}.

Proof. We will use the same argument as in the proof of [AS78, Lemma 3.1]. Let x_{i} ∈ D^{∗} such that (gr(x_{i}) + Γ_{F})_{i=1}^{r} is a basis of Γ_{D}/Γ_{F}. Then, D = D_{0}F [x_{1}, ..., x_{r}]. Let q_{i} = ord(gr(x_{i}) + Γ_{F}), σ_{i} = θ_{D}(gr(x_{i}) + Γ_{F}), where θ_{D} is the group isomorphism of [B95, p.4278] (see also, [HW(2)99, (2.2), p.86]) and I = {m̄ = (m_{1}, ..., m_{r}) ∈ IN^{r} | 0 ≤ m_{i} < q_{i} for all 1 ≤ i ≤ r}. Then, we can write d = axm̄, where a ∈ D_{0}F^{∗} and m̄ ∈ I. Suppose by contradiction that m̄ ≠ 0 and let Dm̄ ; 0 be the subfield of D_{0} fixed by σm̄. Since d^{p} ∈ F^{∗}, then (σm̄)^{p}(= θ_{D}(gr(d^{p}) + Γ_{F}) = id_{D_{0}}. So, [D_{0} : Dm̄ ; 0] = p. Let Nm̄ be the norm of D_{0}/Dm̄ ; 0. Then, (axm̄)^{p} = Nm̄(a)(xm̄)^{p} ∈ F^{∗}. Since rk(Gal(D_{0}/F_{0})) ≥ 2, then there is some 1 ≤ j ≤ r such that ⟨σm̄, σ_{j}⟩ is noncyclic. We have Nm̄(a)(xm̄)^{p} = x_{j}(Nm̄(a)(xm̄)^{p})x_{j}^{−1} = (x_{j}Nm̄(a)x_{j}^{−1})(x_{j}(xm̄)^{p}x_{j}^{−1}) = Nm̄(σ_{j}(a))Nm̄(u_{jm̄})(xm̄)^{p} = Nm̄(σ_{j}(a)u_{jm̄})(xm̄)^{p}. So, Nm̄(σ_{j}(a)a^{−1}u_{jm̄}) = 1. Therefore, by Hilbert 90 Theorem, there is some b ∈ D_{0} such that σ_{j}(a)a^{−1}u_{jm̄} = σm̄(b)b^{−1}. But this is not possible since D is nondegenerate. Hence, m̄ = 0. Remark that D_{0} is separable over F_{0}, so D_{0}F is tame over F. Therefore, if char(F) = p, then d ∈ F^{∗}.

Proposition 4. 2 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. If rk(Γ_{D}/Γ_{E}) ≥ 2, then D has no (non-trivial) totally ramified subfield (over E).

Proof. Suppose first that char(Ē) = p > 0 and deg(D) is a power of p. If K is a totally ramified subfield of D, then GK is a purely wild graded field extension of GE. So, for any x ∈ GK, x^{[GK:GE]} ∈ GE. Hence, by Lemma 4.1, x ∈ GE. So, GK = GE. This proves K = E. Now, assume char(Ē) does not divide deg(D). Suppose that K is a non-trivial totally ramified subfield of D. Let x ∈ GK^{∗} such that gr(x) ∉ Γ_{GE}. If ord(gr(x)+Γ_{GE}) = p^{s} (for some s ≥ 1), then x^{p^{s}} ∈ GE (since GE[x] is totally ramified over GE). Therefore, by Lemma 4.1, x^{p^{s−1}} ∈ D̄GE. A contradiction (since we will have p^{s−1}gr(x) + Γ_{GE} = Γ_{GE}).

Theorem 4. 3 Let E be a Henselian valued filed of residue characteristic p > 0, D a nondegenerate tame semiramified division algebra of degree a power of p with Γ_{D}/Γ_{E} non-cyclic, and K a normal subfield of D. Then, K is a Galois inertial field extension of E. So, there is a subgroup H of Gal(D̄/Ē) such that Gal(K/E) ≅ Gal(D̄/Ē)/H. In particular, if K is a Galois maximal subfield of D, then Gal(K/E) ≅ Gal(D̄/Ē).

Proof. Let K be a normal subfield of D, then by Theorem 1.4, GK is a normal graded field extension of GE. Let L be the graded subfield of GK elementwise invariant by Gal(GK/GE), then by Proposition 1.3 L is a purely wild graded field extension of GE. Therefore, by Lemma 4.1, L = GE. Hence, again by Proposition 1.3, GK is a Galois graded field extension of GE. It follows by Corollary 1.8, that K is a Galois field extension of E. Since [GK : GE] is a power of p and GK is tame over GE, then GK is inertial over GE. So, K is inertial over E. The rest is obvious.

Corollary 4. 4 Let E be a Henselian valued filed of residue characteristic p > 0 and D a nondegenerate tame semiramified division algebra of degree a power of p, then D is cyclic if and only if Γ_{D}/Γ_{E} is cyclic.

Proof. Suppose that D is cyclic and let K be a cyclic maximal subfield of D, then by Theorem 4.3 Gal(K/E) ≅ Gal(D̄/Ē). Moreover, since D is tame semiramified, then Gal(D̄/Ē) ≅ Γ_{D}/Γ_{E}, so Γ_{D}/Γ_{E} is cyclic. Conversely, if Γ_{D}/Γ_{E} is cyclic, then ’the’ inertial lift of D̄ over E in D is a cyclic maximal subfiled of D.

Corollary 4. 5 [S78, Theorem 3.2] Let E be a field of characteristic p > 0, K a non-cyclic abelian field extension of E with dimension a power of p, G the Galois group of K over E, S a basis of G, U a nondegenerate matrix of U_{S}(G, K^{∗}), and L a Galois subfiled of the generic abelian crossed product (K, G, U), then there is a subgroup H of G such that Gal(L/E) ≅ G/H. In particular, if L is a Galois maximal subfield of (K, G, U), then Gal(L/E) ≅ G.

Proposition 4. 6 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Assume rk(Γ_{D}/Γ_{E}) ≥ 2, and let K be an elementary abelian subfield of D. Then, K is inertial over E. Therefore, D is an elementary abelian crossed product if and only if Gal(D̄/Ē) is elementary abelian.

Proof. Since K is an elementary abelian field extension of E, then we can write K = K_{1} ⊗_{E} K_{2} ⊗_{E} ... ⊗_{E} K_{r}, where K_{i} are cyclic field extensions of E with [K_{i} : E] prime. By Proposition 4.2, K_{i} cannot be totally ramified over E. Hence K_{i} is inertial over E. So, K is inertial over E. If K is in addition a maximal subfield of D, then D̄(= K̄) is elementary abelian over Ē. Conversely, suppose that Gal(D̄/Ē) is elementary abelian and let M be the inertial lift of D̄ over E in D, then M is an elementary abelian maximal subfield of D.

Remark 4.7 Let F be a graded field and D a semiramified graded division algebra of degree q^{2} over F, where q is a prime. Suppose Gal(D_{0}/F_{0}) is cyclic and let x ∈ D^{∗} such that gr(x)+Γ_{F} generates Γ_{D}/Γ_{F}, then F [x] is a totally ramified maximal graded subfield of D. Hence, T = F [x^{q}] is a totally ramified graded subfield of D with [T : F] = q. If we assume that F_{0} contains a primitive q^{th} root of unity, then T is a cyclic graded field extension of F. Let σ be the F_{0}-automorphism of D_{0} defined by σ(d) = x^{q}dx^{−q}, and let M_{0} be the subfield of D_{0} fixed by σ, then let M = M_{0}F. Obviously, M[x^{q}] is a graded subfield of D. Since M [resp. T] is inertial [resp. totally ramified] over F, then M[x^{q}] = M ⊗_{F} T. So, D has an elementary abelian maximal graded subfield. Hence HCq(D) has an elementary abelian maximal subfield. This shows that Proposition 4.6 is not true if deg(D) = q^{2} and Γ_{D}/Γ_{E} is cyclic.

Proposition 4. 8 Let E be a Henselian valued field, D a nondegenerate tame semiramified division algebra of prime power degree over E, and K a subfield of D. If Gal(D̄/K̄) is non-cyclic, then K is inertial over E.

Proof. Let M be the inertial lift of K̄ over E in K. Clearly, C_{D}^{M} is a nondegenerate tame semiramified division algebra (over M) and K is a totally ramified subfield of C_{D}^{M}. So, by Proposition 4.2, K = M.

Proposition 4. 9 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Assume rk(Γ_{D}/Γ_{E}) ≥ 3 and let K be an abelian subfield of D, then K is inertial over E.

Proof. Write K = K_{1} ⊗_{E} K_{2} ⊗_{E} ...⊗_{E} K_{r}, where K_{i} are cyclic field extensions of E. Let N_{i} be the inertial lift of K̄_{i} over E in K_{i}. Obviously, N_{i} is cyclic over E. So, K̄_{i} is cyclic over Ē. Therefore, D̄ cannot be cyclic over K̄_{i} (since rk(Gal(D̄/Ē)) ≥ 3). So, by Proposition 4.8 K_{i} is inertial over E.

Lemma 4. 10 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Consider x ∈ D^{∗} such that v(x) ≠ Γ_{E}. Then, Gal(D̄/\overline{E[x]}) is cyclic. In particular, if exp(Γ_{D}/Γ_{E}) is prime, then E[x] is a maximal subfield of D.

Proof. Let M be the inertial lift of \overline{E[x]} over E in E[x]. Clearly, E[x] is a nontrivial totally ramified subfield of C_{D}^{M}. So, by Proposition 4.2, Gal(D̄/M̄) is cyclic. If exp(Γ_{D}/Γ_{E}) = p is a prime, then so is exp(Gal(D̄/Ē)). Hence, card(Gal(D̄/\overline{E[x]})) ≤ p. See that we have [E[x] : E] > [\overline{E[x]} : Ē] = [D̄ : Ē][D̄ : \overline{E[x]}]^{−1} ≥ deg(D)p^{−1}. Hence, [E[x] : E] = deg(D).

Let H be a non-abelian group. We say that H is a quaternion group if H is of order 8 and is generated by two elements a and b satisfying the following conditions a^{4} = b^{4} = 1, a^{2} = b^{2} and ba = a^{−1}b. If K/E is a normal [resp., Galois] field extension with a quaternion Galois group, we say that K is a quaternion (normal [resp., Galois]) field extension of E.

Theorem 4. 11 Let E be a Henselian valued field, p a prime integer and D a nondegenerate tame semiramified division algebra of degree p^{n} (n ∈ IN^{∗}) over E. Assume exp(Γ_{D}/Γ_{E}) = p and let K be a subfield of D, then the following statements hold :

(1) if K is not maximal in D, then K is inertial over E.

(2) if K is a non-quaternion normal maximal subfield of D, then K is either cyclic with dimension ≤ p^{2} or inertial over E.

Remark that in the case of (1) and in the last case of (2) K is elementary abelian over E.

Proof. (1) By Lemma 4.10.

(2) Since K is normal over E, then in the same way as seen in the proof of Theorem 4.3, we prove that K is a Galois tame field extension of E. We will show first that if p ≠ 2 and K is a Galois maximal subfield of D, then Gal(K/E) is abelian. Suppose by contradiction that Gal(K/E) is not abelian, then by Theorem B.11 in comments below (or also by [Ha59, theorem 12.5.4, p. 190]) Gal(K/E) has a non-normal subgroup H. Let L be the subfield of K elementwise fixed by H. Then, L is a non-normal field extension of E. Hence, by Proposition 4.8 both Gal(D̄/K̄) and Gal(D̄/L̄) are non-trivial cyclic groups. Moreover, since exp(Gal(D̄/Ē)) = exp(Γ_{D}/Γ_{E}) = p, then Gal(D̄/K̄) and Gal(D̄/L̄) are of order p. We have L̄ ⊆ K̄, so necessarily L̄ = K̄. Let M be the inertial lift of L̄ over E in L. Then, up to an isomorphism, M is also the inertial lift of K̄ over E in K. So, K is totally ramified over M, and obviously Γ_{K}/Γ_{M}(= Γ_{K}/Γ_{E}) is cyclic of order p. Hence, p ≤ [K : L] ≤ [K : M] = (Γ_{K} : Γ_{M}) = p. So, L = M. But this is not true because L is not normal over E. Therefore, Gal(K/E) is abelian. Assume now that p = 2 and let’s prove that K is an abelian field extension of E. Suppose by contradiction that K is a non-abelian Galois maximal subfield of D. Then, in the same way as above, we show that all subgroups of Gal(K/E) are normal. So, by [Ha59, Theorem 12.5.4, p. 190] Gal(K/E) is the direct product of a quaternion group and an abelian group of exponent 2. Write Gal(K/E) = H_{1} × H_{2}, where H_{1} is a non-trivial quaternion group and H_{2} is a non-trivial abelian group of exponent 2, and let K_{1} [resp., K_{2}] be the subfield of K elementwise invariant by H_{2} [resp., H_{1}], then K = K_{1} ⊗_{E} K_{2}, but by (1) above both K_{1} and K_{2} are inertial over E. Hence, K is an abelian field extension of E. A contradiction. This shows that K is necessarily an abelian field extension of E. Remark that if deg(D) > p^{2} and Gal(K/E) is abelian, then by Proposition 4.9 K is inertial over E. Moreover, if deg(D) ≤ p^{2} and K is a non-cyclic abelian field extension of E, then K = K_{1} ⊗_{E} K_{2}, where K_{i} are cyclic field extension of degree p over E. So, again by (1) above both K_{1} and K_{2} (and hence K) are inertial over E.

(4.12) Let F be a graded field and K a Galois finite-dimensional graded field extension of F. Then, K_{0} is a Galois field extension of F_{0}. Indeed, let σ ∈ Gal(K/K_{0}F) and τ ∈ Gal(K/F). For any a ∈ (K_{0}F)^{∗}, we have gr(τ^{−1}(a)) = gr(a). So, τ^{−1}(a) = ca for some c ∈ K_{0}. Hence, τ ∘ σ ∘ τ^{−1}(a) = a. Therefore, τ ∘ σ ∘ τ^{−1} ∈ Gal(K/K_{0}F). This shows that Gal(K/K_{0}F) is a normal subgroup of Gal(K/F). So, K_{0}F is a Galois graded field extension of F. Hence, by [HW(1)99, Remark 3.1] K_{0} is a Galois field extension of F_{0}. In the same way, we prove that if L is a graded field extension of F in K such that L_{0} = K_{0}, then L is a Galois graded field extension of F. Therefore, by Corollary 1.8 if E is a Henselian valued field, N a tame Galois finite-dimensional field extension of E, and M is a field extension of E in N such that M̄ = N̄, then M is a Galois field extension of E.

Definition 4.13 Let G be an abelian group and H a non-trivial cyclic subgroup of G. We say that H is maximally cyclic in G if there is no cyclic subgroup H′ of G such that H ⊂ H′.

Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. We recall that we have previously seen in the proof of Theorem 4.3 and also in the proof of Theorem 4.11 that if K is a normal field extension of E, then K is a Galois field extension of E.

Proposition 4. 14 Let E be a Henselian valued field, D a nondegenerate tame semiramified division algebra of prime power degree over E, and K a normal subfield of D such that Gal(D̄/K̄) is cyclic. If Gal(D̄/K̄) is maximally cyclic in Gal(D̄/Ē), then the following statements hold :

(1) If deg(D) is odd, then K is an abelian field extension of E.

(2) If deg(D) is a power of 2 and Γ_{D}/Γ_{E} is non-cyclic, then K is either a quaternion or an abelian field extension of E.

Therefore, if rk(Γ_{D}/Γ_{E}) ≥ 3 and K is not a quaternion field extension of E, then K is inertial over E.

Proof. (1) Suppose by contradiction that Gal(K/E) is non-abelian. Then, by Theorem B.11, there is a non-normal subgroup H of Gal(K/E). Let L be the subfield of K elementwise invariant by H. By Proposition 4.8, Gal(D̄/L̄) is cyclic (indeed, otherwise L would be inertial -hence normal- over E). Since Gal(D̄/K̄) is maximally cyclic in Gal(D̄/Ē) and Gal(D̄/K̄) ⊆ Gal(D̄/L̄), then K̄ = L̄. So, by (4.12) Gal(K/L) is a normal subgroup of Gal(K/E). A contradiction.

(2) In the same way as above we prove that all subgroups of Gal(K/E) are normal. Therefore, if K is non-abelian over E, then by [Ha59, Theorem 12.5.4, p.190] Gal(K/E) = H_{1} × H_{2}, where H_{1} is a quaternion group and H_{2} is an abelian group of exponent 2. Let K_{1} [resp., K_{2}] be the subfield of K elementwise invariant by H_{2} [resp., H_{1}]. Then, up to an isomorphism, we have K = K_{1} ⊗_{E} K_{2}. Assume that K̄_{i} ≠ Ē (for both i = 1, 2), then Gal(D̄/K̄_{i}) (i = 1, 2) are non-cyclic (because we have Gal(D̄/K̄) ⊂ Gal(D̄/K̄_{i}) and Gal(D̄/K̄) is maximally cyclic in Gal(D̄/Ē)). Therefore, by Proposition 4.8, K_{1} and K_{2} (and hence K) are inertial over E. But this contradicts the fact that K is non-abelian over E. Therefore, K̄_{1} or K̄_{2} is trivial. This means that K_{1} or K_{2} is totally ramified over E. So, by Proposition 4.2, either K_{1} or K_{2} is trivial. Remark that we cannot have K = K_{1} because K is non-abelian over E. Hence, K(= K_{2}) is a quaternion field extension of E. The rest of the proposition follows by Proposition 4.9.

Corollary 4. 15 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Assume deg(D) is odd and char(Ē) does not divide deg(D) and let K be a Galois maximal subfield of D. If (Γ_{K} : Γ_{E}) = exp(Γ_{D}/Γ_{E}), then K is abelian over E and rk(Γ_{D}/Γ_{E}) ≤ 2.

Proof. Since D is nondegenerate, then Γ_{K}/Γ_{E} is cyclic. Moreover, since K is a maximal subfield of D, then Gal(D̄/K̄) ≅ Γ_{K}/Γ_{E}. Hence, Gal(D̄/K̄) is maximally cyclic in Gal(D̄/Ē). So, by Proposition 4.14 K is abelian over E and rk(Γ_{D}/Γ_{E}) ≤ 2 (because K is not inertial over E).

(4.16) As a consequence of Corollary 4.15, under the hypotheses of Corollary 4.15, if rk(Γ_{D}/Γ_{E}) ≥ 3, x ∈ D^{∗} with ord(v(x) + Γ_{E}) = exp(Γ_{D}/Γ_{E}), and L is a maximal subfield of D that contains x, then L cannot be Galois over E.

Proposition 4. 17 Let F be a graded field, D a nondegenerate semiramified graded division algebra of prime power degree p^{r} over F with rk(Γ_{D}/Γ_{F}) ≥ 3, and K a Galois graded subfield of D such that [K : F] < pdeg(D)(exp(Γ_{D}/Γ_{F}))^{−1}. Then, K is inertial over F.

Proof. We will show that Gal(K/F) is abelian. Our Proposition follows then by Proposition 4.9 -applied to HCq(D)- . Suppose that Gal(K/F) is not abelian and suppose that [K : F] is minimal. By Proposition 4.8 Gal(D_{0}/K_{0}) is cyclic. So, [K_{0} : F_{0}] ≥ [D_{0} : F_{0}](exp(Γ_{D}/Γ_{F}))^{−1}. Remark that necessary (Γ_{K} : Γ_{F}) = p. Indeed, if (Γ_{K} : Γ_{F}) ≠ p, then there exists an intermediate graded field P such that K_{0}F ⊂ P ⊂ K. Since P_{0} = K_{0} and K is Galois over F, then by (4.12) P is also a Galois graded field extension of F. It follows by the minimality of [K : F] that Gal(P/F) is abelian, hence P is inertial over F. But this is not true (since K_{0}F ⊂ P).

So, (Γ_{K} : Γ_{F}) = p. Therefore, [K : F] ≥ pdeg(D)exp(Γ_{D} : Γ_{F})^{−1}. A contradiction.

Corollary 4. 18 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree p^{r} over E. Assume char(Ē) does not divide deg(D) and suppose K is a Galois subfield of D with [K : E] < pdeg(D)exp(Γ_{D} : Γ_{E})^{−1}. Then, K is inertial over E.

Definition 4.19 Let E be a Henselian valued field and K an abelian field extension of E. We say that K is nice over E if there is an inertial field extension M of E and a totally ramified field extension T of E such that K = M ⊗_{E} T. We define nice abelian graded field extensions in a similar way.

Proposition 4. 20 Let E be a Henselian valued field and K a tame abelian field extension of E. Then, K is nice if and only if GK is nice.

Proof. If K is nice, then K = M ⊗_{E} T, where M is inertial and T is totally ramified over E. So, by the graded version of [Mor89, Theorem 1] GK = GM ⊗_{GE} GT. Conversely, suppose that GK is nice. Then, GK = K̄GE ⊗_{GE} T′ where T′ is a totally ramified graded field extension of GE. Let M be the inertial lift of K̄ over E in K and let u_{1}, ..., u_{r} be homogeneous elements of T′ such that Γ_{T′}/Γ_{GE} = ⟨gr(u_{1}) + Γ_{GE}⟩ ⊕ ... ⊕ ⟨gr(u_{r}) + Γ_{GE}⟩. By Proposition 1.5, there are x_{i} ∈ K such that x̃_{i} = u_{i} and G(E[x_{i}]) = GE[u_{i}]. In particular, E[x_{i}] are totally ramified over E, hence by [Mor89, Theorem 1] T := E[x_{1}] ⊗_{E} ... ⊗_{E} E[x_{r}] is a totally ramified field extension of E. Once again, by applying [Mor89, Theorem 1] (and by comparing the dimensions), we have K = M ⊗_{E} T (up to an isomorhism).

Lemma 4. 21 Let E be a Henselian valued field and K a nice cyclic field extension of prime power dimension over E. Then, K is either inertial or totally ramified over E.

Proof. Let M [resp. T] be an inertial [resp. totally ramified] field extension of E such that K = M ⊗_{E} T. Then, Gal(K/E) = Gal(M/E) ⊕ Gal(T/E). Remark that Gal(K/F) is cyclic with prime power order. So, necessaraly Gal(M/E) or Gal(T/E) is a trivial group. Therefore, either M or T is trivial.

Lemma 4. 22 Let E be a Henselian valued field with valuation v and D a semiramified division algebra of degree n over E. Assume Ē contains a primitive n^{th} root of unity. Then, the following statements are equivalent :

(1) D has an abelian maximal subfield K = M ⊗_{E} T, where M [resp. T] is a cyclic inertial [resp. a cyclic totally ramified] field extension of E.

(2) There is a cyclic subfield M of D inertial over E and an element u of C_{D}^{M} such that v(u) + Γ_{E} generates Γ_{C_{D}^{M}} /Γ_{E} and u^{[D̄:M̄]} ∈ E.

Proof. (1) ⇒ (2) By Proposition 2.6 C_{D}^{M} is semiramified (over M), so [D̄ : M̄] = (Γ_{C_{D}^{M}} : Γ_{M}). Moreover, K is a totally ramified maximal subfield of C_{D}^{M}, so [D̄ : Ē] = (Γ_{K} : Γ_{M}) = (Γ_{T} : Γ_{E}) = [T : E]. Since T is cyclic over E, then Γ_{T} /Γ_{E}(≅ Gal(T/E)) is cyclic. Let x ∈ T^{∗} such that Γ_{T} /Γ_{E} = ⟨v(x) + Γ_{E}⟩, then GT = GE[x̃] and the minimal polynomial of x̃ over F rac(GE) is X^{[T:E]} − x̃^{[T:E]}. So, by Proposition 1.5 there is u ∈ T such that T = E[u], u^{[T:E]} ∈ E^{∗} and v(u) + Γ_{E} = v(x) + Γ_{E}.

(2) ⇒ (1) Since u^{[D̄:M̄]} ∈ E^{∗}, then [E[u] : E] ≤ [D̄ : M̄]. Moreover, since C_{D}^{M} is semiramified and v(u) + Γ_{E} generates Γ_{C_{D}^{M}} /Γ_{E}, then D̄ is a cyclic field extension of M̄ and [D̄ : M̄] = card(⟨v(u) + Γ_{E}⟩) ≤ (Γ_{E[u]} : Γ_{E}) ≤ [E[u] : E]. Hence, E[u] is totally ramified over E. Moreover, since Γ_{E[u]}/Γ_{E} is cyclic and Ē contains a primitive n^{th} root of unity, then E[u] is cyclic over E. Therefore, K := M ⊗_{E} E[u] is an abelian maximal subfield of D.

Proposition 4. 23 Let E be a Henselian valued field with valuation v and D a nondegenerate tame semiramified division algebra of prime power degree over E. Then, the following statements are equivalent.

(1) D has a nice abelian maximal subfield.

(2) D has an abelian maximal subfield K = T ⊗_{E} M, where T is cyclic totally ramified over E and M is cyclic inertial over E.

(3) D has a nice cyclic maximal subfield.

(4) D has a cyclic inertial maximal subfield.

(5) D has a non-trivial subfield which is totally ramified over E.

(6) Gal(D̄/Ē) is cyclic.

If Ē contains a primitive deg(D)^{th} root of unity, then the above conditions are equivalent to the following :

(7) There a cyclic subfield M of D inertial over E and an element u of C_{D}^{M} such that v(u) + Γ_{E} generates Γ_{C_{D}^{M}} /Γ_{E} and u^{[D̄:M̄]} ∈ E.

Proof. (1) ⇒ (6) Let K be a nice maximal subfield of D. If K is inertial over E, then K̄ = D̄. So, Gal(D̄/Ē) is cyclic. If K is not inertial over E, then D contains a non-trivial totally ramified subfield. So, by Proposition 4.2, Gal(D̄/Ē)(≅ Γ_{D}/Γ_{E}) is cyclic.

(6) ⇒ (3) Clear (indeed, the inertial lift of D̄ over E in D is a nice cyclic maximal subfield of D).

(3) ⇒ (2) By Lemma 4.21.

(2) ⇒ (1) and (4) ⇔ (6) Obvious.

(5) ⇒ (6) By Proposition 4.2.

(6) ⇒ (5) Let d ∈ D^{∗} such that v(d) + Γ_{E} generates Γ_{D}/Γ_{E}, then E[d] is a non-trivial totally ramified subfield of D. If Ē contains a primitive deg(D)^{th} root of unity, then (2) ⇔ (7) follows by Lemma 4.22.

5 Indecomposable division algebras

Lemma 5. 1 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Assume D = D_{1} ⊗_{E} D_{2}, where D_{i} (i = 1, 2) are non-trivial central division algebras over E, and let L_{i} be a maximal subfield of D_{i}. Then, L̄_{i} = D̄_{i}.

Proof. Since D_{i} are non-trivial, then rk(Γ_{D}/Γ_{E}) ≥ 2. Indeed, if Γ_{D}/Γ_{E} is cyclic, then D is nicely semiramified. So, by [JW90, Lemma 5.15] exp(D) = exp(Γ_{D}/Γ_{E}) = deg(D). Hence, D is indecomposable. A contradiction. Moreover, by Proposition 4.2, D̄_{i} ≠ Ē (for otherwise, all subfields of D_{i} will be totally ramified over E). Let L_{1} be a maximal subfield of D_{1}. Suppose by contradiction that L̄_{1} ≠ D̄_{1}, then rk(Gal(D̄_{1}⊗ĒD̄_{2}/L̄_{1})) ≥ 2 (for if σ ∈ Gal(D̄_{1}/L̄)\{idD̄_{1}} and τ ∈ Gal(D̄_{2}/Ē)\{idD̄_{2}}, then σ ⊗ idD̄_{2} and idD̄_{1} ⊗ τ are non-comparable elements of Gal(D̄_{1} ⊗Ē D̄_{2}/L̄_{1})), so rk(Gal(D̄/L̄_{1})) ≥ 2. Hence, by Proposition 4.8, L_{1} is inertial over E. So, by Proposition 2.6(1) D_{1} is semiramified. Hence, L̄_{1} = D̄_{1}. The contradiction obtained here shows that L̄_{1} = D̄_{1}. In the same way, we prove that if L_{2} is a maximal subfield of D_{2}, then L̄_{2} = D̄_{2}.

Theorem 5. 2 Let E be a Henselian valued field and D a nondegenerate tame semiramified division algebra of prime power degree over E. Then, D is indecomposable.

Proof. As seen above in the proof of Lemma 5.1, if Γ_{D}/Γ_{E} is cyclic, then D is indecomposable. Assume that rk(Γ_{D}/Γ_{E}) ≥ 2 and suppose D = D_{1} ⊗_{E} D_{2}, where D_{1} and D_{2} are non-trivial central division algebras over E. Choose a subfield K of D_{i} with (Γ_{K} : Γ_{E}) maximal (among all subfields of D_{1} and D_{2}). We can suppose i = 1. For an arbitrary element x_{2} of D_{2}, K[x_{2}] is a subfield of D. So, by Proposition 3.2, Γ_{K[x_{2}]}/Γ_{E} is cyclic (with prime power order). Hence, the subgroups of Γ_{K[x_{2}]}/Γ_{E} are totally ordered by inclusion. In particular, we have Γ_{E[x_{2}]}/Γ_{E} ⊆ Γ_{K}/Γ_{E} (we can not have Γ_{K}/Γ_{E} ⊂ Γ_{E[x_{2}]}/Γ_{E} because (Γ_{K} : Γ_{E}) is maximal). So, Γ_{D_{2}}/Γ_{E} is cyclic. Let y ∈ D_{2} such that v(y) + Γ_{E} generates Γ_{D_{2}}/Γ_{E} (v being the extension of the valuation of E to D) and let L_{2} be a maximal subfield of D_{2} that contains y. By Lemma 5.1 L̄_{2} = D̄_{2}. Moreover, we have Γ_{L_{2}}/Γ_{E} = Γ_{D_{2}}/Γ_{E}. Hence, D_{2} = L_{2}. A contradiction.

Corollary 5. 3 Let E be a field, K a finite dimensional non-cyclic abelian field extension of E with Galois group G, S a basis of G, U a nondegenerate matrix of U_{S}(G, K^{∗}), Z the center of the generic abelian crossed product (K, G, U) and L an immediate field extension of Z for its canonical valuation. Assume card(G) = p^{n}, where p is a prime positive integer and n is an arbirary positive integer, then (K, G, U)⊗_{Z} L is an indecomposable division algebra.

Remark 5.4 Recently, Kelly L. Mckinnie has constructed tame semiramified indecomposable division p-algebras of arbitrary degree and of exponent p (for p ≠ 2) [McK05, Corollary 6.15]. Using the same arguments and Theorem 5.2 -instead of [McK05, Corollary 5.4]-, we get examples of tame semiramified indecomposable division algebras of prime exponent q and of degree q^{n} for any prime q ≠ 2 and any positive integer n (even for q ≠ char(Ē)).

Appendix

Claim A. 1 Let E/F be an algebraic graded field extension and g a homogenizable irreducible polynomial of F [X]. Assume g has a root x ∈ E\{0} and let f_{x,F rac(F)} be the minimal polynomial of x over F rac(F). Then, g = af_{x,F rac(F)} for some a ∈ F^{∗}. In particular, x ∈ E^{∗}.

Proof. Since x is a root of g, then f_{x,F rac(F)} divides g in F rac[X]. Let h ∈ F rac(F)[X] such that g = hf_{x,F rac(F)}. Obviously, we can write h = b^{−1}h′ [resp. f_{x,F rac(F)} = c^{−1}f′] with h′, f′ ∈ F [X] and b, c ∈ F. Let λ ∈ Δ_{F} (where Δ_{F} is the divisible hull of Γ_{F}) such that g ∈ (F [X]^{(λ)})^{h}. It follows from the equality (bc)g = h′f′ that (bc)_{m}g = h′_{m}f′_{m}, where (bc)_{m}, h′_{m} and f′_{m} are the components of greater grade of bc, h′ and f′, respectively. Since g is irreducible in F [X], then either h′_{m} or f′_{m} is in F. Assume that h′_{m} ∈ F, then deg(g) = deg(f′_{m}) ≤ deg(f_{x,F rac(F)}). So, deg(g) = deg(f_{x,F rac(F)}) (since f_{x,F rac(F)} divides g). If f′_{m} ∈ F, then deg(g) = deg(h′_{m}) ≤ deg(h). So, deg(f_{x,F rac(F)}) = 0. But this is not true. See that we can write g = a_{0} + a_{1}X + ... + a_{n}X^{n}, where n is a positive integer and a_{i} ∈ F^{∗} for all 0 ≤ i ≤ n (Indeed, write g = d_{0} + d_{1}X + ... + d_{n}X^{n}, where d_{i} ∈ F and write d_{i} = ∑(d_{i})_{γ}, where (d_{i})_{γ} ∈ Fγ̄. Since g is homogeneous (in F [X]^{(λ)}), then we can consider only the (d_{i})_{γ} where gr((d_{i})_{γ}X^{i}) = gr(g)). Let g′ = a_{n}^{−1}g. Then, g′ = f_{x,F rac(F)} (since g′(x) = 0, deg(g′) = deg(f_{x,F rac(F)})). So, g = a_{n}f_{x,F rac(F)}. In particular, f_{x,F rac(F)} is homogeneous in F [X]^{(λ)}. So, x ∈ E^{∗} (by [HW(1)99, Proposition 2.2 and corollary 2.5]).

B Normal subgroups

Lemma B. 1 Let p be a prime positive integer and G a p-group, generated by two noncommutative elements x and y such that ord(x) = p. Then either ⟨x⟩ (i.e. the cyclic subgroup of G generated by x) or ⟨y⟩ is a non normal subgroup of G.

Proof. Let p^{m} = ord(y), then for every 0 < k < p, 0 < k′ < p^{m}, x^{k} ≠ y^{k′}. Indeed, otherwise we will have ⟨x⟩ = ⟨x^{k}⟩ ⊆ ⟨y⟩, which contradicts the noncommutativity of x and y. If ⟨x⟩ and ⟨y⟩ were both normal in G, then there would exist positive integers r and s with 1 < r < p and 1 < s < p^{m} such that yx^{r}y^{−1} = x and x^{−1}y^{s}x = y. Therefore (yx^{r}y^{−1})y = x(x^{−1}y^{s}x), hence yx^{r} = y^{s}x. Thus x^{r−1} = y^{s−1}. But this is not true.

In what follows, we proceed by contradiction to prove that if p is an odd prime integer and G is a non abelian p-group, then necessarily G has a subgroup H which is not normal in G.

B.2. Let G be a non abelian p-group such that all subroups of G are normal (in G). We may assume that G has minimal cardinal. It follows from Lemma B.1, that card(G) > p^{2}. Let Z(G) be the center of G and G^{c} := {αβα^{−1}β^{−1}| α, β ∈ G} be the commutator of G.

Claim B. 3 Under the hypotheses of (B.2), we have G^{c} ⊆ Z(G).

Proof. Since card(G) > 1, then by [L84, theorem 6.4, p. 25], Z(G) ≠ {1}. If G/Z(G) has a non normal subgroup A, then H := {α ∈ G| αZ(G) ∈ A} would be a non normal subgroup of G. A contradiction. Therefore, by the minimality of card(G), G/Z(G) is necessarily abelian. So, G^{c} ⊆ Z(G).

Claim B. 4 Under the hypotheses of (B.2), we have card(G^{c}) = p.

Proof. Assume that card(G^{c}) > p and consider a subgroup G′ of G^{c} with card(G′) = p. Since the cardinal of G is minimal, then either G/G′ is abelian or G/G′ has a subgroup A which is not normal in G/G′. See that G/G′ cannot be abelian since otherwise we would have G^{c} ⊆ G′. Also, if there was a non-normal subgroup A of G/G′, then H := {α ∈ G| αG′ ∈ A} would be a non normal subgroup of G. A contradiction.

Claim B. 5 Under the hypotheses of (B.2), if x and y are two non-commutative elements of G, then every element z of G can be written in the form z = ax^{m}y^{n}, with a ∈ Z(G), 0 ≤ m, n < p. Thus, card(G/Z(G)) = p^{2}. Precisely, G/Z(G) ≅ ZZ/pZZ × ZZ/pZZ.

Proof. Since G/⟨x⟩ and G/⟨y⟩ are abelian, then G^{c} ⊆ ⟨x⟩ ∩ ⟨y⟩. Let p^{r} = card(⟨x⟩/G^{c}) and p^{s} = card(⟨y⟩/G^{c}). Remark that card(G) being minimal, then G = ⟨x, y⟩ (i.e. G is generated by x and y). Let u = xyx^{−1}y^{−1}, then x^{p}y = u^{p}yx^{p} = yx^{p} ( since u ∈ G^{c} and card(G^{c}) = p.) So x^{p} ∈ Z(G). By symmetry, we have also y^{p} ∈ Z(G). Therefore, every element of G can be written in the form ax^{m}y^{n}, with a ∈ Z(G) and 0 ≤ m, n < p. Thus, the elements of G/Z(G) can be written as x̄^{m}ȳ^{n}, with 0 ≤ m, n < p (where x̄ = xZ(G) and ȳ = yZ(G).) So, card(G/Z(G)) ≤ p^{2}. Remark that G/Z(G) can not be cyclic since otherwise G would be abelian. Hence card(G/Z(G)) = p^{2} and G/Z(G) ≅ ZZ/pZZ × ZZ/pZZ.

Claim B. 6 Under the hypotheses of (B.2), if p is an odd prime integer and if we consider the map ϕ : G → G, defined by ϕ(z) = z^{p}, then ϕ is a group homomorphism and G^{p} := ϕ(G) ⊆ Z(G).

Proof. Let x, y ∈ G and let u = xyx^{−1}y^{−1}. It is straightforword that (xy)^{p} = uu^{2}..u^{p−1}x^{p}y^{p} = (u^{\frac{p−1}{2}} )^{p}x^{p}y^{p} = x^{p}y^{p} (since u^{\frac{p−1}{2}} ∈ G^{c} and card(G^{c}) = p). So, ϕ is a group homomorphism. Since x^{p}y = u^{p}yx^{p} = yx^{p}, then ϕ(G) ⊆ Z(G).

Obviously, Under the hypotheses of Claim B.6, G/ker(ϕ) ≅ G^{p}.

Claim B. 7 under the hypotheses of Claim B.6, we have ker(ϕ) ⊆ Z(G).

Proof. If ker(ϕ) ⊈ Z(G), then one can choose two non-commutative elements x and y of G with x ∈ ker(ϕ). Since G = ⟨x, y⟩ and ord(x) = p, then by Lemma B.1, either ⟨x⟩ or ⟨y⟩ is not normal in G. A contradiction.

Claim B. 8 Under the hypotheses of (B.2), Z(G) is cyclic.

Proof. Let x and y be two non-commutative elements of G, and consider w ∈ G with ord(w) ≥ p. Then x̄ȳ = ȳx̄ in G/⟨w⟩. Hence u := xyx^{−1}y^{−1} ∈ ⟨w⟩. In particular, this is true for 1 ≠ w ∈ Z(G). Therefore, Z(G) is cyclic (for if Z(G) were not cyclic, then it would have a decomposition Z(G) = ⟨w_{1}⟩ × ⟨w_{2}⟩ × .. × ⟨w_{k}⟩ with ord(w_{i}) ≥ p (1 ≤ i ≤ k) and k > 1. In particular, we have ⟨w_{1}⟩ ∩ ⟨w_{2}⟩ = {1}, hence necessarily u = 1. But this is not true since x and y are non-commutative.)

Claim B. 9 Under the hypotheses of Claim B.6, ker(ϕ) = G^{c}.

C CYCLIC GRADED P -ALGEBRAS 60

Proof. Since G/ker(ϕ) is abelian (otherwise G would have a non normal subgroup), then G^{c} ⊆ ker(ϕ). Moreover, by Claim B.7, we have ker(ϕ) ⊆ Z(G). So ker(ϕ) is cyclic (of exponent p). So ker(ϕ) = G^{c}.

Claim B. 10 Under the hypotheses of Claim B.6, G/G^{p} is cyclic with cardinal p.

Proof. Since G^{p} ≅ G/ker(ϕ), then card(G/G^{p}) = p, so G/G^{p} is cyclic.

Since G^{p} ⊆ Z(G) and G/G^{p} is cyclic, then G is abelian. But this contradicts the hypotheses of Claim B.6. Therefore, we have the following Theorem.

Theorem B. 11 Let p > 2 be a prime integer and G a non abelian p-group. Then G has a non normal subgroup.

C Cyclic graded p-algebras

We recall first the following facts concerning cyclic graded simple algebras. All these results follow from analogous ones proved in the ungraded case (it suffices to use the canonical embedding GBr(F) → Br(F rac(F)), defined by [D] ↦ [D ⊗_{F} F rac(F)]) :

(C.1) Let F be a graded field and L/F a finite-dimensional cyclic graded field extension of degree n, then the following assertions hold (for arbitrary a, b ∈ F^{∗}) :

(1) (L/F, σ, a) = (L/F, σ^{t}, a^{t}), for any positive integer t coprime to n.

(2) (L/F, σ, a) ⊗_{F} (L/F, σ, b) is similar to (L/F, σ, ab) with respect to GBr(F).

(C.2) (1) Let L/F be a finite-dimensional cyclic graded field extension with Galois group generated by σ, then the mapping a ↦ [(L/F, σ, a)] (a ∈ F^{∗}) induces a group isomorphism F^{∗}/N_{L/F} (L^{∗}) → GBr(L/F).

(2) Let N/F be a cyclic graded field extension with Galois group generated by τ such that F ⊆ L ⊆ N and let s = [N : L] and a ∈ F^{∗}, then (N/F, τ, a^{s}) is similar to (L/F, σ, a) with respect to GBr(F).

Theorem C. 3 Let F be a graded field with characteristic p > 0 and a a non-zero homogeneous element of F such that a^{1/p} ∉ F. If D is a non-trivial graded central division algebra split by F [a^{1/p}], then there is a cyclic graded field extension L of dimension p over F such that D = (L/F, σ, a), where σ is a generator of Gal(L/F).

Proof. Since deg(D) = p and F [a^{1/p}] splits D, then we can assume F [a^{1/p}] is a maximal graded subfield of D. Let x := a^{1/p} and consider the graded F -automorphism Φ : D ↦ D, defined by z ↦ xzx^{−1}. Since x ∉ F, then Φ ≠ id_{D}. So, ϕ = Φ − id_{D} is a non-zero graded F -space homomorphism of D. Clearly, we have ϕ^{p} = 0. Let r be the greater positive integer such that ϕ^{r} ≠ 0 and let z ∈ D^{∗} such that ϕ^{r}(z) ≠ 0. Put v = ϕ^{r−1}(z), w = ϕ^{r}(z) and b = w^{−1}v. One can easily check that ϕ(b) = b + 1. Therefore, L := F [b] is a cyclic graded field extension of F with Galois group generated by σ := ϕ_{/L}, and D = (L/F, σ, a).

Theorem C. 4 Let F be a graded field with characteristic p > 0, a a non-zero homogeneous element of F and A a non-trivial graded central simple algebra which satisfies [A] ∈ GBr(F [a^{1/p^{e}}]/F) and [A] ∉ GBr(F [a^{1/p^{f}}]/F) for all f < e, then [A] = [(N/F, σ, a)], where N/F is a cyclic graded field extension of dimension p^{e} with Galois group generated by σ.

Proof. Assume first that F [a^{1/p^{e}}] has dimension p^{e} over F. We will proceed by induction on e. The case e = 1 was proved in Theorem C.3 above. Assume this is true for e < n and let e = n. By the induction hypothesis there is a cyclic graded field extension M of dimension p^{n−1} over F [a^{1/p}] such that A ⊗_{F} F [a^{1/p}] ∼ (M/F [a^{1/p}], σ, a^{1/p}), where σ is a generator of the Galois group of M/F [a^{1/p}]. Clearly, F rac(M) is a normal field extension of F rac(F) with Gal(F rac(M)/F rac(F)) = Gal(F rac(M)/F rac(F [a^{1/p}])). Moreover if N′ is the tame closure of F in M, then by [HW(1)99, Proposition 3.7 and Definition 3.8] F rac(N′) is the separable closure of F rac(F) in F rac(M). Therefore by [Karp89, Proposition 7.7, p.283] F rac(M) = F rac(N′) ⊗_{F rac(F)} F rac(F [a^{1/p}]) (up to an isomorphism). This shows that F rac(N′) is a cyclic field extension of F rac(F), hence by [HW(1)99, Theorem 3.11] N′ is a cyclic graded field extension of F (with Galois group generated by σ_{/N′}). Let K be a cyclic field extension of F_{0} that contains N′_{0} such that [K : N′_{0}] = p (see [D82, Lemma 2, p. 107]) and let N = KF (i.e. the graded field extension of F generated by K). By, [HW(1)99, Remark 3.1] N is a cyclic inertial graded field extension of F with Galois group generated by some τ such that τ_{/N′} = σ_{/N′}. We have A ⊗_{F} F [a^{1/p}] = [(M/F [a^{1/p}], σ, a^{1/p})] = [(NF [a^{1/p}]/F [a^{1/p}], τ, a)] = [(N/F, τ, a) ⊗_{F} F [a^{1/p}]]. So, [A] − [(N/F, τ, a)] ∈ GBr(F [a^{1/p}]/F) and a^{1/p} ∉ F. Hence, again by Theorem C.3, [A] = [(N/F, τ, a)] + [(T/F, θ, a)], where T/F is a cyclic graded field extension of dimension p with galois group generated by θ. We have either, N ∩ T = F or T ⊆ N. For the first case, we consider the graded subfield L of N ⊗_{F} T elementwise fixed by α := τ^{p^{n−1}} ⊗ θ^{−1}. By [D82, Lemma 9, p. 75] we have [(F rac(N)/F rac(F), τ, a)] + [(F rac(T)/F rac(F), θ, a)] = [(F rac(L)/F rac(F), α, a)]. So, [(N/F, τ, a)]+[(T/F, θ, a)] = [(L/F, α, a)] (this follows from the canonical embedding GBr(F) → Br(F rac(F)) defined by [D] ↦ [D ⊗_{F} F rac(F)]).

Now, if we suppose that T ⊆ N, then there exists a positive integer t coprime to p such that τ_{/T} = θ^{t}. Since 1 + tp^{n−1} and p^{n} are coprime, then there is a positive integer s such that s(1 + tp^{n−1}) ≡ 1(mod p^{n}). Hence, by (C.1 and C.2) above [(T/F, θ, a)] = [(T/F, θ^{t}, a^{t})] = [(N/F, τ, a^{tp^{n−1}})]. So, [A] = (1 + tp^{n−1})[(N/F, τ, a)] = (1 + tp^{n−1})[(N/F, τ^{s}, a^{s})] = [(N/F, τ^{s}, a)].

If F [a^{1/p^{e}}] has dimension p^{f} < p^{e} over F, then F [a^{1/p^{e}}] = F [b^{1/p^{f}}], where a = b^{p^{e−f}}. It follows by the above that [A] = [(N′/F, σ′, b)], where N′/F is a cyclic graded field extension with dimension p^{f}. We can then apply (C.2(2)) to conclude.

Corollary C. 5 Let F be a graded field with characteristic p > 0, A a graded central division algebra of degree p^{e} over F. Assume that A has a simple purely wild maximal graded subfield, then A is cyclic.

Theorem C. 6 Let F be a graded field with characteristic p > 0 and D a graded central division algebra of degree a power of p over F. Then, D is cyclic if and only if it has a simple purely wild maximal graded subfield.

Proof : If D is cyclic, then we can write D = (L/F, σ, a), where L is a a cyclic maximal graded subfield of D, σ is a generating element of the Galois group Gal(L/F) and a is a nonzero homogeneous element of F. Let n be the degree of D, then by definition, there is a nonzero homogeneous element x of D such that D = ⊕_{i=1}^{n}Lx^{i}, with x^{n} = a and xyx^{−1} = σ(y) for all y ∈ L. Hence F [x] is a simple purely wild maximal graded subfield of D. The converse follows by Corollary C.5.

The author would like to thank Pr. J.P. Tignol and Pr. A. Wadsworh for their help.

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