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\title{ Nondegenerate semiramified valued and graded  division 
algebras\footnote{Mathematics subject classification (MSC2000):  16K50,   
16W50, 16W60 and 16W70.}\footnote{Key Words: (Graded) Brauer group, Valued 
division algebras,  Henselization, Graded division algebras,  Generic 
abelian crossed products.}}
\author{Karim Mounirh}
\date {D\'epartement des Math\'ematiques \\
Facult\'e des Sciences\\
Universit\'e Catholique de Louvain\\
Belgium\\
e.mail: akamounirh@hotmail.com}

\begin {document}
\maketitle

\begin{abstract}
  {\normalsize 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.\\\\}
\end {abstract}


\newpage
{\Large\bf Preliminaries}

\large

Let $F$ be an associative ring with a unit and $\Gamma$ a totally ordered 
abelian group. We say  that $F$ is   a graded ring of type $\Gamma$ if there 
are subgroups $F_{\gamma}$ ($\gamma \in \Gamma$) of $F$ such that 
$F=\oplus_{\gamma \in \Gamma}F_{\gamma}$ and $F_{\gamma}F_{\delta} \subseteq 
F_{\gamma +\delta}$, for all $\gamma,$ $ \delta \in \Gamma$. In this case, 
the set $\Gamma_F=\{\gamma \in \Gamma$ $|$ $ F_{\gamma} \neq 0 \}$ is called 
the support of $F$.

If $F$ is a graded ring of type $\Gamma$ and $x \in F_{\gamma}$ for some 
$\gamma\in \Gamma_F$, we say that $x$ is a homogeneous element of $F$. In 
particular, if $x$ is a nonzero element of $F_{\gamma}$, we say that $x$ has 
grade $\gamma$ and we write $gr(x)=\gamma$. 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 $\Gamma$ and there is a ring isomorphism $\sigma : F\mapsto 
F_0[\Gamma]$ such that $\sigma(F_0)=F_0$ and $\sigma(F_{\gamma})=F_0\gamma$ 
for any $\gamma\in \Gamma - \{ 0\}$, then $F$ is called a graded field  of 
group-ring type.

Let $F$ be a commutative graded ring of type $\Gamma$. A (left) algebra $A$ 
over $F$  is called a graded  algebra over $F$ (of type $\Gamma$) if $A$ is  
a graded ring of type $\Gamma$ and  $F_{\gamma}\subseteq A_{\gamma}$, for 
all $\gamma \in \Gamma$. 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 $[K : F]=(\Gamma_K : \Gamma_F)$. 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 
$\Gamma_K/\Gamma_F$ 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 
$\Gamma_K/\Gamma_F$ is a $p$-group. By [HW(1)99, Lemma 3.6], $K/F$ is purely 
wild if and only if $Frac(E)/Frac(F)$ is purely inseparable.

Let $F$ be  a graded field and $Frac(F)_{alg}$ an algebraic closure of 
$Frac(F)$. We recall that for any element $\lambda$ of the divisible hull 
$\Delta_F$($=\Gamma_F\otimes_{{\it\mathsf Z\!\!\mathsf 
Z}}Q\!\!\!\!\Huge\prime\;$)   of $\Gamma_F$, 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 $gr(X)=\lambda$. We denote this grading by 
$F[X]^{(\lambda)}$. A polynomial $f \in F[X]$  is called homogenizable if 
there is $\lambda \in \Delta_F$ such that $ f \in (F[X]^{(\lambda)})^h$.
Let $x \in Frac(F)_{alg}$ and denote by $f_{x, Frac(F)}$ its minimal 
polynomial over $Frac(F)$. We say that $x$ is algebraic  over $F$ if $f_{x, 
Frac(F)}$ 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 $F_{alg}=F[\{x \in Frac(F)_{alg}$ $|$ $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 $Frac(F)_{alg}$ 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 $a\in A^*$ (See [B98, \S4] or [HW(2)99, 
\S4]).   We denote by $HCq(A)$ the algebra $Cq(A)\otimes_{Frac(F)}HFrac(F)$, 
where $HFrac(F)$ is the Henselization of $Frac(F)$ with respect to its 
canonical valuation (see [E72, \S16]).

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 $E^{\gamma}=\{x\in E$ $|$ $ v(x)\geq \gamma\}$, $E^{>\gamma}$ = 
$\{x\in E$ $|$ $ v(x)> \gamma\}$. Obviously, $E^{>\gamma}$  is a subgroup of 
the additive group $E^{\gamma}$. So, we can define the quotient group 
$GE_{\gamma}=E^{\gamma}/E^{>\gamma}$. For $x\in E\backslash\{0\}$, we denote 
by $\tilde x$ the element $x+E^{>v(x)}$ of $GE_{v(x)}$. One can easily see 
that the additive group $GE=\oplus_{\gamma\in \Gamma}GE_{\gamma}$ with the 
multiplication law defined by $\tilde x\tilde y=\widetilde{xy}$ 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, \S4] or [HW(2)99, \S4]).  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 $x\mapsto \tilde x$, where $x$ is an arbitrary homogeneous 
element of $A$. Indeed, we have $A_0=(GHCq(A))_0$ and 
$\Gamma_A=\Gamma_{GHCq(A)}$. We write $A\cong_g GHCq(A)$.

A valued central division algebra $D$ over a  field $E$ is called defectless 
(over $E$) if $[D : E]=[\bar D : \bar E](\Gamma_D : \Gamma_E)$, where $[\bar 
D : \bar E]$ [resp. $(\Gamma_D : \Gamma_E)$] is the residue  degree [resp. 
ramification index] of $D$ over $E$. If in addition, $\bar D$ is a field and 
$[\bar D :\bar E]=(\Gamma_D : \Gamma_E)$, 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 $$\theta_D:   \Gamma_D/\Gamma_E  
\rightarrow  Gal(Z(\bar D)/\bar E)$$   $$  \gamma+\Gamma_E  \mapsto  
\theta_D(\gamma+\Gamma_E): \bar a \mapsto \overline {dad^{-1}}$$ ($d$ being 
an arbitrary element of $D$ such that $v(d)=\gamma$) 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 $Z(\bar D)$ of $\bar D$ is 
separable  over $\bar E$ and the characteristic of $\bar E$ does not divide 
the cardinality of the kernel of $\theta_D$.

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 
$HFrac(F)$ (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 $[D_0 : 
E_0]=(\Gamma_D : \Gamma_E)$ [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 $GBr(F)\rightarrow TBr(HFrac(F))$, where 
$TBr(HFrac(F))$ is the tame part of $Br(HFrac(F))$ [HW(2)99, Theorem 5.1]. 
Conversely, for any Henselian valued field $E$, there is a canonical group 
isomorphism $TBr(E)\rightarrow GBr(GE)$ [HW(2)99, Theorem 5.3].

Now, let $M/E$ be a finite-dimensional abelian field extension with Galois 
group $G$; $S=(\sigma_i)_{1\leq i \leq r}$ be a basis of $G$ (i.e. such that 
$G=\langle \sigma_1\rangle \oplus ...\oplus\langle \sigma_r\rangle$); 
$q_i=ord(\sigma_i)$ for $1\leq i \leq r$; $M_i$ (resp. $M_{ij}$) be the 
subfield of $M$ fixed by $\sigma_i$ (resp. $\sigma_i$ and $\sigma_j$); $N_i$ 
(resp. $N_{ij}$) be the norm of $M/M_i$ (resp. $M/M_{ij}$); and  
$I=\{(m_1,..., m_r) \in I\!\!N^r$ $|$ $ 0\leq m_i < q_i$ for $ 1\leq i \leq 
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) $A=\oplus_{\bar m \in I}Mx^{\bar m}$, where $x^{\bar 
m}=x_1^{m_1}...x_r^{m_r}$ ;\\
(0.2) $x_ia=\sigma_i(a)x_i$ for  $a \in M$ and $1\leq i \leq r$.\\
Let $u_{ij}=x_ix_jx_i^{-1}x_j^{-1}$ and $b_i=x_i^{q_i}$ for $1\leq i, j \leq 
r$, then the matrices $U=(u_{ij})_{1\leq i, j \leq r}$ and $B=(b_i)_{1\leq i 
\leq r}$ satisfy the following conditions (for $1 \leq i, j, k \leq r$) :\\
(0.3) $u_{ij}$ and $b_i$ are non-zero elements of $M$ ;\\
(0.4) $u_{ji}=u_{ij}^{-1}$ ;\\
(0.5) $\sigma_i(u_{jk})\sigma_j(u_{ki})\sigma_k(u_{ij})=u_{jk}u_{ki}u_{ij}$ 
;\\
(0.6) $N_{ij}(u_{ij})=1$ ;\\
(0.7) $\sigma_j(b_i)b_i^{-1}=N_i(u_{ji})$.

Conversely, for any pair of matrices $(U, B)$ satisfying conditions 0.3 to 
0.7, we can construct a central simple $E$-algebra $(M, G, U, 
B)=\oplus_{\bar m\in I}Mz^{\bar m}$, where $z_i$ are independent 
indeterminates that satisfy the  following  conditions : 
$z_ia=\sigma_i(a)z_i$, $z_iz_j=u_{ij}z_jz_i$ 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, \S 1, p. 421]. 
Hence he denoted $(U, B)\in Z^2_S(G, M^*)$. 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 $U\in U_S(G, M^*)$ is said to be {\it 
degenerate} (with respect to the field extension $M/E$) if there exist $\bar 
m, \bar n \in I$ and $a, b \in M^*$ such that the subgroup 
$\langle\sigma^{\bar m}, \sigma^{\bar n}\rangle$ of $G$, generated by 
$\sigma^{\bar m}$ and $\sigma^{\bar n}$ is non-cyclic, and $u_{\bar m, \bar 
n}=\sigma^{\bar m}(a)a^{-1}\sigma^{\bar n}(b)b^{-1}$,  where $u_{\bar m, 
\bar n}=x^{\bar m}x^{\bar n}(x^{\bar m})^{-1}(x^{\bar n})^{-1}$ [AS78, 
Definition, p. 81].

Now, for a matrix $U=(u_{ij})_{1\leq i, j\leq r}$ of $U_S(G, M^*)$, consider 
the amalgamated sum $M.E\langle z_1,..., z_r\rangle$ over $E$ of $M$ and the 
free $E$-algebra on the independent indeterminates $z_i$ ($1\leq i\leq r$); 
then let $I$ be the ideal of $M.E\langle z_1,..., z_r\rangle$, generated by 
the elements $z_iz_j-u_{ij}z_jz_i$ and $z_ia-\sigma_i(a)z_i$, for $1\leq i, 
j\leq r$ and $a\in M$. We recall  that the generic abelian crossed product 
$(M, G, U)$ is the algebra of central quotients of $M.E\langle z_1,..., 
z_r\rangle/I$.  Let $B=(b_i)_{1\leq i\leq r}$ be a matrix of $M$ such that 
$(U, B)\in Z^2_S(G, M^*)$ 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, \S 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 $E(Z_1,..., Z_r)=Frac(F)$ 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 $\Gamma_F$ and  $MF$ the graded 
subfield  of $F_{alg}$ generated by $M$ and  $F$.  Assume there is a totally 
ordered abelian group $\Delta$ containing $\Gamma_F$ with ordering extending 
that of $\Gamma_F$ and suppose $\epsilon : \Delta \rightarrow G$ is a group 
homomorphism with kernel $\Gamma_F$. Let $\delta_i\in \Delta$ such that 
$\epsilon(\delta_i)=\sigma_i$,   and consider a family 
$(x_{q_i\delta_i})_{1\leq i\leq r}$ of elements of $F^*$ with $x_0=1$ and 
$gr(x_{q_i\delta_i})=q_i\delta_i$. For a cocycle $(U, B)\in Z^2_S(G, M^*)$, 
consider the iterated twisted polynomial ring $A:=MF[X_1, ..., X_r ;
\sigma_1, ..., \sigma_r]$ defined by $X_iX_j=u_{ij}X_jX_i$, 
$X_ia=\sigma_i(a)X_i$ for all $1\leq i, j\leq r$ and $a\in MF$ (see [AS, p. 
78]), and let  $I$ be the left ideal of $A$, generated by the elements 
$X_i^{q_i}-b_ix_{q_i\delta_i}$. Then, $I$ is a two-sided ideal of $A$. 
Indeed, for any $1\leq i, j\leq r$, we have 
$(X_i^{q_i}-b_ix_{q_i\delta_i})X_j=N_i(u_{ij})X_jX_i^{q_i} - 
X_j\sigma_j^{-1}(b_i)x_{q_i\delta_i}=X_j(\sigma_j^{-1}(N_i(u_{ij}))X_i^{q_i}-\sigma_j^{-1}(b_i)x_{q_i\delta_i})=X_j\sigma_j^{-1}(N_i(u_{ij}))(X_i^{q_i}-b_ix_{q_i\delta_i})$. 
So, we can consider the quotient ring $(MF, \epsilon, G, U, B):=A/I$. One 
can easily see that $(MF,\epsilon, G, U, B)$ 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 $E$, $g=\sum_{i=0}^n\tilde{a_i}X^i$ is 
a homogenizable irreducible polynomial of $GE[X]$ ($a_i \in E$) with a  
simple root $\alpha $ in $GM$ and  $f=\sum_{i=0}^na_iX^i$, then $f$ is 
irreducible and has a simple root $a$ in $M$ such that $\tilde a= \alpha$ 
[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 $D_0Frac(F)$ 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 
$\Gamma_D/\Gamma_E$  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 
$\Gamma_D/\Gamma_E$ 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 
$Gal(\bar D/\bar E)$ such that $Gal(K/E)\cong Gal(\bar D/\bar E)/H$. In 
particular, if $K$ is a Galois maximal subfield of $D$, then $Gal(K/E)\cong 
Gal(\bar D/\bar E)$ [Proposition 4.3]. This Proposition generalizes [S78, 
Theorem 3.1] and [McK05, Theorem 2.3]. We prove also that if $char(\bar E)$ 
is arbitrary, $deg(D)$ is a power of a prime integer $p$  and 
$rk(\Gamma_D/\Gamma_E)\geq 3$, then all abelian subfield of $D$ are inertial 
over $E$ [Proposition 4.9]. Then, we show that if $rk(\Gamma_D/\Gamma_E)$ is 
arbitrary but $exp(\Gamma_D/\Gamma_E)=p$, 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 $\leq p^2$ or 
inertial over $E$ [Theorem 4.11]. More results for the case where  
$char(\bar E)$ 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).

\section {Graded and valued field extensions}
\setcounter{Def}{0}
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 $x\in 
E\backslash\{0\}$ and let $f_{x, Frac(F)}$ be the minimal polynomial of $x$ 
over $Frac(F)$.  By  Claim A.1  comments below, we have $g=af_{x, Frac(F)}$ 
for some $a\in F^*$. So, by [HW(1)99, Proposition 2.2 and corollary 2.5] 
$x\in E^*$ . Therefore, $E$ is normal over $F$ if and only if for any $x\in 
E^*$, $f_{x, Frac(F)}$ splits into one degree polynomials in $E[X]$.

\begin {Lem}Let $E/F$ be an algebraic graded field extension. Then, $E$ is 
normal over $F$ if and only if $Frac(E)$ is a normal field extension of 
$Frac(F)$.
\end {Lem}

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


\begin {Pro} 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 $F$.
\end {Pro}

{\it 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 $E/F$.

\begin {Pro} Let $E/F$ be a finite-dimensional normal graded field extension 
  of Galois group $G$ and let $Fix_G(E)$ be the graded subfield of $E$, 
elementwise invariant under the action of $G$. Then, $Fix_G(E)$ is purely 
wild over $F$ and $E$ is Galois over $Fix_G(E)$. Moreover, if $T$ is the 
tame closure of $F$ in $E$, then $E=TFix_G(E)\cong_gT\otimes_FFix_G(E)$.
\end {Pro}

{\it Proof.} Since $E$ is normal over $F$, then by Lemma 1.1 
$Frac(E)/Frac(F)$ is a normal field extension. Moreover, we have 
$Frac(E)\cong E \otimes_FFrac(F)$, so  $Gal(E/F)=Gal(Frac(E)/Frac(F))$ -up 
to a group isomorphism- (See that by  [HW(1)99, Corollary 2.5(d)] the 
restriction to $E$ of any $\sigma\in Gal(Frac(E)/Frac(F))$  is in 
$Gal(E/F)$). By identification of $Frac(E)$ and $E\otimes_FFrac(F)$, for any 
element $x \in Frac(E)$ there are $\alpha \in E$ and $\beta \in Frac(F)$ 
such that $x=\alpha \otimes \beta$ (indeed, we can write $x=\sum a_i\otimes 
b_i$ where $a_i\in E$ and $b_i\in Frac(F)$. Let $0\neq b\in F$ such that 
$h_i:=b_ib\in F$ for all $i$, then we have $x=(\sum a_ih_i)\otimes b^{-1}$). 
So, for any $\sigma\in Gal(Frac(E)/Frac(F))$, $\sigma(x)=x$ if and only if 
$\sigma(\alpha)=\alpha$. Accordingly, $Fix_G(Frac(E))=Fix_G(E) 
\otimes_FFrac(F)=Frac(Fix_G(E))$. Therefore, by [Karp89, Proposition 7.7, 
p.283] and [HW(1)99, Theorem 3.11(b) and Lemma 3.6], $E/Fix_G(E)$ is Galois 
and $Fix_G(E)/F$ is purely wild. By [HW(1)99, (3.8)], $Frac(T)$ is the 
separable closure of $Frac(F)$ in $Frac(E)$. Furthermore, we have 
$TFix_G(E)\otimes_F Frac(F)=Frac(T)Fix_G(Frac(E))$. So, by [Karp89, 
Proposition 7.7, p.283] $TFix_G(E)\otimes_F Frac(F)=Frac(E)$. Therefore, 
$TFix_G(E)=E$. Remark that $Frac(T)$ is separable over $Frac(F)$ and 
$Frac(Fix_G(E))$ is purely inseparable over $Frac(F)$. Therefore,  $T$ and 
$Fix_G(E)$ are linearly disjoint over $F$. Hence, 
$E\cong_gT\otimes_FFix_G(E)$.\\

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 $1\leq i, j\leq n$   and such that 
$f:=\prod_{i=1}^n(X-x_i)$ is an element of $E[X]$. We aim to show here that 
$\prod_{i=1}^n(X-\tilde x_i)$ is an element of $GE[X]$. Let for $\bar 
m=(m_1,..., m_n)\in  I\!\! N^n$, $supp(\bar m)=\{i $ $|$  such that $m_i\neq 
0\}$, then let $I^n_k=\{\bar m=(m_1,..., m_n)\in  I\!\! N^n$ $|$ 
$card(supp(\bar m))=k$ and $m_i=0$ or $1$ for all $1\leq i\leq n\}$. For any 
$\bar m=(m_1, ..., m_n)\in I^n_k$, denote $x^{\bar m}=\prod_{i\in supp(\bar 
m)} x_i$. Then, $\prod_{i=1}^n(X-x_i)=\sum_{i=0}^nd_kX^{n-k}$ [resp. 
$\prod_{i=1}^n(X-\tilde x_i)=\sum_{i=0}^nd'_kX^{n-k}$] , where $d_0=1$ 
[resp. $d'_0=1$] and $d_k=\sum_{\bar m \in I^n_k}(-1)^k x^{\bar m}$ [resp. 
$d'_k=\sum_{\bar m \in I^n_k}(-1)^k \tilde x^{\bar m}$] (for $1\leq k\leq 
n$). See that $v(x^{\bar m})=kv(x)$ for any $\bar m\in I^n_k$. So, in  
$GE_{kv(x)}(=E^{kv(x)}/E^{>kv(x)}$) we have  $\bar d_k=\overline{\sum_{\bar 
m \in I^n_k}(-1)^k x^{\bar m}}=\sum_{\bar m \in I^n_k}(-1)^k \tilde x^{\bar 
m}=d'_k$. So, $\prod_{i=1}^n(X-\tilde x_i)\in GE[X]$.

\begin {Th} 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 $E$.
\end {Th}

{\it Proof.}  Let $v$ be the valuation of $M$ and let $HM$ be the 
henselization of $M$ with respect to $v$ [E72, \S16]. Since $\bar 
M\otimes_{\bar E}\overline {HE}=\bar M$ and $\Gamma_M\cap 
\Gamma_{HE}=\Gamma_E$, then by [Mor89,  Theorem 1] $M\otimes_EHE$ is a 
field. So, $HM=MHE=M\otimes_EHE$. Moreover, because $M\subseteq HM$, 
$(GM)_0=\bar M=(GHM)_0$ and $\Gamma_{GM}=\Gamma_{GHM}$, 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 
$f_{x, E}=\prod_{i=1}^n(X-x_i)$, where $x_i=\sigma_i(x)$ for some 
$\sigma_i\in Gal(M/E)$. Since $E$ is Henselian, then $v(x_i)=v(x)$. 
Therefore, by the above the polynomial $\prod_{i=1}^n(X-\tilde x_i)$ is in 
$GE[X]$.  Hence, the minimal polynomial $f_{\tilde x, Frac(GE)}$ of $\tilde 
x$ over $Frac(GE)$ splits into one degree polynomials in $GM[X]$.

\begin {Pro} Let $E$ be a Henselian valued field, $M$ a finite-dimensional 
field extension of $E$, $g=\sum_{i=0}^n\tilde{a_i}X^i$ a homogenizable 
irreducible polynomial of $GE[X]$, with $a_i \in E$, and 
$f=\sum_{i=0}^na_iX^i$. If $g$ has a simple root $\alpha $ in $GM$, then $f$ 
is irreducible and has a simple root $a$ in $M$ such that $\tilde a= 
\alpha$.
\end {Pro}

{\it Proof.}   Since $X-\alpha$ divides $g$ and $g$ is homogenizable, then 
$\alpha \in GM^h$ and $gr(\alpha)=gr(X)$. Therefore, $\alpha=\tilde{e}$ for 
some $e \in M$. Let $f_{\alpha, Frac(GE)}$ be the minimal polynomial of 
$\alpha $ over $Frac(GE)$.  Then, by Claim A.1 comments below,  $g=\beta 
f_{\alpha, Frac(GE)}$ for some $\beta \in GE^h$. Let $\Gamma_M$ be the value 
group of $M$ (for the extension of the valuation of $E$ to $M$), $\Delta_M= 
\Gamma_M \otimes_{{\large \it\mathsf Z\!\!\mathsf Z}}Q\!\!\!\!\Huge\prime 
\;$  (the divisible hull of $\Gamma_M$) and $\mu=gr(g)(\in \Delta_M)$. Then, 
$ gr(\tilde {a_i})+igr(\tilde e)=\mu$ for all $1\leq i \leq n$ when $a_i\neq 
0$.  See that   $\tilde a_0\neq 0$  (because $g$ is irreducible). Hence 
$e\neq 0$.\\
We distinguish the following two cases:\\
i) If $\mu=0$, then setting $Y=\tilde{e}^{-1}X$, we get 
$g=\sum_{i=0}^n\overline{a_ie^i}Y^i$, with $Y=1$ as a simple root in $\bar 
M$. Therefore, by Hensel Lemma [E72, Corollary 16.6(iv)], there is a $b \in 
V_M$ (where $V_M$ is the valuation ring of the extension of the valuation of 
$E$ to $M$) such that $\bar b =1$ and $\sum_{i=0}^na_i(eb)^i=0$. So, $eb$ is 
a root of $f$ and $\widetilde {eb}=\tilde e=\alpha$.\\
ii) If $\mu \neq 0$, then setting $c=a_ke^k$, where $a_k$ is a nonzero 
coefficient of $f$, $Y=\tilde{e}^{-1}X$ and $h=\tilde c^{-1}g$, we get 
$h=\sum_{i=0}^n\overline {c^{-1}a_ie^i}Y^i$. So again by Hensel Lemma, there 
is $b \in V_M$ such that $c^{-1}\sum_{i=0}^n a_i(eb)^i=0$ and $\bar b=1$. 
Hence, $eb$ is a root of $f$ with $\widetilde{eb}=\alpha$.\\
Now to prove that $f$ is irreducible,  remark that because $g=\beta 
f_{\alpha, Frac(GE)}$, then $[GE[\tilde a] : GE]=deg(g)=deg(f)$. Moreover, 
since $f(a)=0$, then the minimal polynomial $f_{a, E}$ of $a$ over $E$ 
divides $f$, hence $[E[a] : E]\leq deg(f)$. On the other hand, we have 
$[E[a] : E]\geq [G(E[a]) : GE]\geq [GE[\tilde a] : GE]$. Therefore, $[E[a] : 
E]=[G(E[a]) : GE]=[GE[\tilde a] : GE]=deg(g)=deg(f)$. So, $f$ is 
irreducible. (The reader can see that $E[a]$ is necessaraly defectless over 
$E$).\\
$\alpha$ 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 
$\delta$ of $g$ there is a root $d$ of $f$ in $N$ such that $\tilde 
d=\delta$. Therefore,  $f$ has $deg(f)$ distinct roots. This shows that $a$ 
is a simple root of $f$.\\\\
{\bf Remark 1.6}   Let $E$ be a Henselian valued field,  
$g=\sum_{i=0}^n\tilde{a_i}X^i$ a homogenizable irreducible polynomial of 
$GE[X]$, with $a_i \in E$, and $f=\sum_{i=0}^na_iX^i$. If $f$  has a root 
$a$ such that $\tilde a$ is a root of $g$, then using the same arguments as 
in Proposition 1.5, $f$ is irreducible in $E[X]$ and $G(E[a])=GE[\tilde a]$.

\setcounter{Def}{6}

\begin{Pro} 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 $E$.
\end {Pro}

{\it Proof.} Let $N$ be a normal field extension of $E$ such that 
$L\subseteq GN$. Assume first that $L=GE[\tilde a]$ (for some $a\in N$)  and 
let $p^n=[L : GE]$. Since $Frac(L)$ is  purely inseparable over $Frac(GE)$, 
then the minimal polynomial of $\tilde a$ over $Frac(GE)$ is  
$g:=X^{p^n}-\tilde a^{p^n}$. We have $L\cap Frac(GE)\subseteq GN\cap 
Frac(GE)=GE$. So,  there is $b\in E$ such that $\tilde a^{p^n}=\tilde b$. 
Let $f=X^{p^n}-b$ and let  $x$ be a root of $f$ in $N$. Clearly, $\tilde x$ 
is a root of $g$. So, $\tilde x=\tilde a$ (because $\tilde x^{p^n}=\tilde 
a^{p^n}$ in $GN$). By remark 1.6, we have  $G(E[x])=GE[\tilde x]=GE[\tilde 
a]=L$. \\
Now, let $L$ be an arbitrary finite-dimensional purely wild graded field 
extension  of $GE$. Then, we can write $L=GE[\tilde a_1, ..., \tilde a_r]$. 
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.

\begin {Cor} $[$HW(1)99, Theorem 5.2$]$ Let $E$ be a Henselian valued field. 
  Then, the map $K\mapsto GK$ 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 $Gal(GK/GE)$.
\end {Cor}

{\it 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 $K'\cong K$. Since $E$ is Henselian, then clearly 
$GK\cong_gGK'$.
Conversely, if $L$ is a tame finite-dimensional graded field extension of 
$GE$, then we can write $L=GE[\tilde x_1, ..., \tilde x_r]$, where $x_i\in 
E_{alg}$ with $\tilde x_i$ separable over $Frac(GE)$. Assume  first that 
$r=1$ and let $M=E[x_1]$. By Proposition 1.5, there is $a_1\in M$ such that 
$\tilde a_1=\tilde x_1$ and $[E[a_1] : E]=[L : GE]$. Hence, $G(E[a_1])=L$. 
By induction, assume  there are $a_1, ..., a_{r-1}\in E[x_1,..., x_{r-1}]$ 
such that $G(E[a_1,..., a_{r-1}])=GE[\tilde x_1, ..., \tilde x_{r-1}]$. 
Then, again by Proposition 1.5, there is $a_r\in E[x_1,..., x_r]$ 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 $GK'\cong_gL$, then by Proposition 1.5, there is 
$a'_1\in K'$ such that $G(E[a'_1])\cong_gG(E[a_1])$ and both $a'_1$ and 
$a_1$ have the same minimal polynomial over $E$. So, $E[a'_1]\cong E[a_1]$. 
By induction, we prove that $K'\cong K$. \\
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 $\sigma\in Gal(K/E)$, we can 
define a graded $GE$-automorphism $\tilde \sigma : GK \rightarrow GK$ by 
extending the mapping $\tilde x\mapsto \widetilde{\sigma(x)}$. Let $\phi : G 
\rightarrow Gal(GK/GE)$ be the group homomorphism defined by 
$\phi(\sigma)=\tilde\sigma$, $B$ be the valuation ring of the extension of 
$v$ to $K$ and let $G^v$ the ramification group of $B$ over $E$. See that 
$\phi(\sigma)=id_{GK}$ if and only if $v(\frac{\sigma(x)}{x}-1)>0$, for any 
$x\in K^*$. Otherwise said, $\phi(\sigma)=id_{GK}$ if and only if $\sigma 
\in G^v$. But $G^v=\{id_K\}$ (since $K$ is tame over $E$). Therefore, $\phi$ 
is injective. Remark that $Gal(K/E)$ and $Gal(GK/GE)$ have the same 
cardinality (since $Gal(GK/GE)\cong Gal(Frac(GE)/Frac(GE))$). So, $\phi$ 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 
$Gal(N/E)\cong Gal(GN/GE)$ [resp., $Gal(N/M)\cong Gal(GN/GM)$]. 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 
$A=\Gamma_K/\Gamma_F$- 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 
$\langle a_1, a_2,.., a_r\rangle$ of A, generated by $a_1$, $a_2$..., $a_r$, 
equals $\oplus_{i=1}^r\langle a_i\rangle$. We recall that $K$ is called 
totally ramified of radical type ($TRRT$) over $F$ if there are homogeneous 
elements $t_1,..., t_r \in F^*$ 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 \\
$(2)$ $gr(t_i^{1/n_i})+\Gamma_F$ $(1\leq i\leq r)$ are independent elements 
of $\Gamma_K/\Gamma_F$, with order $n_i$, respectively.

One can see that in the same way as for $TRRT$ valued field extensions (See 
[JW90, Lemma 4.1]), a totally ramified finite-dimensional  graded field 
extension $K$ of $F$  is $TRRT$ over $F$ if and only if there is a subgroup 
$G$ of $K^*/F^*$ such that the mapping $G\rightarrow \Gamma_K/\Gamma_F$, 
defined by $xF^*\mapsto gr(x)+\Gamma_F$, is a group isomorphism.

\begin {Lem} Let $F$ be a graded field and $K$ a totally ramified 
finite-dimensional  graded field extension of $F$. Then, $K$ is $TRRT$ over 
$F$.
\end {Lem}

{\it Proof.} [M05, Lemma 1].

\begin {Cor} $[$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.
\end {Cor}

{\it Proof.} By  Lemma 1.9 $GK$ is totally ramified of radical type. Write 
$GK=GE[\alpha_1]\otimes_{GE}...\otimes_{GE}GE[\alpha_r]$ for some 
$\alpha_1,...,\alpha_r \in GK^*$ where $gr(\alpha_i)+\Gamma_{GE}$ are 
independent elements of $\Gamma_{GK}/\Gamma_{GE}$. Assume  first that $r=1$. 
By Proposition 1.5 there is $x_1\in K$ such that $\tilde x_1=\alpha_1$ and 
$[E[x_1] : E]=[GE[\alpha_1] : GE]=[GK : GE]$. Hence, $E[x_1]=K$ and 
$v(E[x_1])=\langle gr(\alpha_1)+\Gamma_{GE}\rangle=\langle 
v(x_1)+\Gamma_E\rangle$. More generally, for an arbitrary positive integer 
$r$, by the above there are $x_i\in K$ such that $[E[x_i] : E]=[GE[\alpha_i] 
: GE]$ and $v(x_i)+\Gamma_E$ generates $\Gamma_{E[x_i]}/\Gamma_E$. Hence, by 
[Mor89, Theorem 1] $K=E[x_1]\otimes_E...\otimes_EE[x_r]$.\\

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. 
$N_{E/F}=\prod_{\sigma\in Gal(E/F)}\sigma(x)$ for all $x\in E$). Then, the 
following lemma is a direct consequence of [L84, Theorem 6.1, p323].

\begin {Lem} Let $E/F$ be a finite-dimensional cyclic  graded field 
extension with dimension $n$ and with Galois group generated by $\sigma$ and 
$x \in E^*$. Then,  $N_{E/F}(x)=1 $ if and only if there exists $y \in E^*$ 
such that $x=y\sigma(y)^{-1}$
\end {Lem}
{\it Proof.} Assume that $N_{E/F}(x)=1 $. Then,  
$N_{Frac(E)/Frac(F)}(x)(=N_{E/F}(x))=1$. So,  by [L84, Theorem 6.1, p. 323] 
there is $z \in Frac(E)^*$ such that $x=z\sigma(z)^{-1}$. We may assume $z 
\in E^*$. Write $z=z_1+...+z_r$, where $z_i$ ($1\leq i \leq r$) are nonzero 
homogeneous elements of $E$ with $gr(z_i)<gr(z_{i+1})$, for all $1 \leq 
i<r$. Since $\sigma(z)x=z$ and $x$ is homogeneous, then for all $1\leq i 
\leq r$, $\sigma(z_i)x=z_i$.
The converse is clear.

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


\begin {Pro} 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 $\zeta$. Then, the following statements hold :\\
(1) If $E$ is a cyclic graded field extension of $F$ with dimension $n$, 
then there is $x \in E^*$ such that $E=F[x]$, $x^n \in F^*$ and $Gal(E/F)$ 
is generated by the graded $F$-isomorphism $\sigma$ defined by 
$\sigma(x)=\zeta x$.\\
(2) Conversely, if $a \in F^*$ 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  $x^m \in F$.
\end {Pro}
{\it Proof.}  $(1)$ Let  $\sigma$ a generator of $Gal(E/F)$. We have  
$N_{E/F}(\zeta ^{-1})=1$, so by Lemma 1.11 there is $x \in E^*$ such that 
$\sigma (x)=\zeta x$. Accordingly, $\sigma(x^n)=\sigma(x)^n=(\zeta 
x)^n=x^n$. Hence, $x^n \in F^*$. Since $\sigma^i(x)(=\zeta^ix)$ ($1 \leq i 
\leq n$) are pairwise distinct, then the minimal polynomial of $x$ over 
$Frac(F)$ is $X^n-x^n$. So, $F[x]=E$.\\
$(2)$  By  [L84, Theorem 6.2, p. 324] $Frac(F)(x)/Frac(F)$ is cyclic with 
dimension $m$ dividing $n$ and with $x^m \in Frac(F)^*$. Hence,  $F[x]/F$ is 
cyclic of dimension $m$ and by [HW(1)99, Corollary 2.5(b)] $x^m \in F[x]^* 
\cap Frac(F)^*=F^*$.\\

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  $KFF$, 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 $x=(x_1,..., x_n)\in W_n(E_0)$ and 
$(x_1^p,..., x_n^p)-(x_1,..., x_n)\in W_n(F_0)$ (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 $a\in F_0$ with $x\notin F_0$.\\


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 $KUM(K/F)=\{x\in K^*$ $|$ 
$ x^m\in 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 $Frac(F)$ does (indeed, if $\xi$ is a primitive $n^{th}$ root of unity in 
$Frac(F)$, then clearly $F[\xi]$ is a graded field extension of $F$. So, 
$[F[\xi] : F]=[Frac(F[\xi]) : Frac(F)]=1$. Hence $\xi\in F_0$).  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 $Frac(K)$ [resp. 
$HFrac(K)$] is a Kummer field extension of $Frac(F)$ [resp. $HFrac(F)$]. 
Moreover, if  $K$ is a Kummer graded field extension of $F$, then  
$kum(K/F)$ embeds canonically in $kum(Frac(K)/Frac(F))$ [resp. in 
$kum(HFrac(K)/HFrac(F))$].  Since both groups have the same cardinality, 
then -up to a group isomorphism- $kum(Frac(K)/Frac(F))=kum(K/F)$ [resp. 
$kum(HFrac(K)/HFrac(F))=kum(K/F)$]. Obviously, this isomorphism can be 
deduced easily from the fact that $Gal(K/F)\cong_gGal(Frac(K)/Frac(F))$ 
[resp., $Gal(K/F)\cong_gGal(HFrac(K)/HFrac(F))$], but the above shows   that 
  every class of $kum(Frac(K)/Frac(F))$ [resp. $kum(HFrac(K)/HFrac(F))$] can 
be represented by an element of $KUM(K/F)$.

\begin {Pro} Let $F$ be a graded field and $K$ a totally ramified graded 
field extension of $F$. Let $m$ be the exponent of $\Gamma_K/\Gamma_F$. 
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 $Gal(K/F)\cong kum(K/F)\cong 
\Gamma_K/\Gamma_F$.
\end {Pro}

{\it Proof.} By  [HW(1)99, Proposition 3.3].


\section {Nicely semiramified division algebras over
Henselian valued fields}
\setcounter{Def}{0}

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 
$Frac(F)$) ([B98, Proposition 6.4]). Moreover, $Cq(D)$ is $NSR$ (over 
$Frac(F)$) if and only if $HCq(D)$ is $NSR$ (over $HFrac(F)$). Indeed, 
assume that $Cq(D)$ is $NSR$  and let $K$ [resp. $L$] be an inertial [resp. 
a $TRRT$] maximal subfield of $Cq(D)$. Then, by [Mor89, Theorem 1], 
$HK(=K\otimes_{Frac(F)}HFrac(F))$ [resp. $HL(=L\otimes_{Frac(F)}HFrac(F))$] 
is an inertial [resp. a $TRRT$] maximal subfield of $HCq(D)$. Conversely, if 
$HCq(D)$ is $NSR$, then it has an inertial [resp. a $TRRT$] maximal subfield 
$K'$ [resp. $L'$]. So $GK'$ [resp. $GL'$] is an inertial [resp. a totally 
ramified] maximal graded subfield of  $D(\cong_gGHCq(D))$. 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  
$L^{(1)}\otimes_F....\otimes_F L^{(k)}$ is a graded field.

\begin {Lem} 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) $D\cong_g (L^{(1)}/F, \sigma_1, t_1)\otimes_F...\otimes_F(L^{(k)}/F, 
\sigma_k, t_k)$, where $L^{(i)}$, $\sigma_i$ 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 $\sigma_i$ ($1\leq i\leq k$), $\prod_{i=1}^kn_i=n$,

$(ii)$  $t_i$ are nonzero homogeneous elements of $F$ such that 
$gr(t_i^{n/n_i})+n\Gamma_F$ are independent elements of 
$\Gamma_F/n\Gamma_F$, with order $n_i$ ($1\leq i\leq k$), respectively.

In such a case, $\Gamma_D/\Gamma_F=\oplus_{i=1}^k\langle 
gr(t_i)+\Gamma_F\rangle$.
\end {Lem}

{\it Proof.} [M05, Lemma 2].

\begin {Lem} 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 $GE$.
\end {Lem}

{\it Proof.} [M05, Lemma 3].

\begin {Th} 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 $E$.
\end {Th}

{\it Proof.} [M05, Theorem 4].



\begin {Cor} 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 $E$.
\end {Cor}

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

\begin {Pro} 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.
\end {Pro}

{\it Proof.} $(1) \Rightarrow (2)$ This follows by [JW90, Theorem 4.4].\\
$(2) \Rightarrow (3)$ Assume that $D=D_1\otimes_E...\otimes_ED_r$, 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,  
$K:=K_1\otimes_E...\otimes_EK_r$ is a purely inseparable maximal subfield of 
$D$.\\
$(3) \Rightarrow (1)$ Let $K$ be a purely inseparable maximal subfield of 
$D$. Since $D$ is defectless over $E$ and $\bar D$ is separable over $\bar 
E$, then $K$ is totally ramified over $E$. So,  $D$ is nicely semiramified 
(by Corollary 2.4).



\begin {Pro} 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 $\bar D$ 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 $D=D_1\otimes_ED_2$ with $D_1$ and $D_2$ 
inertially split. Then, each $D_i$ is semiramified, $\bar D=\bar 
D_1\otimes_{\bar E}\bar D_2$, and 
$\Gamma_D/\Gamma_E=\Gamma_{D_1}/\Gamma_E\times\Gamma_{D_2}/\Gamma_E$ (via 
the canonical inclusion $\Gamma_{D_i}/\Gamma_E\rightarrow 
\Gamma_D/\Gamma_E$). 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 $\Gamma_D/\Gamma_E$ is cyclic.
\end {Pro}

{\it Proof. } $(1)$ This follows easily from the equality $(\Gamma_D : 
\Gamma_E)=[Z(\bar D) : \bar E]$ obtained from the isomorphism $\theta_D$, 
since $Z(\bar D)$ is separable over $\bar E$ (see [JW90, Lemma 5.1]).\\
$(2)$ Clearly, $C$ is also inertially split. If $D$ is semiramified, then by 
$(1)$ $C$ is also semiramified (because $\bar C\subseteq \bar D$). 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], $K\otimes_EL$ 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 $\bar D_i$ over $E$ in $D_i$ -see [JW90, Theorem 2.8]-, and let 
$L=L_1\otimes_EL_2$, which is a subfield of $D$. Clearly, $L$ is inertial 
over $E$, since it is a compositum of two inertial extensions of $E$. Hence, 
$\bar L=\bar D_1\otimes_{\bar E}\bar D_2$, which by dimension count must be  
all of $\bar D$. Now, $\Gamma_{D_1}/\Gamma_E$ acts trivially on $\bar D_2$ 
via $\theta_D$, as $D_1$ centralizes $D_2$; likewise $\Gamma_{D_2}/\Gamma_E$ 
acts trivially on $\bar D_1$. Hence, $(\Gamma_{D_1}/\Gamma_E)\cap 
(\Gamma_{D_2}/\Gamma_E)$ must be trivial, since it acts trivially on all of 
$\bar D_1\otimes_{\bar E}\bar D_2=\bar D$ and $\theta_D$ is injective. 
Hence, $(\Gamma_{D_1}/\Gamma_E)\times (\Gamma_{D_2}/\Gamma_E)\subseteq 
\Gamma_D/\Gamma_E$, 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] $K_1\otimes_EK_2$ 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 $\bar K(=\bar 
D)$ is a cyclic field extension of $\bar E$. So,  $\Gamma_D/\Gamma_E$ is 
cyclic (since $\theta_D$ is an isomorphism). Let $\gamma+\Gamma_E$ be a 
generator of $\Gamma_D/\Gamma_E$ and let $x$ be an element of $D$ with 
valuation $\gamma$. Then, $E[x]$ is a totally ramified maximal subfield of 
$D$. Hence, $D$ is nicely semiramified.

\begin{Cor}  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 $\bar D$ is a field.
\end {Cor}

{\it 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.\\\\
{\bf 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.\\
\setcounter{Def}{8}

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 $A\cong B\otimes_EC$, necessarily 
$B$ or $C$ is trivial.

\begin {Pro} 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 $Frac(F)$ for its canonical valuation (i.e., such 
that $\bar K=\overline{Frac(F)}$ and $\Gamma_K=\Gamma_{Frac(F)}$). Then, the 
following statements are equivalent :\\
(1) $D$ is cyclic.\\
(2) $D_0$ is a cyclic field extension of $F_0$.\\
(3) $\Gamma_D/\Gamma_F$ 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) $Cq(D)\otimes_{Frac(F)}K$ is nicely semiramified and 
$exp(Cq(D)\otimes_{Frac(F)}K)=deg(Cq(D)\otimes_{Frac(F)}K)$.\\
(8) $Cq(D)\otimes_{Frac(F)}K$ is cyclic and indecomposable.\\
Furthermore, if $HFrac(F)\subseteq K$, then the above statements are 
equivalent to the following condition :\\
(9) $Cq(D)\otimes_{Frac(F)}K$ is cyclic and t-indecomposable.
\end {Pro}

{\it Proof.}
$(1)\Rightarrow (2)$ 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 $F_0$. 
\\
$(2)\Leftrightarrow (3)$ By [B98, Proposition 6.1].\\
$(3) \Rightarrow (4)$  Let $x\in D^*$ such that $gr(x)+\Gamma_F$ generates 
$\Gamma_D/\Gamma_F$. Then,  $F[x]$ is a simple totally ramified -hence a 
purely wild- maximal graded subfield of $D$. \\
$(4)\Rightarrow (6)$ Clearly $D_0F$ 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  $\Gamma_D/\Gamma_F(=\Gamma_L/\Gamma_F)$ is 
cyclic (because $L$ is simple over $F$). Therefore, by [B98, Proposition 
6.5] $exp(D)=exp(\Gamma_D/\Gamma_F)=(\Gamma_D : \Gamma_F)=deg(D)$. \\
$(6)\Rightarrow(5)$  Clear.\\
$(5)\Rightarrow (1)$ By  Lemma 2.1.\\
$(6)\Rightarrow (7)$  We have  $G(Cq(D)\otimes_{Frac(F)}K)\cong_gD$. 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 :
$$\begin {array}{ccc}
TBr(HFrac(F)) &\stackrel{Ext}{\rightarrow}& TBr(HK)\\
\cong \downarrow  & & \cong \downarrow \\
GBr(F) &\stackrel{Ext}{\rightarrow}& GBr(GHK)
\end {array}$$ (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 $Frac(F)$, then $GK\cong_g GHK\cong_g F$. 
Moreover, we have $G(Cq(D)\otimes_{Frac(F)}K)=G(Cq(D)\otimes_{Frac(F)}HK)$. 
Hence, $G(Cq(D)\otimes_{Frac(F)}K)\cong_gD$. Therefore, 
$Cq(D)\otimes_{Frac(F)}K$ is nicely semiramified and  
$deg(Cq(D)\otimes_{Frac(F)}K)=deg(D)=exp(D)=exp(Cq(D)\otimes_{Frac(F)}K)$. 
\\
$(7) \Rightarrow (8)$ Obviously, $Cq(D)\otimes_{Frac(F)}K$ is 
indecomposable, so $D$ is indecomposable. Moreover, as seen above, we have 
$D\cong_g G(Cq(D)\otimes_{Frac(F)}HK)$. So, $D$ is  nicely semiramified. 
Hence, by Lemma 2.1, $D$ (so $Cq(D)$) is cyclic. Therefore, 
$Cq(D)\otimes_{Frac(F)}K$ is cyclic.\\
Now, if $HFrac(F)\subseteq K$, then  $(8)\Rightarrow (9)$ is evident. \\
$(9)\Rightarrow (1)$ Since $Cq(D)\otimes_{Frac(F)}K$ is t-indecomposable, 
$GK\cong_g F$ and $G(Cq(D)\otimes_{Frac(F)}K)\cong_gD$, then $D$ is 
indecomposable (it suffices to use the canonical group isomorphism 
$TBr(K)\rightarrow GBr(GK)$). Moreover, by  [A61, Theorem 26, p. 107] 
$Cq(D)\otimes_{Frac(F)}K$  has a purely inseparable maximal subfield $L$. 
See that  $Cq(D)\otimes_{Frac(F)}K$ is defectless over $K$ and 
$\overline{Cq(D)\otimes_{Frac(F)}K}(\cong D_0)$ is separable over $\bar 
K(\cong F_0)$, so $L$ is totally ramified over $K$. Therefore, 
$Cq(D)\otimes_{Frac(F)}K$ (hence $D$) is nicely semiramified. So, by Lemma 
2.1 $D$ is cyclic.\\\\
{\bf 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 $G=\langle\sigma_1\rangle\oplus ...\oplus 
\langle\sigma_r\rangle$, where $\langle\sigma_i\rangle$ is the cyclic group 
generated by $\sigma_i$. We recall that the number $r$ is called the rank of 
$G$. We write $r=rk(G)$. \\\\
{\bf (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  $v(t_i^{1/n_i})+\Gamma_E$ 
$(1\leq i\leq r)$ are independent elements of $\Gamma_L/\Gamma_E$, with 
order $n_i$, respectively. Then, the integer $r$  depends only on $D$ 
(because $\oplus_{i=1}^r\langle 
v(t_i^{1/n_i})+\Gamma_E\rangle=\Gamma_L/\Gamma_E=\Gamma_D/\Gamma_E$). We 
call $r$ the radical length of $D$ and we write $r=rl(D)$. By definition, we 
have $rl(D)=rk(\Gamma_D/\Gamma_E)$.\\


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

\setcounter{Def}{11}

\begin {Cor} 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.
\end{Cor}

{\it Proof.} If  $D=D_1\otimes_F...\otimes_FD_r$, 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 
$F$) ;  $\Gamma_D/\Gamma_F=\Gamma_{D_1}/\Gamma_F 
\oplus...\oplus\Gamma_{D_r}/\Gamma_F$ and  $\Gamma_{D_i}/\Gamma_F$ are 
cyclic. So, $rl(D)=r$. The converse follows  by  Lemma 2.1.

\section{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 
$D\otimes_EM$ with respect to $Br(M)$. We recall that if $M$ is a subfield 
of $D$, then up to an algebra isomorphism, we have $D_M=C^M_D$.\\\\
{\bf 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 $r\geq 
2$.

\setcounter{Def}{1}

\begin {Pro} 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 
$\Gamma_K/\Gamma_E$ is non-cyclic.
\end{Pro}

{\it Proof.} Assume $D$ is degenerate and let $M$ be an inertial  subfield 
of $D$ such that $D_M$ is nicely semiramified with $rl(D_M)\geq 2$.   Let 
$K$ be a totally ramified maximal   subfield of $C^M_D$, then 
$\Gamma_{C^M_D}/\Gamma_M=\Gamma_K/\Gamma_E$. So $\Gamma_K/\Gamma_E$ is 
non-cyclic.\\
Conversely, suppose that there is a  subfield $K$ of $D$ such that 
$\Gamma_K/\Gamma_E$ is non-cyclic.  Let $L$ be a maximal  subfield of $D$ 
that contains $K$ and let $M$ be 'the' inertial lift of $\bar L$ over $E$ in 
$L$, then $L$ is a totally ramified maximal  subfield of the inertially 
split division algebra $C^M_D$. So, $C^M_D$ is  nicely semiramified  with 
$rl(C^M_D)\geq 2$.

\begin {Cor} Let $E$ be a Henselian valued field such that $\bar E$ is 
finite. Then, any inertially split division algebra $D$ of prime power 
degree over $E$ is nondegenerate cyclic nicely semiramified (over $E$) with 
$exp(D)=deg(D)$.
\end {Cor}

{\it Proof.} Since $\bar E$ is finite, then $\bar D$ is a cyclic field 
extension of $\bar E$. Moreover, since  $D$ is inertially split over $E$, 
then  $\Gamma_D/\Gamma_E$ is isomorphic to $Gal(\bar D/\bar E)$.  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 $x\in D$ such that 
$v(x)+\Gamma_E$ generates $\Gamma_D/\Gamma_E$. 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 
$exp(D)=exp(\Gamma_D/\Gamma_E)=(\Gamma_D : \Gamma_E)=deg(D)$.\\\\
{\bf 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.

\begin {Cor} 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) $\Gamma_D/\Gamma_E$ 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 $x\in D^*$)
\end {Cor}

{\it Proof.}  $(1)\Rightarrow (2)$  Clear.\\
$(2)\Rightarrow (3)$ By Proposition 2.5.\\
$(3)\Rightarrow (4)$ Obvious (since $D$ is nondegenerate).\\
$(4)\Rightarrow (1)$ Since $D$ is tame semiramified, then $\bar D$ is a 
Galois field extension of $\bar E$ with  $Gal(\bar D/\bar E)$ isomorphic to 
$\Gamma_D/\Gamma_E$. Therefore, the inertial lift of $\bar D$ over $E$ in 
$D$ is a cyclic maximal subfield of $D$. \\
Assume these conditions hold, then by [JW90, Lemma 5.15] 
$exp(D)=exp(\Gamma_D/\Gamma_E)=(\Gamma_D : \Gamma_E)=deg(D)$. 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  $\bar 
D$ is separable over $\bar E$).   $D$ being nondegenerate, then by 
Proposition 3.2 $\Gamma_K/\Gamma_E$ is cyclic. Let   $x\in K^*$ with 
$v(x)+\Gamma_E$ generating $\Gamma_K/\Gamma_E$, then  $K=E[x]$.

\begin {Cor} 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.
\end{Cor}

{\it Proof.} $(1)\Leftrightarrow (2)$ This follows by Corollary 3.4 and  
Proposition 2.9.\\
$(1)\Leftrightarrow (3)$ By Corollary 3.4.

\begin {Pro} 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 (non-trivial) nicely semiramified  division algebras.\\
$(3)$ There is an inertial subfield $M$ of $D$ such $D_M$ is the tensor 
product of two (non-trivial) cyclic nicely semiramified  division 
algebras.\\
(4) There is an inertial subfield $M$ of $D$ such $D_M$ is the tensor 
product of two (non-trivial) cyclic nicely semiramified  division algebras 
$D_1$ and $D_2$ with $\Gamma_{D_i}/\Gamma_M$ cyclic.
\end {Pro}

{\it Proof.} $(2)\Rightarrow (1)$ Assume there is an inertial  subfield $M$ 
of $D$ such $D_M=D_1\otimes_MD_2$, where $D_i$ are non-trivial nicely 
semiramified  division algebras.  Then, by  Proposition 2.6(3), 
$\Gamma_{D_M}/\Gamma_M=\Gamma_{D_1}/\Gamma_M\oplus \Gamma_{D_2}/\Gamma_M$. 
Let $L_i$ be a totally ramified maximal  subfield of $D_i$ ($i=1, 2$). Since 
$\Gamma_{L_i}/\Gamma_M=\Gamma_{D_i}/\Gamma_M$, then by [Mor89, Theorem 1] 
$L_1\otimes_ML_2$ is a totally ramified maximal  subfield of $D_M$. So, 
$D_M$ is nicely semiramified with $rl(D_M)\geq 2$. \\
$(1)\Rightarrow (4)$ If $M$ is an inertial  subfield of $D$ such that $D_M$ 
is nicely semiramified with radical length $r\geq 2$, then we can write 
$C^M_D=(M_1/M, \sigma_1, a_1)\otimes_M...\otimes_M(M_r/M, \sigma_r, a_r)$, 
where $M_i$, $\sigma_i$ and $a_i$ satisfy the conditions of [JW90, Theorem 
4.4(iii)]. We have $C^{MM_3...M_r}_D=C^{M_3...M_r}_{C^M_D}$, so  
$C^{MM_3...M_r}_D\cong(M_1M_3...M_r/M_3...M_r, \sigma_1, 
a_1)\otimes_{MM_3..M_r}(M_2M_3...M_r/\\
M_3...M_r, \sigma_2, a_2)$.   Let $D_i:=(M_iM_3...M_r/M_3...M_r, \sigma_i, 
a_i)$ ($1\leq i\leq 2$),  then by [JW90, Theorem 4.4] $D_i$ are nicely 
semiramified and  $\Gamma_{D_i}/\Gamma_M$ are cyclic. \\
$(4)\Rightarrow (3)$ and $(3)\Rightarrow (2)$ Obvious.\\\\
{\bf 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. \\\\
{\bf (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 
$I\otimes_EN\otimes_E T$, 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 $\bar 
N$ splits $\bar I$. Indeed, if $D$ is semiramified, then $T$ will be 
inertially split, hence trivial. Moreover, in this case, by [JW90, Lemma 
5.15] $\bar N$ splits $\bar I$. 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.

\setcounter{Def}{8}
\begin {Pro} 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 $I\otimes_EN$ 
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 
$\bar N$, $\bar N/K$ is cyclic or $\bar I_{K}\notin Dec(\bar N/K)$.
\end {Pro}

{\it Proof.} Let $M$ be an inertial  subfield of $D$. By [JW90, Lemma 5.15] 
we have $\bar D\cong \bar N$, 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  $\bar I_{\bar M}\in Dec(\bar N/\bar M)$ (see that $\bar N_{\bar 
M}=\bar N$). So, $D$ is nondegenerate if and only if for any subfield $K$ of 
$\bar N$, $\bar N/K$ is cyclic or $\bar I_{K}\notin Dec(\bar N/K)$. \\\\
{\bf (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 
$\langle\sigma_1\rangle$ and $\langle\sigma_2\rangle$, and \\
(2) the centralizer $C^{K}_A$ decomposes into a tensor product of two cyclic 
algebras, split by the subfields $K_1$ and $K_2$ of $L$ fixed respectively 
by $\sigma_2$ and $\sigma_1$ (i.e. $C^{K}_A=(K_1/K, \sigma_1, 
a_1)\otimes_{K}(K_2/K, \sigma_2, a_2)$ for some $a_1$, $a_2\in K^*$) [BM00,  
Definition 0.12].

\setcounter{Def}{10}

\begin {Pro} 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 $\bar N$ is a maximal subfield of $\bar 
I$, and $D$ the central division algebra over $E$ similar to $I\otimes_EN$ 
with respect to $Br(E)$. Then, $D$ is semiramified. Furthermore, $D$ is 
degenerate if and only if $\bar N$ is degenerate in $\bar I$.
\end {Pro}

{\it Proof.} Since $\bar N$ is a maximal subfield of $\bar I$, 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 $rl(D_M)\geq 2$. 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 $I_M\otimes_MN_M$, so again  by [JW90, Lemma 1.8 and Lemma 
5.15] $\bar I_{\bar M}\in Dec(\bar N/\bar M)$.  Let $r=rk(Gal(\bar N/\bar 
M))(=rk(Gal(\overline {D_M}/\bar M))=rl(D_M))$, and write $\bar 
N=K_1\otimes_{\bar M}...\otimes_{\bar M}K_r$, where $K_i/\bar M$ are cyclic 
field extensions ($1\leq i\leq r$). Since  $\bar I_{\bar M}\in Dec(\bar 
N/\bar M)$, then   $\bar I_{\bar M}=(K_1/\bar M, \sigma_1, a_1)\otimes_{\bar 
M}...\otimes_{\bar M}(K_r/\bar M, \sigma_r, a_r)$, where $\sigma_i$ is a 
generator of the cyclic field extension $K_i/\bar M$ and $a_i\in \bar M^*$. 
Let $L=K_3...K_r$, then $\bar I_{L}=(K_1L/L, \sigma_1, a_1)\otimes_L(K_2L/L, 
\sigma_2, a_2)$. So, $\bar N$ is degenerate in $\bar I$.\\
Conversely, assume that $\bar N$ is degenerate in $\bar I$ and let $L$ be an 
intermediate field of $\bar N/\bar E$ such that $\bar I_L=(L_1/L, \tau_1, 
b_1)\otimes_L(L_2/L, \tau_2, b_2)$, where $L_1\otimes_LL_2=\bar N$, $\tau_i$ 
is a generator of the cyclic field extension $L_i/L$ and $b_i\in L^*$. Let 
$M$ be the inertial lift of $L$ over $E$ in $D$, then $\bar I_{\bar M}\in 
Dec(\bar N/\bar M)$ and $rl(D_M)=rk(Gal(\bar N/\bar M))=2$. So, $D$ is 
degenerate.

\begin {Pro} 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 $\bar D$ over $E$ in $D$. Then, $D$ is degenerate if and only if $L$ is 
degenerate in $D$.
\end {Pro}

{\it Proof.} If $D$ is degenerate, then by Proposition 3.6 there is an 
inertial  subfield $M$ of $D$ such that $C^M_D$ is a tensor product of two 
cyclic nicely semiramified  division algebras $D_1$ and $D_2$ with 
$\Gamma_{D_i}/\Gamma_M$ cyclic. Write $D_i=(M_i/M, \sigma_i, a_i)$ ($1\leq 
i\leq 2$), where $M_i$ are inertial cyclic  field extensions of $E$ with 
Galois group generated by $\sigma_i$, and $a_i\in M^*$. Clearly, this 
implies that $M_1\otimes_MM_2$ is a degenerate maximal subfield of  $C_D^M$. 
Moreover, by comparing the cardinalities, we have  $\bar M_1\otimes_{\bar 
M}\bar M_2=\bar D$.  So,  $M_1\otimes_MM_2$ is isomorphic to $L$ (because, 
$\bar M=\bar M_1\otimes_{\bar M}\bar M_2=\bar D$), 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 $C^K_D=(K_1/K, \sigma_1, 
a_1)\otimes_K(K_2/K, \sigma_2, a_2)$, where $K_1$ and $K_2$ are the 
subfields of $L$ fixed by $\sigma_2$ and $\sigma_1$ respectively and $a_1$, 
$a_2\in K^*$, and  let $D_i=(K_i/K, \sigma_i, a_i)$ ($i=1,2$). By 
Proposition 2.6(2 and 3) $C^K_D$ 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 $\Gamma_{D_i}/\Gamma_K$ 
is cyclic. Hence, by Proposition 2.6(3) $C^K_D$ is nicely semiramified with 
radical length $r=2$. Therefore, $D$ is degenerate.


\begin {Pro} 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 $(\Gamma_K : \Gamma_E)$ 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.
\end {Pro}

{\it Proof.} By Proposition 2.6(2) $C^K_D$ is semiramified. Since $(\Gamma_K 
: \Gamma_F)$ is maximal, then $(\Gamma_{C^K_D} : \Gamma_K)=1$. Hence, 
$C^K_D=K$. 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. 
\\\\
{\bf 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 $rl(D_M)\geq 2$. 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 $\Gamma_K/\Gamma_F$ is 
non-cyclic. One can also see that there are analogous graded versions for 
all the results  we are given in this work.

\setcounter{Def}{14}

\begin {Lem} 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.
\end {Lem}

{\it Proof.} Suppose that $GD$ is degenerate and let $K$ be an inertial 
graded subfield of $GD$ such that $GD_{K}$ is nicely semiramified with 
$rl(GD_{K})\geq 2$. Let $M$ be the inertial lift of $K_0$ over $E$ in $D$. 
Then, $G(D_M)\cong GD_K$. So, by  Lemma 2.2, $D_M$ is nicely semiramified. 
Moreover, we have  $rl(D_M)=rl(GD_K)\geq 2$. 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 
$rl(D_L)\geq 2$. Then, clearly $GD_{GL}$ is nicely semiramified and 
$rl(GD_{GL})=rl(D_L)\geq 2$.


\begin {Pro}  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 $Frac(F)$ (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) $Cq(D)\otimes_{Frac(F)}HL$ is degenerate.\\
(3) $D_0HL$ is degenerate in $Cq(D)\otimes_{Frac(F)}HL$.\\
(4) $D_0L$ is degenerate in $Cq(D)\otimes_{Frac(F)}L$.\\
(5) $D_0Frac(F)$ is degenerate in $Cq(D)$.
\end {Pro}

{\it Proof.} By [Mor89, Theorem 1] $Cq(D)\otimes_{Frac(F)}HL$ is a division 
algebra (of center $HL$). Since $L$ is an immediate valued field extension 
of $Frac(F)$, then so is $HL$. Therefore, $GHL\cong F$ and     
$G(Cq(D)\otimes_{Frac(F)}HL)\cong_g D$. So, $(1)\Leftrightarrow (2)$ follows 
by Lemma 3.15.\\
$(2)\Leftrightarrow (3)$ By Proposition 3.12.\\
$(1)\Rightarrow (5)$ By the graded version of  Proposition 3.6 there is an 
inertial graded subfield $M$ of $D$ such that $C^M_D$ is a tensor product of 
two cyclic nicely semiramified graded division algebras $(M_i/M, \sigma_i, 
a_i)$,  where $M_i$ are inertial cyclic graded field extensions of $M$ with 
Galois group generated by $\sigma_i$, and $a_i\in M^*$.
So, up to an isomorphism, we have :
$$\begin {array}{ccl}
C^{Frac(M)}_{Cq(D)}& = & C^M_D\otimes_MFrac(M)=((M_1/M, \sigma_1, 
a_1)\otimes_M(M_2/M, \sigma_2, a_2))\otimes_MFrac(M)\\
& = & ((M_1/M, \sigma_1, a_1)\otimes_MFrac(M))\otimes_{Frac(M)}((M_2/M, 
\sigma_2, a_2)\otimes_MFrac(M))\\
& = & (Frac(M_1)/Frac(M), \sigma_1, a_1)\otimes_{Frac(M)}(Frac(M_2)/Frac(M), 
\sigma_2, a_2)
\end {array}$$
By the graded version of Proposition 2.6(3), we have $M_1\otimes_MM_2=D_0M$, 
so  $Frac(M_1)\otimes_{Frac(M)}Frac(M_2)=D_0Frac(F)$ (see that 
$D_0Frac(F)=D_0Frac(M)$). Moreover, it is clear that  $Frac(M_1)$ [resp. 
$Frac(M_2)$] is the subfield of $D_0Frac(F)$ fixed by $\sigma_2$ [resp. by 
$\sigma_1$]. This shows that $D_0Frac(F)$ is  degenerate in $Cq(D)$.\\
$(5)\Rightarrow (4)$ and $(4)\Rightarrow (3)$ Obvious.


\begin {Cor}   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  
  $(K, G, U)\otimes_ZL$ is semiramified but not nicely semiramified. In 
particular,  if $char(E)=p$ and $G$ is a $p$-group, then  $(K, G, 
U)\otimes_ZL$ is not a tensor product of cyclic algebras. If $p=2$ and   
$exp((K, G, U)\otimes_ZL)=2$, then  $(K, G, U)\otimes_ZL$ has an involution 
of the first kind but is not a tensor product of 2-symbols.
\end {Cor}

{\it 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 
$D_0Frac(F)/Frac(F)$) if and only if $D_0Frac(F)$ is nondegenerate in 
$Cq(D)$. So,  by  Proposition 3.16 $D$ is nondegenerate. Therefore, $D$ 
cannot be nicely semiramified (because $\Gamma_D/\Gamma_F(\cong G)$ is 
non-cyclic). The rest follows by Corollary 2.5.


\begin {Cor} $[$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.
\end {Cor}
{\bf Remark 3.19}  (1) Let $F$ be a graded field with support $\Gamma_F$ ; 
$E=F_0$ ; $M$ a finite-dimensional abelian field extension of $E$ with 
Galois group $G$ ; $S$ a basis of $G$ ; $\Delta$ a totally ordered abelian 
group that contains $\Gamma_F$ with ordering extending that of $\Gamma_F$ ; 
$\epsilon :\Delta \rightarrow G$ a group epimorphism with kernel $\Gamma_F$ 
and $(U, B)$ a cocycle of $Z^2_S(G, M^*)$. We have seen in the preliminaries 
how to construct the semiramified graded division algebra $(MF, \epsilon, G, 
U, B)$. Assume that $U$ is nondegenerate in $M/E$ and let $L$ be an 
immediate field extension of $Frac(F)$. One can easily see that Corollary 
3.17 is still true if we replace the generic abelian crossed product $(K, G, 
U)$ by $Cq((MF, \epsilon, G, U, B))$. \\
More generally, let $F$ be a graded field and $D$ a semiramified graded 
division algebra over $F$. Let $x_i\in D^*$ such that 
$(gr(x_i)+\Gamma_F)_{i=1}^r$ is a basis of $\Gamma_D/\Gamma_F$ and let 
$u_{ij}=x_ix_jx_i^{-1}x_j^{-1}$ and $U=(u_{ij})_{1\leq i,j\leq r}$. Then, by 
[BM00, Proposition 0.13] $U$ is nondegenerate with respect to $D_0/F_0$ if 
and only if $D_0Frac(F)$ 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 
$D\otimes_EHE$ 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 
$GD=G(D\otimes_EHE)$. 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.

\section {Subfields of nondegenerate tame semiramified division algebras }

\setcounter{Def}{0}
\begin {Lem} Let $F$ be a graded field and $D$ a nondegenerate semiramified 
graded division algebra of prime power degree $p^n$ over $F$ with 
$rk(Gal(D_0/F_0))\geq 2$. Let $d\in D^*$ such that $d^p\in F$. Then, $d\in 
D_0F^*$. If in addition $p=char(F)$, then $d\in F^*$.
\end {Lem}

{\it Proof.} We will use the same argument as in the proof of [AS78, Lemma 
3.1]. Let $x_i\in D^*$ such that $(gr(x_i)+\Gamma_F)_{i=1}^r$ is a basis of 
$\Gamma_D/\Gamma_F$. Then, $D=D_0F[x_1,...,x_r]$. Let 
$q_i=ord(gr(x_i)+\Gamma_F)$, $\sigma_i=\theta_D(gr(x_i)+\Gamma_F)$, where 
$\theta_D$ is the group isomorphism of [B95, p.4278] (see also, [HW(2)99, 
(2.2), p.86]) and $I=\{\bar m=(m_1,...,m_r)\in I\!\!N^r$ $|$ $0\leq m_i<q_i$ 
for all $1\leq i\leq r\}$. Then, we can write $d=ax^{\bar m}$, where $a\in 
D_0F^*$ and $\bar m\in I$. Suppose by contradiction that $\bar m\neq 0$ and 
let $D_0^{\bar m}$ be the subfield of $D_0$ fixed by $\sigma^{\bar m}$. 
Since $d^p\in F^*$, then $(\sigma^{\bar 
m})^p(=\theta_D(gr(d^p)+\Gamma_F)=id_{D_0}$. So, $[D_0 : D_0^{\bar m}]=p$. 
Let $N_{\bar m}$ be the norm of $D_0/D_0^{\bar m}$. Then, $(ax^{\bar 
m})^p=N_{\bar m}(a)(x^{\bar m})^p\in F^*$. Since $rk(Gal(D_0/F_0))\geq 2$, 
then there is some $1\leq j \leq r$ such that $\langle\sigma^{\bar m}, 
\sigma_j\rangle$ is non-cyclic. We have $N_{\bar m}(a)(x^{\bar 
m})^p=x_j(N_{\bar m}(a)(x^{\bar m})^p)x_j^{-1}=(x_jN_{\bar 
m}(a)x_j^{-1})(x_j(x^{\bar m})^px_j^{-1})=N_{\bar m}(\sigma_j(a))N_{\bar 
m}(u_{j\bar m})(x^{\bar m})^p=N_{\bar m}(\sigma_j(a)u_{j\bar m})(x^{\bar 
m})^p$. So, $N_{\bar m}(\sigma_j(a)a^{-1}u_{j\bar m})=1$. Therefore, by 
Hilbert 90 Theorem, there is some $b\in D_0$ such that 
$\sigma_j(a)a^{-1}u_{j\bar m}=\sigma^{\bar m}(b)b^{-1}$. But this is not 
possible since $D$ is nondegenerate. Hence, $\bar m=0$. Remark that $D_0$ is 
separable over $F_0$, so $D_0F$ is tame over $F$. Therefore, if $char(F)=p$, 
then $d\in F^*$.


\begin {Pro} Let $E$ be a Henselian valued field and $D$ a nondegenerate 
tame semiramified division algebra of prime power degree over $E$. If 
$rk(\Gamma_D/\Gamma_E)\geq 2$, then $D$ has no (non-trivial) totally 
ramified subfield (over $E$).
\end  {Pro}

{\it Proof.} Suppose first that $char(\bar E)=p>0$ 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 $x\in GK$, $x^{[GK : 
GE]}\in GE$. Hence, by Lemma 4.1, $x\in GE$. So, $GK=GE$. This proves 
$K=E$.\\
Now, assume $char(\bar E)$ does not divide $deg(D)$. Suppose that  $K$ is a 
non-trivial totally ramified subfield of $D$. Let $x\in GK^*$ such that 
$gr(x)\notin \Gamma_{GE}$. If $ord(gr(x)+\Gamma_{GE})=p^s$ (for some $s\geq 
1$), then $x^{p^s}\in GE$ (since $GE[x]$ is totally ramified over $GE$). 
Therefore, by Lemma 4.1, $x^{p^{s-1}}\in \bar DGE$. A contradiction (since 
we will have $p^{s-1}gr(x)+\Gamma_{GE}=\Gamma_{GE}$).

\begin {Th} 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 $\Gamma_D/\Gamma_E$ 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 $Gal(\bar D/\bar E)$ such that $Gal(K/E)\cong Gal(\bar 
D/\bar E)/H$. In particular, if $K$ is a Galois maximal subfield of $D$, 
then $Gal(K/E)\cong Gal(\bar D/\bar E)$.
\end {Th}

{\it 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.

\begin{Cor}  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 $\Gamma_D/\Gamma_E$ is 
cyclic.
\end {Cor}

{\it Proof.} Suppose that $D$ is cyclic and let $K$ be a cyclic maximal 
subfield of  $D$, then by Theorem 4.3 $Gal(K/E)\cong Gal(\bar D/\bar E)$. 
Moreover, since $D$ is tame semiramified, then $Gal(\bar D/\bar E)\cong 
\Gamma_D/\Gamma_E$, so $\Gamma_D/\Gamma_E$ is cyclic. Conversely, if 
$\Gamma_D/\Gamma_E$ is cyclic, then 'the' inertial lift of $\bar D$ over $E$ 
in $D$ is a cyclic maximal subfiled of $D$.

\begin {Cor} $[$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 $Gal(L/E)\cong G/H$. In particular, if $L$ is a Galois 
maximal subfield of $(K, G, U)$, then $Gal(L/E)\cong G$.
\end {Cor}


\begin {Pro} Let $E$ be a Henselian valued field and $D$ a nondegenerate 
tame semiramified division algebra of prime power degree over $E$. Assume  
$rk(\Gamma_D/\Gamma_E)\geq 2$, 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 $Gal(\bar D/\bar E)$ is elementary 
abelian.
\end  {Pro}

{\it Proof.}  Since $K$ is an elementary abelian field extension of $E$, 
then we can write  $K=K_1\otimes_EK_2\otimes_E...\otimes_EK_r$, 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 $\bar D(=\bar K)$ is elementary abelian over $\bar E$. 
Conversely, suppose that $Gal(\bar D/\bar E)$ is elementary abelian and let 
$M$ be the inertial lift of $\bar D$ over $E$ in $D$, then $M$ is an 
elementary abelian maximal subfield of $D$.\\\\
{\bf 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 $x\in D^*$ such that $gr(x)+\Gamma_F$ 
generates $\Gamma_D/\Gamma_F$, 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 $\sigma$ be the $F_0$-automorphism of $D_0$ defined by 
$\sigma(d)=x^qdx^{-q}$, and let $M_0$ be the subfield of $D_0$ fixed by 
$\sigma$, then let $M=M_0F$. Obviously, $M[x^q]$ is a graded subfield of 
$D$. Since $M$ [resp. $T$] is inertial [resp. totally ramified] over $F$, 
then $M[x^q]=M\otimes_FT$. 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  
$\Gamma_D/\Gamma_E$ is cyclic.

\setcounter{Def}{7}


\begin {Pro} 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 $Gal(\bar D/\bar K)$ is non-cyclic, then $K$ is inertial 
over $E$.
\end {Pro}

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

\begin{Pro} Let $E$ be a Henselian valued field and $D$ a nondegenerate tame 
semiramified division algebra of prime power degree over $E$. Assume 
$rk(\Gamma_D/\Gamma_E)\geq 3$ and let $K$ be an abelian subfield of $D$, 
then $K$ is inertial over $E$.
\end {Pro}

{\it Proof.} Write $K=K_1\otimes_EK_2\otimes_E...\otimes_EK_r$, where $K_i$  
are cyclic   field extensions of $E$. Let $N_i$ be the inertial lift of 
$\bar K_i$ over $E$ in $K_i$. Obviously, $N_i$ is cyclic over $E$. So, $\bar 
K_i$ is cyclic over $\bar E$. Therefore, $\bar D$ cannot be cyclic over 
$\bar K_i$ (since $rk(Gal(\bar D/\bar E))\geq 3$). So, by Proposition 4.8 
$K_i$ is inertial over $E$.

\begin {Lem} Let $E$ be a Henselian valued field and $D$ a nondegenerate 
tame semiramified division algebra of prime power degree over $E$. Consider 
$x\in D^*$ such that $v(x)\neq \Gamma_E$. Then, $Gal(\bar 
D/\overline{E[x]})$ is cyclic. In particular, if $exp(\Gamma_D/\Gamma_E)$ is 
prime, then $E[x]$ is a maximal subfield of $D$.
\end {Lem}

{\it Proof.}  Let $M$ be the inertial lift of $\overline{E[x]}$ over $E$ in 
$E[x]$. Clearly, $E[x]$ is a non-trivial totally ramified subfield of 
$C^M_D$. So, by Proposition 4.2, $Gal(\bar D/\bar M)$ is cyclic. If 
$exp(\Gamma_D/\Gamma_E)=p$ is a prime, then so is $exp(Gal(\bar D/\bar E))$. 
Hence, $card(Gal(\bar D/\overline{E[x]}))\leq p$. See that we have $[E[x] : 
E]>[\overline{E[x]} : \bar E]=[\bar D : \bar E][\bar D : \overline 
{E[x]}]^{-1}\geq deg(D)p^{-1}$. 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 
$E$.


\begin {Th} Let $E$ be a Henselian valued field,  $p$ a prime integer and 
$D$ a nondegenerate tame semiramified division algebra of  degree  $p^n$ 
($n\in I\!\!N^*$) over $E$. Assume $exp(\Gamma_D/\Gamma_E)=p$ 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 $\leq p^2$ or inertial over $E$.

Remark that in the case of (1) and in the last case of (2) $K$ is elementary 
abelian over $E$.
\end {Th}

{\it 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 $p\neq 2$ 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 $Gal(\bar D/\bar K)$ 
and $Gal(\bar D/\bar L)$ are non-trivial cyclic groups.  Moreover, since  
$exp(Gal(\bar D/\bar E))=exp(\Gamma_D/\Gamma_E)=p$, then  $Gal(\bar D/\bar 
K)$ and $Gal(\bar D/\bar L)$ are of order $p$. We have $\bar L\subseteq \bar 
K$, so necessarily $\bar L=\bar K$. Let $M$ be the inertial lift of $\bar L$ 
over $E$ in $L$. Then, up to an isomorphism, $M$ is also the inertial lift 
of $\bar K$ over $E$ in $K$. So, $K$ is totally ramified over $M$, and 
obviously $\Gamma_K/\Gamma_M (=\Gamma_K/\Gamma_E)$ is cyclic of order $p$. 
Hence, $p\leq [K : L]\leq [K : M]=(\Gamma_K : \Gamma_M)=p$. 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 $Gal(K/E)=H_1\times H_2$, 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 $K=K_1\otimes_EK_2$, 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 $deg(D)\leq p^2$ and $K$ is a 
non-cyclic abelian field extension of $E$, then $K=K_1\otimes_EK_2$, 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 $E$.\\\\
{\bf (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 $\sigma\in Gal(K/K_0F)$ and $\tau\in Gal(K/F)$. For any 
$a\in (K_0F)^*$, we have $gr(\tau^{-1}(a))=gr(a)$. So, $\tau^{-1}(a)=ca$ for 
some $c\in K_0$. Hence, $\tau \circ \sigma \circ \tau^{-1}(a)=a$. Therefore, 
$\tau \circ \sigma \circ \tau^{-1}\in Gal(K/K_0F)$. This shows that 
$Gal(K/K_0F)$ is a normal subgroup of $Gal(K/F)$. So, $K_0F$ 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 $\bar 
M=\bar N$, then $M$ is a Galois field extension of $E$.\\\\
{\bf 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 $H\subset H'$.

\setcounter{Def}{13}

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 $E$.

\begin{Pro} 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 $Gal(\bar D/\bar K)$ is cyclic. If 
$Gal(\bar D/\bar K)$ is maximally cyclic in $Gal(\bar D/\bar E)$, 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 $\Gamma_D/\Gamma_E$ is non-cyclic, 
then $K$ is either a quaternion or an abelian field extension of $E$.

Therefore, if $rk(\Gamma_D/\Gamma_E)\geq 3$ and $K$ is not a quaternion 
field extension of $E$, then $K$ is inertial over $E$.
\end {Pro}

{\it 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, 
$Gal(\bar D/\bar L)$ is cyclic (indeed, otherwise $L$ would be inertial 
-hence normal- over $E$). Since $Gal(\bar D/\bar K)$ is maximally cyclic in 
$Gal(\bar D/\bar E)$ and $Gal(\bar D/\bar K)\subseteq Gal(\bar D/\bar L)$, 
then $\bar K=\bar L$. 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] $Gal(K/E)=H_1\times H_2$, 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 $K=K_1\otimes_EK_2$. Assume that $\bar K_i\neq 
\bar E$ (for both $i=1, 2$), then $Gal(\bar D/\bar K_i)$ ($i=1, 2$) are 
non-cyclic (because we have $Gal(\bar D/\bar K)\subset Gal(\bar D/\bar K_i)$ 
and $Gal(\bar D/\bar K)$ is maximally cyclic in $Gal(\bar D/\bar E)$). 
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, $\bar K_1$ or $\bar K_2$ 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.

\begin {Cor} 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  $char(\bar E)$ does not divide $deg(D)$ and let  $K$ be 
  a Galois maximal subfield of $D$. If $(\Gamma_K : 
\Gamma_E)=exp(\Gamma_D/\Gamma_E)$, then $K$ is abelian over $E$ and 
$rk(\Gamma_D/\Gamma_E)\leq 2$.
\end {Cor}

{\it Proof.} Since $D$ is nondegenerate, then $\Gamma_K/\Gamma_E$ is cyclic. 
Moreover, since $K$ is a maximal subfield of $D$, then $Gal(\bar D/\bar 
K)\cong \Gamma_K/\Gamma_E$. Hence, $Gal(\bar D/\bar K)$ is maximally cyclic 
in $Gal(\bar D/\bar E)$. So, by Proposition 4.14 $K$ is abelian over $E$ and 
  $rk(\Gamma_D/\Gamma_E)\leq 2$ (because $K$ is not inertial over $E$).\\\\
{\bf (4.16)} As a consequence of Corollary 4.15, under the hypotheses of 
Corollary 4.15, if $rk(\Gamma_D/\Gamma_E)\geq 3$, $x\in D^*$ with 
$ord(v(x)+\Gamma_E)=exp(\Gamma_D/\Gamma_E)$, and $L$ is  a maximal subfield  
of $D$ that contains $x$, then $L$ cannot be Galois over $E$.

\setcounter{Def}{16}

\begin{Pro}  Let $F$ be a graded field, $D$ a nondegenerate semiramified 
graded division algebra of prime power degree $p^r$ over $F$ with 
$rk(\Gamma_D/\Gamma_F)\geq 3$, and $K$  a Galois graded subfield of $D$ such 
that  $[K : F]<pdeg(D)(exp(\Gamma_D/\Gamma_F))^{-1}$. Then, $K$ is inertial 
over $F$.
\end {Pro}

{\it 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, $[K_0 : F_0]\geq [D_0 : 
F_0](exp(\Gamma_D/\Gamma_F))^{-1}$. Remark  that necessary $(\Gamma_K : 
\Gamma_F)=p$. Indeed, if $(\Gamma_K : \Gamma_F)\neq p$, then  there exists 
an intermediate graded field $P$ such that $K_0F\subset P \subset K$. 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 $K_0F\subset P$). So, $(\Gamma_K : \Gamma_F)=p$. Therefore, $[K 
: F]\geq pdeg(D)exp(\Gamma_D:\Gamma_F)^{-1}$. A contradiction.

\begin {Cor} Let $E$ be a Henselian valued field and $D$ a nondegenerate 
tame semiramified division algebra of prime power degree $p^r$ over $E$. 
Assume $char(\bar E)$ does not divide $deg(D)$ and suppose $K$ is a Galois 
subfield of $D$ with $[K : E]<pdeg(D)exp(\Gamma_D:\Gamma_E)^{-1}$. Then, $K$ 
is inertial over $E$.
\end {Cor}
{\bf 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 $K=M\otimes_ET$. We define nice abelian graded field 
extensions in a similar way.

\setcounter{Def}{19}

\begin {Pro} 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.
\end {Pro}

{\it Proof.} If $K$ is nice, then $K=M\otimes_ET$, where $M$ is inertial and 
$T$ is totally ramified over $E$. So, by the graded version of [Mor89, 
Theorem 1] $GK=GM\otimes_{GE}GT$.\\
Conversely, suppose that $GK$ is nice. Then, $GK=\bar KGE\otimes_{GE}T'$ 
where $T'$ is a totally ramified graded field extension of $GE$. Let $M$ be 
the inertial lift of $\bar K$ over $E$ in $K$ and let $u_1,..., u_r$ be 
homogeneous elements of $T'$ such that $\Gamma_{T'}/\Gamma_{GE}=\langle 
gr(u_1)+\Gamma_{GE}\rangle\oplus...\oplus\langle 
gr(u_r)+\Gamma_{GE}\rangle$. By Proposition 1.5, there are $x_i\in K$ such 
that $\tilde x_i=u_i$ and $G(E[x_i])=GE[u_i]$. In particular, $E[x_i]$ are 
totally ramified over $E$, hence by [Mor89, Theorem 1] 
$T:=E[x_1]\otimes_E...\otimes_EE[x_r]$ is a totally ramified field extension 
of $E$. Once again, by applying [Mor89, Theorem 1] (and by comparing the 
dimensions), we have $K=M\otimes_ET$ (up to an isomorhism).


\begin {Lem} 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 $E$.
\end {Lem}
{\it Proof.} Let $M$ [resp. $T$] be an inertial [resp. totally ramified]  
field extension of $E$ such that $K=M\otimes_ET$. Then, 
$Gal(K/E)=Gal(M/E)\oplus Gal(T/E)$. 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.

\begin {Lem} Let $E$ be a Henselian valued field with valuation $v$ and $D$ 
a semiramified division algebra of degree $n$ over $E$. Assume  $\bar E$ 
contains a primitive $n^{th}$ root of unity. Then, the following statements 
are equivalent : \\
(1) $D$ has an abelian maximal subfield $K=M\otimes_ET$, 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  $C^M_D$ such that $v(u)+\Gamma_E$ generates 
$\Gamma_{C^M_D}/\Gamma_E$ and $u^{[\bar D : \bar M]}\in E$.
\end{Lem}

{\it Proof.} $(1) \Rightarrow (2)$  By  Proposition 2.6 $C^M_D$ is 
semiramified (over $M$), so $[\bar D : \bar M]=(\Gamma_{C^M_D} : \Gamma_M)$. 
Moreover,  $K$ is a  totally ramified maximal subfield of $C^M_D$,  so 
$[\bar D : \bar E]=(\Gamma_K : \Gamma_M)=(\Gamma_T : \Gamma_E)=[T : E]$. 
Since $T$ is cyclic over $E$, then $\Gamma_T/\Gamma_E(\cong Gal(T/E))$ is 
cyclic. Let $x\in T^*$ such that $\Gamma_T/\Gamma_E=\langle 
v(x)+\Gamma_E\rangle$, then $GT=GE[\tilde x]$ and the minimal polynomial of 
$\tilde x$ over $Frac(GE)$ is $X^{[T  : E]}-\tilde {x}^{[T : E]}$. So, by  
Proposition 1.5 there is $u\in T$ such that $T=E[u]$, $u^{[T : E]}\in E^*$ 
and $v(u)+\Gamma_E=v(x)+\Gamma_E$.\\
$(2) \Rightarrow (1)$ Since $u^{[\bar D : \bar M]}\in E^*$, then $[E[u] : 
E]\leq [\bar D : \bar M]$. Moreover, since $C^M_D$ is semiramified and 
$v(u)+\Gamma_E$ generates $\Gamma_{C^M_D}/\Gamma_E$, then $\bar D$ is a 
cyclic field extension of $\bar M$ and $[\bar D : \bar M]=card(\langle 
v(u)+\Gamma_E\rangle)\leq (\Gamma_{E[u]} : \Gamma_E)\leq [E[u] : E]$. Hence, 
  $E[u]$ is totally ramified over $E$. Moreover, since 
$\Gamma_{E[u]}/\Gamma_E$ is cyclic and $\bar E$ contains a primitive 
$n^{th}$ root of unity, then $E[u]$ is cyclic over $E$. Therefore, 
$K:=M\otimes_EE[u]$ is an abelian maximal  subfield of $D$.
\begin {Pro} 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 $K=T\otimes_EM$, 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) $Gal(\bar D/\bar E)$ is cyclic.

If $\bar E$ 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  $C^M_D$ such that $v(u)+\Gamma_E$ generates $\Gamma_{C^M_D}/\Gamma_E$ 
and $u^{[\bar D : \bar M]}\in E$.
\end {Pro}
{\it Proof.} $(1) \Rightarrow (6)$ Let $K$ be a nice maximal  subfield of 
$D$. If $K$ is inertial over $E$, then $\bar K=\bar D$. So, $Gal(\bar D/\bar 
E)$ is cyclic.  If $K$ is not inertial over $E$, then $D$ contains a 
non-trivial totally ramified subfield. So, by Proposition 4.2, $Gal(\bar 
D/\bar E)(\cong \Gamma_D/\Gamma_E)$ is cyclic.\\
$(6) \Rightarrow (3)$ Clear (indeed,  the inertial lift of $\bar D$ over $E$ 
in $D$ is a nice cyclic maximal subfield of $D$).\\
$(3) \Rightarrow (2)$  By Lemma 4.21.\\
$(2) \Rightarrow (1)$ and $(4) \Leftrightarrow (6)$  Obvious.\\
$(5)\Rightarrow (6)$ By Proposition 4.2.\\
$(6)\Rightarrow (5)$ Let $d\in D^*$ such that $v(d)+\Gamma_E$ generates 
$\Gamma_D/\Gamma_E$, then $E[d]$ is a non-trivial totally ramified subfield 
of $D$.\\
If $\bar E$ contains a  primitive $deg(D)^{th}$ root of unity, then $(2) 
\Leftrightarrow (7)$ follows by Lemma 4.22.



\section {Indecomposable division algebras}
\setcounter {Def}{0}

\begin {Lem} Let $E$ be a Henselian valued field and $D$ a nondegenerate 
tame semiramified division algebra of prime power degree over $E$. Assume 
$D=D_1\otimes_ED_2$, 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, $\bar 
L_i=\bar D_i$.
\end {Lem}

{\it Proof.} Since $D_i$ are non-trivial, then $rk(\Gamma_D/\Gamma_E)\geq 
2$. Indeed, if $\Gamma_D/\Gamma_E$ is cyclic, then $D$ is nicely 
semiramified. So, by [JW90, Lemma 5.15] 
$exp(D)=exp(\Gamma_D/\Gamma_E)=deg(D)$. Hence, $D$ is indecomposable. A 
contradiction. Moreover, by Proposition 4.2, $\bar D_i\neq \bar E$ (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  $\bar 
L_1\neq \bar D_1$, then $rk(Gal(\bar D_1\otimes_{\bar E}\bar D_2/\bar 
L_1))\geq 2$  (for if  $\sigma\in Gal(\bar D_1/\bar L)\backslash \{id_{\bar 
D_1}\}$ and $\tau\in Gal(\bar D_2/\bar E)\backslash\{id_{\bar D_2}\}$, then 
$\sigma\otimes {id_{\bar D_2}}$ and $id_{\bar D_1}\otimes\tau$ are 
non-comparable elements of $Gal(\bar D_1\otimes_{\bar E}\bar D_2/\bar 
L_1)$), so $rk(Gal(\bar D/\bar L_1))\geq 2$. Hence, by Proposition 4.8, 
$L_1$ is inertial over $E$.  So, by Proposition 2.6(1) $D_1$ is 
semiramified. Hence, $\bar L_1=\bar D_1$.  The contradiction obtained here 
shows that $\bar L_1=\bar D_1$. In the same way, we prove that if $L_2$ is a 
maximal subfield of $D_2$, then $\bar L_2=\bar D_2$.

\begin {Th} 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.
\end {Th}

{\it Proof.} As seen above in the proof of Lemma 5.1, if $\Gamma_D/\Gamma_E$ 
is cyclic, then $D$ is indecomposable. Assume that 
$rk(\Gamma_D/\Gamma_E)\geq 2$ and suppose $D=D_1\otimes_ED_2$, where $D_1$ 
and $D_2$ are non-trivial central division algebras over $E$. Choose a 
subfield $K$ of $D_i$ with $(\Gamma_K : \Gamma_E)$ 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, $\Gamma_{K[x_2]}/\Gamma_E$ is cyclic (with prime power order). Hence, 
the subgroups of $\Gamma_{K[x_2]}/\Gamma_E$ are totally ordered by 
inclusion. In particular, we have $\Gamma_{E[x_2]}/\Gamma_E\subseteq 
\Gamma_K/\Gamma_E$ (we can not have $\Gamma_K/\Gamma_E\subset 
\Gamma_{E[x_2]}/\Gamma_E$ because $(\Gamma_K : \Gamma_E)$ is maximal). So, 
$\Gamma_{D_2}/\Gamma_E$ is cyclic. Let $y\in D_2$ such that $v(y)+\Gamma_E$ 
generates $\Gamma_{D_2}/\Gamma_E$ ($v$ 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 $\bar L_2=\bar D_2$. Moreover, we have 
$\Gamma_{L_2}/\Gamma_E=\Gamma_{D_2}/\Gamma_E$. Hence, $D_2=L_2$. A 
contradiction.

\begin {Cor}  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 $(K, G, 
U)\otimes_ZL$ is an indecomposable division algebra.
\end{Cor}
{\bf Remark 5.4} Recently, Kelly L. Mckinnie has constructed tame 
semiramified indecomposable division $p$-algebras  of arbitrary degree and 
of exponent $p$ (for $p\neq 2$) [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 $q\neq 2$ and any positive 
integer $n$ (even for $q\neq char(\bar E)$).\\
{\Large\bf\center Appendix}
\appendix
\setcounter{section}{1}
\setcounter{Def}{0}
\begin {C} Let $E/F$ be an algebraic graded field extension and $g$ a 
homogenizable irreducible polynomial of $F[X]$. Assume $g$ has a root $x\in 
E\backslash \{0\}$ and let $f_{x, Frac(F)}$ be  the minimal polynomial of 
$x$ over $Frac(F)$. Then, $g=af_{x, Frac(F)}$ for some $a\in F^*$. In 
particular, $x\in E^*$.
\end {C}

{\it Proof.}  Since $x$ is a root of $g$, then $f_{x, Frac(F)}$ divides $g$ 
in $Frac[X]$. Let $h\in Frac(F)[X]$ such that $g=hf_{x, Frac(F)}$. 
Obviously, we can write $h=b^{-1}h'$ [resp. $f_{x, Frac(F)}=c^{-1}f'$] with 
$h', f'\in F[X]$ and $b, c\in F$. Let $\lambda \in \Delta_F$ (where 
$\Delta_F$ is the divisible hull of $\Gamma_F$) such that $g\in 
(F[X]^{(\lambda)})^h$. It follows from the equality $(bc)g=h'f'$ that 
$(bc)_mg=h'_mf'_m$, 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 $h'_m\in F$, 
then $deg(g)=deg(f'_m)\leq deg(f_{x, Frac(F)})$. So, $deg(g)=deg(f_{x, 
Frac(F)})$ (since $f_{x, Frac(F)}$ divides $g$). If $f'_m \in F$, then 
$deg(g)=deg(h'_m)\leq deg(h)$. So, $deg(f_{x, Frac(F)})=0$. But this is not 
true. See that we can write $g=a_0+a_1X+...+a_nX^n$, where $n$ is a positive 
integer and $a_i\in F^*$ for all $0\leq i\leq n$ (Indeed, write 
$g=d_0+d_1X+...+d_nX^n$, where $d_i\in F$ and write $d_i=\sum 
(d_i)_{\gamma}$, where $(d_i)_{\gamma}\in F_{\bar \gamma}$. Since $g$ is 
homogeneous (in $F[X]^{(\lambda)}$), then we can consider only the 
$(d_i)_{\gamma}$ where $gr((d_i)_{\gamma}X^i)=gr(g)$). Let $g'=a_n^{-1}g$. 
Then, $g'=f_{x, Frac(F)}$  (since $g'(x)=0$, $deg(g')=deg(f_{x, Frac(F)})$). 
So, $g=a_nf_{x, Frac(F)}$. In particular, $f_{x, Frac(F)}$ is homogeneous in 
  $F[X]^{(\lambda)}$. So, $x\in E^*$ (by [HW(1)99, Proposition 2.2 and 
corollary 2.5]).

\section {Normal subgroups}



\setcounter{Def}{0}

\begin {Lem} 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 $\langle x\rangle$ (i.e. the cyclic subgroup of $G$ generated by 
$x$)  or $\langle y\rangle$ is a non normal subgroup of $G$.
\end {Lem}

{\it Proof.} Let $p^m=ord(y)$, then for every $0<k<p$, $0<k'<p^m$, $x^k \neq 
y^{k'}$. Indeed, otherwise we will have $\langle x\rangle=\langle x^k\rangle 
\subseteq \langle y\rangle$, which contradicts the noncommutativity of $x$ 
and $y$. If $\langle x\rangle$ and $\langle y\rangle$ 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 $yx^ry^{-1}=x$ and $x^{-1}y^sx=y$. Therefore 
$(yx^ry^{-1})y=x(x^{-1}y^sx)$, hence $yx^r=y^sx$. 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 $G$.\\\\
{\bf 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 
$G^c:=\{\alpha \beta \alpha^{-1}\beta^{-1}|$ $ \alpha, \beta \in G\}$ be the 
commutator of $G$.
\setcounter {Def}{2}
\begin {C} Under the hypotheses of (B.2), we have $G^c \subseteq Z(G)$.
\end {C}

{\it Proof.} Since $card(G)>1$, then by [L84, theorem 6.4, p. 25], $Z(G) 
\neq \{1\}$. If $G/Z(G)$ has a non normal subgroup $A$, then $H:=\{\alpha 
\in G|$ $ \alpha Z(G) \in 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, $G^c \subseteq Z(G)$.

\begin {C} Under the hypotheses of (B.2), we have $card(G^c)=p$.
\end {C}

{\it 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 $G^c \subseteq G'$.  
Also, if there was  a non-normal subgroup $A$ of $G/G'$, then $H:=\{\alpha 
\in G|$ $ \alpha G' \in A\}$ would be a non normal subgroup of $G$. A 
contradiction.

\begin {C} 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 $z=ax^my^n$, with $a \in Z(G)$, $ 0 \leq m, n <p$. Thus, 
$card(G/Z(G))=p^2$. Precisely, $G/Z(G) \cong {\it\mathsf Z\!\!\mathsf 
Z}/p{\it\mathsf Z\!\!\mathsf Z}\times{\it\mathsf Z\!\!\mathsf 
Z}/p{\it\mathsf Z\!\!\mathsf Z}$.
\end {C}

{\it Proof.} Since $G/\langle x\rangle$ and $G/\langle y\rangle$ are 
abelian, then $G^c \subseteq \langle x\rangle \cap \langle y\rangle$. Let 
$p^r=card(\langle x\rangle/G^c)$ and $p^s=card(\langle y\rangle/G^c)$. 
Remark that  $card(G)$ being minimal, then $G=\langle x,y\rangle$ (i.e. $G$ 
is generated by $x$ and $y$). Let $u=xyx^{-1}y^{-1}$, then 
$x^py=u^pyx^p=yx^p$ ( since $u \in G^c$ and $ card(G^c)=p$.) So $x^p \in 
Z(G)$. By symmetry, we have also $y^p \in Z(G)$. Therefore, every element of 
$G$ can be written in the form $ax^my^n$, with $a \in Z(G)$ and $ 0 \leq m, 
n <p$.  Thus, the elements of $G/Z(G)$ can be written as $\bar x^m \bar 
y^n$, with  $ 0 \leq m, n <p$ (where $\bar x=xZ(G)$ and $\bar y= yZ(G)$.) 
So, $card(G/Z(G)) \leq p^2$. Remark that
$G/Z(G)$ can not be cyclic since otherwise $G$ would be abelian. Hence 
$card(G/Z(G))=p^2$ and $G/Z(G) \cong  {\it\mathsf Z\!\!\mathsf 
Z}/p{\it\mathsf Z\!\!\mathsf Z}\times{\it\mathsf Z\!\!\mathsf 
Z}/p{\it\mathsf Z\!\!\mathsf Z}$.

\begin {C}Under the hypotheses of (B.2), if $p$ is an odd prime integer and 
if we consider the map $\phi: G \rightarrow G$, defined by $\phi(z)=z^p$, 
then $\phi$ is a group homomorphism and $G^p:=\phi(G) \subseteq Z(G)$.
\end {C}

{\it Proof.} Let $x, y\in G$ and let $u=xyx^{-1}y^{-1}$. It is 
straightforword that 
$(xy)^p=uu^2..u^{p-1}x^py^p=(u^{\frac{p-1}{2}})^px^py^p=x^py^p$ (since 
$u^{\frac{p-1}{2}}\in G^c$ and $card(G^c)=p$). So, $\phi$ is a group 
homomorphism. Since $x^py=u^pyx^p=yx^p$, then $\phi(G)\subseteq Z(G)$. \\

Obviously, Under the hypotheses of Claim B.6, $G/ker(\phi)\cong G^p$.

\begin {C} under the hypotheses of Claim B.6, we have $ker(\phi) \subseteq 
Z(G)$.
\end {C}

{\it Proof.} If $ker(\phi)\not \subseteq Z(G)$, then one can choose two 
non-commutative elements $x$ and $y$ of $G$ with $x \in ker(\phi)$. Since 
$G=\langle x, y\rangle$ and $ord(x)=p$, then by Lemma B.1, either $\langle 
x\rangle$ or $\langle y\rangle$ is not normal in $G$. A contradiction.

\begin {C} Under the hypotheses of (B.2), $Z(G)$ is cyclic.
\end {C}

{\it Proof.} Let $x$ and $y$ be two non-commutative elements of $G$, and 
consider $w \in G$ with $ord(w)\geq p$. Then $\bar x \bar y=\bar y \bar x$ 
in $G/\langle w\rangle$. Hence $u:=xyx^{-1}y^{-1} \in \langle w\rangle$. In 
particular, this is true for $1 \neq w \in Z(G)$. Therefore, $Z(G)$ is 
cyclic (for if $Z(G)$ were not cyclic, then it would have a decomposition 
$Z(G)=\langle w_1\rangle\times \langle w_2\rangle\times ..\times \langle 
w_k\rangle$ with $ord(w_i) \geq p$ $(1 \leq i \leq k)$ and $k>1$. In 
particular, we have $\langle w_1\rangle \cap \langle w_2\rangle=\{1\}$, 
hence necessarily $u=1$. But this is not true since $x$ and $y$ are 
non-commutative.)

\begin {C} Under the hypotheses of Claim B.6, $ker(\phi)=G^c$.
\end {C}

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

\begin {C} Under the hypotheses of Claim B.6, $G/G^p$ is cyclic with 
cardinal $p$.
\end {C}

{\it Proof.}  Since $G^p \cong G/ker(\phi)$, then $card(G/G^p)=p$, so 
$G/G^p$ is cyclic.\\

Since $G^p\subseteq Z(G)$ 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.

\begin {Th} Let $p>2$ be a prime integer and $G$ a non abelian $p$-group. 
Then $G$ has a non normal subgroup.
\end {Th}

\section {Cyclic graded $p$-algebras}
\setcounter{Def}{0}

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 $GBr(F) 
\rightarrow Br(Frac(F))$, defined by $[D]\mapsto [D\otimes_FFrac(F)]$) :\\\\
{\bf(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 $a$, $b\in F^*$) :\\
(1) $(L/F, \sigma, a)=(L/F, \sigma^t, a^t)$, for any positive integer $t$ 
coprime to $n$.\\
(2) $(L/F, \sigma, a)\otimes_F(L/F, \sigma, b)$ is similar to $(L/F, \sigma, 
ab)$ with respect to $GBr(F)$.\\\\
{\bf(C.2)} (1) Let $L/F$ be a finite-dimensional cyclic graded field 
extension with Galois group generated by $\sigma$, then the mapping 
$a\mapsto [(L/F, \sigma, a)]$ ($a\in F^*$) induces a group isomorphism 
$F^*/N_{L/F}(L^*)\rightarrow GBr(L/F)$.\\
(2) Let $N/F$ be a cyclic graded field extension with Galois group generated 
by $\tau$ such that $F\subseteq L\subseteq N$ and let $s=[N : L]$ and $a\in 
F^*$, then $(N/F, \tau, a^s)$ is similar to  $(L/F, \sigma, a)$ with respect 
to $GBr(F)$.
\setcounter {Def}{2}
\begin {Th} Let $F$ be a graded field with characteristic $p>0$ and $a$ a 
non-zero homogeneous element of $F$ such that $a^{1/p}\notin F$.  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 $D=(L/F, \sigma, a)$, where $\sigma$ is a generator of $Gal(L/F)$.
\end {Th}

{\it 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 $\Phi : D \mapsto D$, defined by 
$z\mapsto xzx^{-1}$. Since $x\notin F$, then $\Phi\neq id_D$. So, $\phi=\Phi 
- id_D$ is a non-zero graded $F$-space homomorphism of $D$. Clearly, we have 
$\phi^p=0$. Let $r$ be the greater positive integer such that $\phi^r\neq 0$ 
and let $z\in D^*$ such that $\phi^r(z)\neq 0$. Put $v=\phi^{r-1}(z)$, 
$w=\phi^r(z)$ and $b=w^{-1}v$. One can easily check that $\phi(b)=b+1$. 
Therefore, $L:=F[b]$ is a cyclic graded field extension of $F$ with Galois 
group generated by $\sigma:=\phi_{/L}$, and $D=(L/F, \sigma, a)$.

\begin {Th} 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  $[A]\in GBr(F[a^{1/p^e}]/F)$ and $[A]\notin 
GBr(F[a^{1/p^f}]/F)$ for all $f<e$, then $[A]=[(N/F, \sigma, a)]$, where 
$N/F$ is a cyclic graded field extension of dimension $p^e$ with Galois 
group generated by $\sigma$.
\end {Th}

{\it 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 $A\otimes_FF[a^{1/p}]\sim 
(M/F[a^{1/p}], \sigma, a^{1/p})$, where $\sigma$ is a generator of the 
Galois group of $M/F[a^{1/p}]$. Clearly, $Frac(M)$ is a normal field 
extension of $Frac(F)$ with 
$Gal(Frac(M)/Frac(F))=Gal(Frac(M)/Frac(F[a^{1/p}]))$. Moreover if $N'$ is 
the tame closure of $F$ in $M$, then by [HW(1)99, Proposition 3.7 and 
Definition 3.8] $Frac(N')$ is the separable closure of $Frac(F)$ in 
$Frac(M)$. Therefore by [Karp89, Proposition 7.7, p.283]  
$Frac(M)=Frac(N')\otimes_{Frac(F)}Frac(F[a^{1/p}])$ (up to an isomorphism). 
This shows that $Frac(N')$ is a cyclic field extension of $Frac(F)$, hence 
by [HW(1)99, Theorem 3.11] $N'$ is a cyclic graded field extension of $F$ 
(with Galois group generated by $\sigma_{/N'}$). 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 $\tau$ such that 
$\tau_{/N'}=\sigma_{/N'}$. We  have  $A\otimes_FF[a^{1/p}]=[(M/F[a^{1/p}], 
\sigma, a^{1/p})]=[(NF[a^{1/p}]/F[a^{1/p}], \tau, a)]=[(N/F, \tau, 
a)\otimes_FF[a^{1/p}]]$. So,  $[A]-[(N/F, \tau, a)]\in GBr(F[a^{1/p}]/F)$ 
and $a^{1/p}\notin F$. Hence, again by Theorem C.3,  $[A]=[(N/F, \tau, 
a)]+[(T/F, \theta, a)]$, where  $T/F$ is a cyclic graded field extension of 
dimension $p$ with galois group generated by  $\theta$. We have either, 
$N\cap T=F$ or $T\subseteq N$. For the first case, we consider the graded 
subfield $L$ of $N\otimes_FT$ elementwise fixed by 
$\alpha:=\tau^{p^{n-1}}\otimes \theta^{-1}$. By [D82, Lemma 9, p. 75] we 
have $[(Frac(N)/Frac(F), \tau, a)]+[(Frac(T)/Frac(F), \theta, 
a)]=[(Frac(L)/Frac(F), \alpha, a)]$. So, $[(N/F, \tau, a)]+[(T/F, \theta, 
a)]=[(L/F, \alpha, a)]$ (this follows from  the canonical embedding 
$GBr(F)\rightarrow Br(Frac(F))$ defined by $[D]\mapsto 
[D\otimes_FFrac(F)]$). \\
Now, if we suppose that $T\subseteq N$, then there exists a positive integer 
$t$ coprime to $p$ such that $\tau_{/T}=\theta^t$. Since $1+tp^{n-1}$ and 
$p^n$ are coprime, then there is a positive integer $s$ such that 
$s(1+tp^{n-1})\equiv 1 (mod$ $p^n)$. Hence, by (C.1 and C.2) above $[(T/F, 
\theta, a)]=[(T/F, \theta^t, a^t)]=[(N/F, \tau, a^{tp^{n-1}})]$. So, 
$[A]=(1+tp^{n-1})[(N/F, \tau, a)]=(1+tp^{n-1})[(N/F, \tau^s, a^s)]=[(N/F, 
\tau^s, a)]$.

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 $[A]=[(N'/F, \sigma', b)]$, where $N'/F$ is a cyclic graded field 
extension with dimension $p^f$. We can then apply (C.2(2)) to conclude.

\begin {Cor} 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.
\end {Cor}

\begin {Th} 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.
\end {Th}

{\it Proof :} If $D$ is cyclic, then we can write $D=(L/F, \sigma, a)$, 
where $L$ is a a cyclic maximal graded subfield of $D$, $\sigma$ 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 
$D=\oplus_{i=1}^{n}Lx^i$, with $x^n=a$ and $xyx^{-1}=\sigma(y)$ for all 
$y\in L$. 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.
\newpage
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