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\title[Essential dimension of simple algebras] % colontitle
{Essential dimension of simple algebras}

%\keywords { {\em 2000 Mathematical Subject
%Classifications:} }


\date
%{\today}
{}


\author
[A. Merkurjev]
{Alexander S. Merkurjev}



\address
{Department of Mathematics\\
University of California\\
Los Angeles, CA 90095-1555\\
 USA}

\email
{merkurev@math.ucla.edu}

\thanks
{}




\begin{document}



\maketitle



\section{Introduction}

The essential dimension of an ``algebraic structure" is a numerical
invariant that measures its complexity. Informally, the essential
dimension of an algebraic structure over a field $F$ is the smallest
number of algebraically independent parameters required to define
the structure over a field extension of $F$ (see \cite{BerhuyFavi03}
or \cite{Merkurjev09}).



Let $\cF:\cat{Fields}/F\to\cat{Sets}$ be a functor (an ``algebraic
structure") from the category $\cat{Fields}/F$ of field extensions
of $F$ and field homomorphisms over $F$ to the category of sets. Let
$K\in\cat{Fields}/F$, $\alpha\in \cF(K)$ and $K_0$ a subfield of $K$
over $F$. We say that $\alpha$ is \emph{defined over $K_0$} (and
$K_0$ is called a \emph{field of definition of $\alpha$}) if there
exists an element $\alpha_0\in\cF(K_0)$ such that the image
$(\alpha_0)_K$ of $\alpha_0$ under the map $\cF(K_0)\to\cF(K)$
coincides with $\alpha$. The \emph{essential dimension of $\alpha$},
denoted $\ed^{\cF}(\alpha)$, is the least transcendence degree
$\td_F(K_0)$ over all fields of definition $K_0$ of $\alpha$. The
\emph{essential dimension of the functor $\cF$} is
\[
\ed(\cF)=\sup \{\ed^{\cF}(\alpha)\},
\]
where the supremum is taken over fields $K\in\cat{Fields}/F$ and all
$\alpha\in \cF(K)$.





Let $p$ be a prime integer and $\alpha\in \cF(K)$. The
\emph{essential $p$-dimension $\ed_p^{\cF}(\alpha)$ of $\alpha$} is
the minimum of $\ed^{\cF}(\alpha_{K'})$ over all finite field
extensions $K'/K$ of degree prime to $p$. The \emph{essential
$p$-dimension $\ed_p(\cF)$ of $\cF$} is the supremum of
$\ed_p^{\cF}(\alpha)$ over all fields $K\in\cat{Fields}/F$ and all
$\alpha\in \cF(K)$ (see \cite[\S 6]{Reichstein00a}). Clearly,
$\ed(\cF)\geq\ed_p(\cF)$ for all $p$.

Let $G$ be an algebraic group over $F$. The \emph{essential
dimension $\ed(G)$} (resp. \emph{essential $p$-dimension
$\ed_p(G)$}) of $G$ is the essential dimension (resp. essential
$p$-dimension) of the functor $G\text{-}\tors$ taking a field $K$ to
the set of isomorphism classes of all $G$-torsors (principal
homogeneous $G$-spaces) over $K$.

If $G=\gPGL_n$ over $F$, the functor $G\text{-}\tors$ is isomorphic
to the functor $\Alg_F(n)$ taking a field $K$ to the set of
isomorphism classes of central simple $K$-algebras of degree $n$.
Let $p$ be a prime integer and let $p^r$ be the highest power of $p$
dividing $n$. Then $\ed_p\(\Alg_F(n)\)=\ed_p\(\Alg_F(p^r)\)$
\cite[Lemma 8.5.5]{Reichstein00a}. Every central simple $E$-algebra
of degree $p$ is cyclic over a finite field extension of degree
prime to $p$, hence $\ed_p\(\Alg_F(p)\)=2$ \cite[Lemma
8.5.7]{Reichstein00a}. It was proven in \cite{Merkurjev09a} that
$\ed_p\(\Alg_F(p^2)\)=p^2+1$ and in general,
$\ed_p\(\Alg_F(p^r)\)\geq 2r$ for all $r$ in \cite[Th.
8.6]{Reichstein00a}.







We prove the following:

\noindent{\bf Theorem.} \emph{Let $F$ be a field and $p$ an integer
different from $\ch(F)$. Then}
\[
\ed_p\(\Alg_F(p^r)\)\geq (r-1)p^r+1.
\]

In other words, we have the following lower bound for the essential
dimension of $\gPGL_F(p^r)$:
\[
\ed\(\gPGL_F(p^r)\)\geq\ed_p\(\gPGL_F(p^r)\)\geq (r-1)p^r+1.
\]


\section{Preliminaries}

\subsection{Characters}\label{cycalg}
Let $F$ be a field, $F_{\sep}$ a separable closure of $F$ and
$\Gamma=\Gal(F_{\sep}/F)$ the \emph{absolute Galois group} of $F$.
For a $\Gamma$-module $M$ we write $H^n(F,M)$ for the cohomology
group $H^n(\Gamma,M)$.


The \emph{character group} $\cat{\Ch}(F)$ of $F$ is defined as
\[
\Hom_{cont}(\Gamma,\Q/\Z)=H^1(F,\Q/\Z)\simeq H^2(F,\Z).
\]
For a character $\chi\in\Ch(F)$, set
$F(\chi)=(F_{\sep})^{\Ker(\chi)}$. Then $F(\chi)/F$ is a cyclic
field extension of degree $\ord(\chi)$. If $\Phi\subset \Ch(F)$ is a
finite subgroup, we set
\[
F(\Phi)=(F_{\sep})^{\cap\Ker(\chi)},
\]
where the intersection is taken over all $\chi\in\Phi$. The Galois
group $G=\Gal\(F(\Phi)/F\)$ is abelian and $\Phi$ is canonically
isomorphic to the character group $\Ch(G)=\Hom(G,\Q/\Z)$ of $G$.


If $F'\subset F$ is a subfield and $\chi\in\Ch(F')$, we write
$\chi_F$ for the image of $\chi$ under the natural map $\Ch(F')\to
\Ch(F)$ and $F(\chi)$ for $F(\chi_F)$. If $\Phi\subset \Ch(F)$ is a
finite subgroup, then the character $\chi_{F(\Phi)}$ is trivial if
and only if $\chi\in\Phi$.

%Let $K/F$ be cyclic field extension. Choose a character
%$\chi\in\Ch(F)$ such that $K=F(\chi)$.

%In fact, every element of $\Br(K/F)$ is of the form
%$\chi\tens a$ for some $a\in F^\times$.



\begin{lemma}\label{charup}
Let $\Phi,\Phi'\subset \Ch(F)$ be two finite subgroups. Suppose that
for a field extension $K/F$, we have $\Phi_K=\Phi'_K$ in $\Ch(K)$.
Then there is a finite subextension $K'/F$ in $K/F$ such that
$\Phi_{K'}=\Phi'_{K'}$ in $\Ch({K'})$.
\end{lemma}

\begin{proof}
Choose a set of characters $\{\chi_1,\dots,\chi_m\}$ generating
$\Phi$ and a set of characters $\{\chi'_1,\dots,\chi'_m\}$
generating $\Phi'$ such that $(\chi_i)_K=(\chi'_i)_K$ for all $i$.
Let $\eta_i=\chi_i-\chi'_i$. As all $\eta_i$ vanish over $K$, the
finite field extension $K':=F(\eta_1,\dots, \eta_m)$ of $F$ can be
viewed as a subextension in $K/F$. As
$(\chi_i)_{K'}=(\chi'_i)_{K'}$, we have $\Phi_{K'}=\Phi'_{K'}$.
\end{proof}

\subsection{Brauer group}

We write $\Br(F)$ for the \emph{Brauer group}
$H^2(F,F_{\sep}^\times)$ of a field $F$. If $a\in \Br(F)$ and $K/F$
is a field extension, then we write $a_K$ for the image of $a$ under
the natural homomorphism $\Br(F)\to\Br(K)$. We write $\Br(K/F)$ for
the \emph{relative Brauer group} $\Ker\(\Br(F)\to \Br(K)\)$. We say
that $K$ is a splitting field of $a$ if $a_K=0$, i.e.,
$a\in\Br(K/F)$. The \emph{index} $\ind(a)$ of $a$ is the smallest
degree of a splitting field of $a$.




The cup-product
\[
\Ch(F)\tens F^\times=H^2(F,\Z)\tens H^0(F,F_{\sep}^\times)\to
H^2(F,F_{\sep}^\times)=\Br(F)
\]
takes $\chi\tens a$ to the class $\chi\cup(a)$ in $\Br(F)$ that is
split by $F(\chi)$.

For a finite subgroup $\Phi\subset \Ch(F)$ write
$\Br_{\dec}\(F(\Phi)/F\)$ for the \emph{subgroup of decomposable
elements} in $\Br\(F(\Phi)/F\)$ generated by the elements
$\chi\cup(a)$ for all $\chi\in\Phi$ and $a\in F^\times$. The
\emph{indecomposable relative Brauer group}
$\Br_{\ind}\(F(\Phi)/F\)$ is the factor group
$\Br\(F(\Phi)/F\)/\Br_{\dec}\(F(\Phi)/F\)$.



\subsection{Complete fields}
Let $E$ be a complete field with respect to a discrete valuation $v$
and $K$ its residue field.

Let $p$ be a prime integer different from $\ch(K)$. There is a
natural injective homomorphism $\Ch(K)\{p\}\to \Ch(E)\{p\}$ of the
$p$-primary components of the character groups that identifies
$\Ch(K)\{p\}$ with the character group of an unramified field
extension of $E$. For a character $\chi\in\Ch(K)\{p\}$, we write
$\widehat\chi$ for the corresponding character in $\Ch(E)\{p\}$.

By \cite[\S 7.9]{GMS}, there is an exact sequence
\begin{equation}\label{brp}
0\to \Br(K)\{p\}\xra{i} \Br(E)\{p\}\xra{\partial_v} \Ch(K)\{p\}\to
0.
\end{equation}

If $a\in\Br(K)\{p\}$, then we write $\widehat a$ for the element
$i(a)$ in $\Br(E)\{p\}$. For example, if $a=\chi\cup(\bar u)$ for
some $\chi\in\Ch(K)\{p\}$ and a unit $u\in E$, then $\widehat
a=\widehat\chi\cup(u)$.

The following proposition was proved in \cite[Th. 5.15(a)]{JW90},
\cite[Prop. 2.4]{Tignol78}) and \cite[Prop. 8.2]{GMS}.

\begin{proposition}\label{propbr}
Let $E$ be a complete field with respect to a discrete valuation $v$
and $K$ its residue field of characteristic different from $p$. Then
\begin{enumerate}
  \item $\ind(\widehat a)=\ind(a)$ for any $a\in \Br(K)\{p\}$.
  \item Let $b=\widehat a+\(\chi\cup(x)\)$ for an element $a\in
  \Br(K)\{p\}$, $\chi\in\Ch(K)\{p\}$ and $x\in E^\times$ such
  that $v(x)$ is not divisible by $p$. Then
  \[
  \ind(b)=\ind(a_{K(\chi)}\)\cdot\ord(\chi).
  \]
  \item Let $E'/E$ be a finite field extension and $v'$ the discrete valuation
   on $E'$ extending $v$ with residue field $K'$. Then for any $b\in
   \Br(E)\{p\}$, one has
\[
\partial_{v'}(b_{E'})=e\cdot\partial_v(b)_{K'},
\]
where $e$ is the ramification index of $E'/E$.
\end{enumerate}
\end{proposition}







The choice of a prime element $\pi$ in $E$ provides with a splitting
of the sequence (\ref{brp}) by sending a character $\chi$ to the
class $\widehat\chi\cup (\pi)$ in $\Br(E)\{p\}$. Thus, any
$b\in\Br(E)\{p\}$ we can written in the form:
\begin{equation}\label{spec}
b=\widehat a+\(\widehat\chi\cup (\pi)\)
\end{equation}
for $\chi=\partial_v(b)$ and a unique $a\in \Br(K)\{p\}$.

The homomorphism
\[
s_\pi:\Br(E)\{p\}\to \Br(K)\{p\},
\]
defined by $s_\pi(b)=a$, where $a$ is given by (\ref{spec}), is
called a \emph{specialization} map. For example, $s_\pi(\widehat
a)=a$ for any $a\in \Br(K)\{p\}$ and
$s_\pi\(\widehat\chi\cup(x)\)=\chi\cup(\bar u)$, where
$\chi\in\Ch(K)\{p\}$, $x\in E^\times$ and $u$ is the unit in $E$
such that $x=u\pi^{v(x)}$.

Moreover, if $v$ is trivial on a subfield $F\subset E$ and
$\Phi\subset\Ch(F)\{p\}$ a finite subgroup, then
\begin{equation}\label{miau}
s_\pi\(\Br_{\dec}(E(\Phi)/E)\)\subset \Br_{\dec}(K(\Phi)/K).
\end{equation}


\smallskip

We shall need the following technical Lemma. For an abelian group
$A$ we write $_p A$ for the subgroup of all elements in $A$ of
exponent $p$.

\begin{lemma}\label{base}
Let $(E, v)$ be a complete discrete valued field with the residue
field $K$ of characteristic different from $p$ containing a
primitive $p^2$-th root of unity. Let $\eta\in \Ch(E)$ be a
character of order $p^2$ such that $p\cdot\eta$ is unramified, i.e.,
$p\cdot\eta=\widehat\nu$ for some $\nu\in \Ch(K)$ of order $p$. Let
$\chi\in_p\!\Ch(K)$ be a character linearly independent from $\nu$.
Let $a\in\Br(K)$ and set $b=\widehat
a+\(\widehat\chi\cup(x)\)\in\Br(E)$, where $x\in E^\times$ is an
element such that $v(x)$ is not divisible by $p$. Then:
\begin{enumerate}
  \item If $\eta$ is unramified, , i.e.,
$\eta=\widehat\mu$ for some $\mu\in \Ch(K)$ of order $p^2$, then
$\ind(b_{E(\eta)})=p\cdot\ind(a_{K(\mu,\chi)})$.
  \item If $\eta$ is ramified, then there exists a unit $u\in
  E^\times$ such that $K(\nu)=K(\bar u^{1/p})$ and
  $\ind(b_{E(\eta)})=\ind\(a-(\chi\cup(\bar u^{1/p}))\)_{K(\nu)}$.
\end{enumerate}
\end{lemma}

\begin{proof}
$(1)$ If $\eta=\widehat\mu$ for some $\mu\in \Ch(K)$, then $K(\mu)$
is the residue field of $E(\eta)$ and we have
\[
b_{E(\eta)}=\widehat a_{K(\mu)}+\(\widehat{\chi}_{K(\mu)}\cup(x)\).
\]
As $\chi$ and $\nu$ are linearly independent, the character
$\chi_{K(\mu)}$ is nontrivial. The first statement follows from
Proposition \ref{propbr}(2).

\smallskip

$(2)$ Since $p\cdot\eta$ is unramified, the ramification index of
$E(\eta)/E$ is equal to $p$, hence $E(\eta)=E\((ux^p)^{1/p^2}\)$ for
some unit $u\in E$. Note that $K(\nu)=K(\bar u^{1/p})$ is the
residue field of $E(\eta)$. As $u^{1/p}x$ is a $p$-th power in
$E(\eta)$, the class
\[
b_{E(\eta)}=\widehat
a_{K(\nu)}-\(\widehat{\chi}_{K(\nu)}\cup(u^{1/p})\)=
\widehat{a}_{K(\nu)}-\(\widehat{\chi_{K(\nu)}\cup(\bar u^{1/p})\)}
\]
is unramified. It follows from Proposition \ref{propbr}(1) that the
elements $b_{E(\eta)}$ in $\Br\(E(\eta)\)$ and
$a_{K(\nu)}-\(\chi_{K(\nu)}\cup(\bar u^{1/p})\)$ in $\Br\(K(\nu)\)$
have the same indices.
\end{proof}



\section{Brauer group and algebraic tori}\label{brtori}

\subsection{Torsors}
Let $G$ be an algebraic groups over $F$ and let $K/F$ be a field
extension. The set of isomorphism classes of $G$-torsors (principal
homogeneous spaces) over $K$ is bijective to $H^1(K,G)$ (see
\cite{Serre97}).

\begin{example}\label{algtorsors}
Let $A$ be a central simple $F$-algebra of degree $n$ and
$G=\gAut(A)$. Then $H^1(K,G)$ is the set of isomorphism classes of
central simple $K$-algebras of degree $n$, or equivalently, the set
of elements in $\Br(K)$ of index dividing $n$. If $A=M_n(F)$ is the
split algebra, then $G=\gPGL_{n,F}$.
\end{example}

\begin{example}\label{toritorsors}
Let $L$ be an \'etale $F$-algebra of dimension $n$. Consider the
algebraic torus $U=R_{L/F}(\gml)/\gm$ over $F$. The exact sequence
\[
1\to \gm\to R_{L/F}(\gml)\to U\to 1
\]
and Hilbert Theorem 90 yield an isomorphism
$\theta:H^1(F,U)\iso\Br(L/F)$. Note that if $L$ is a subalgebra of a
central simple $F$-algebra $A$ of degree $n$, then $U$ is a maximal
torus in the group $\gAut(A)$.
\end{example}

Let $\alpha:G\to\gGL(W)$ be a finite dimensional representation over
$F$. Suppose that $\alpha$ is \emph{generically free}, i.e., there
is a non-empty open subset $W'\subset W$ and a $G$-torsor
$\beta:W'\to X$ for a variety $X$ over $F$. The torsor $\beta$ is
\emph{versal}, i.e., every $G$-torsor over a field extension $K/F$
is the pull-back of $\beta$ with respect to a $K$-point of $X$. The
generic fiber of $\beta $ is called a \emph{generic} $G$-torsor. It
is a torsor over the function field $F(X)$ (see \cite{GMS} and
\cite{Reichstein00}).

\begin{example}\label{toritorsors2}
Let $S$ be an algebraic torus over $F$. We embed $S$ into the
quasi-trivial torus $P=R_{L/F}(\gml)$, where $L$ is an \'etale
$F$-algebra (see \cite{Colliot77}). Then $S$ acts on the vector
space $L$ by multiplication, so that the action on the open subset
$P$ is regular. If $T$ is the factor torus $P/S$, then the
$S$-torsor $P\to T$ is versal.
\end{example}


\subsection{The tori $P^\Phi$, $S^\Phi$, $T^\Phi$, ${U}^\Phi$ and
$V^\Phi$}\label{toruss}

Let $F$ be a field, $\Phi$ a subgroup of $_p\Ch(F)$ of rank $r$ and
$L=F(\Phi)$. Let $G=\Gal(L/F)$. Choose a basis
$\chi_1,\chi_2,\dots,\chi_r$ for $\Phi$. We can view each $\chi_i$
as a character of $G$, i.e., as a homomorphism $\chi_i:G\to\Q/\Z$.
Let $\sigma_1,\sigma_2,\dots,\sigma_r$ be the dual basis for $G$,
i.e.,
\[
\chi_i(\sigma_j)= \left\{
  \begin{array}{ll}
    (1/p)+\Z, & \hbox{if $i=j$;} \\
    0, & \hbox{otherwise.}
  \end{array}
\right.
\]

Let $R$ be the group ring $\Z[G]$. Consider the surjective
homomorphism of $G$-modules $k:R^r\to R$ taking the $i$-th basis
element $e_i$ of $R^r$ to $\sigma_i-1$. The image of $k$ is the
\emph{augmentation ideal} $I=\Ker(\varepsilon)$ in $R$, where
$\varepsilon: R\to\Z$ is defined by $\varepsilon(\rho)=1$ for all
$\rho\in G$.


Write $N_i=1+\sigma_i+\sigma_i^2+\dots +\sigma_i^{p-1}\in R$.

Set $N:=\Ker(k)$. Consider the following elements in $N$:
\[
e_{ij}:=(\sigma_i-1)e_j-(\sigma_j-1)e_i\quad\text{and}\quad
f_i=N_ie_i,\ \ i,j=1,\dots r.
\]

\begin{lemma}\label{gene}
The $G$-module $N$ is generated by $e_{ij}$ and $f_i$.
\end{lemma}
\begin{proof}
Let $\overline R=\Z[t_1,\dots,t_r]$ be the polynomial ring.
Acyclicity of the Koszul complex for the homomorphism $\bar k:
(\overline R)^r\to \overline R$, taking the $i$-th basis element
$\bar e_i$ to $t_i-1$ (see \cite[Th. 43]{Matsumura}) implies that
$\Ker(\bar k)$ is generated by $\bar e_{ij}:=(t_i-1)\bar
e_j-(t_j-1)\bar e_i$.


The kernel $J$ of the surjective homomorphism $\overline R\to R$,
taking $t_i$ to $\sigma_i$, is generated by $t_i^p-1$.

Let $x:=\sum x_ie_i\in\Ker(k)$. Lift every $x_i$ to a polynomial
$\bar x_i\in \overline R$ and consider $\bar x:=\sum \bar x_i\bar
e_i\in (\overline R)^r$. We have $\bar k(\bar x)\in J$, hence
\[
\bar k(\bar x)=\sum (t_i-1)\bar x_i =\sum (t_i^p-1)h_i=\sum
(t_i-1)\overline N_ih_i
\]
for some polynomials $h_i\in \overline R$, where $\overline
N_i=1+t_i+t_i^2+\dots +t_i^{p-1}\in R$. Hence the element $\sum
(\bar x_i-h_i\overline N_i)\bar e_i$ belongs to the kernel of $\bar
k$ and therefore is a linear combination of $\bar e_{ij}$. It
follows that $\bar x$ is a linear combination of $\bar e_{ij}$ and
$\overline N_i\bar e_i$, hence $x$ is a linear combination of
$e_{ij}$ and $f_i$.
\end{proof}


Let $\varepsilon_i:R^r\to\Z$ be the $i$-th projection followed by
the augmentation map $\varepsilon$. It follows from Lemma \ref{gene}
that $\varepsilon_i(N)=p\Z$ for every $i$. Moreover, the
$G$-homomorphism
\[
l:N\to\Z^r,\quad m\mapsto
(\varepsilon_1(m)/p,\dots,\varepsilon_r(m)/p)
\]
is surjective. Set $M=\Ker(l)$ and $Q=R^r/M$.

\begin{lemma}\label{genforn}
The $G$-module $M$ is generated by $e_{ij}$.
\end{lemma}

\begin{proof}
Let $M'$ be the submodule of $N$ generated by $e_{ij}$. Clearly,
$M'\subset M$. Note also that $(\sigma_j-1)f_i=N_i e_{ij}\in M'$,
hence $If_i\subset M'$.



Suppose that $m\in M$. By Lemma \ref{gene}, modifying $m$ by an
element in $M'$ we can assume that $m=\sum_{i=1}^r x_if_i$ for some
$x_i\in R$. As $l(m)=0$, we have $\varepsilon(x_i)=0$, i.e., $x_i\in
I$ for all $i$, hence $m\in\sum If_i\subset M'$.
\end{proof}



Let $P^\Phi$, $S^\Phi$, $T^\Phi$, ${U}^\Phi$ and $V^\Phi$ be the
algebraic tori over $F$ with the character $G$-modules $R^r$, $Q$,
$M$, $I$ and $N$,  respectively. The diagram of homomorphisms of
$G$-modules with exact columns and rows
\begin{equation}\label{diagmod}
\xymatrix{
M  \ar@{_{(}->} [d]\ar@{=}[r] & M \ar@{_{(}->}[d] &  \\
N\ar@{->>}_l[d]\ar@{_{(}->}[r]  &  R^r \ar@{->>}[d]\ar@{->>}^k[r] & I \ar@{=}[d]\\
\Z^r \ar@{_{(}->}[r] & Q \ar@{->>}[r] & I  \\
}
\end{equation}
yields the following diagram of homomorphisms of the tori
\begin{equation}\label{diag}
\xymatrix{
U^\Phi  \ar@{=} [d]\ar@{_{(}->}[r] & S^\Phi \ar@{_{(}->}[d] \ar@{->>}[r]&\gm^r \ar@{_{(}->}[d]  \\
U^\Phi  \ar@{_{(}->}[r]  &  P^\Phi \ar@{->>}[d]\ar@{->>}[r] & {V}^\Phi \ar@{->>}[d]\\
&  T^\Phi \ar@{=}[r] & T^\Phi   \\
}
\end{equation}


\begin{comment}
\begin{equation}\label{diag}
\begin{CD}
U^\Phi  @>>> S^\Phi @>>> \gm^r       \\
@|    @VVV  @VVV    \\
U^\Phi  @>>> P^\Phi @>>>  {V}^\Phi \\
@.    @VVV   @VVV   \\
  @. T^\Phi @= T^\Phi        \\
\end{CD}
\end{equation}
\end{comment}

\medskip

Let $K/F$ be a field extension. Set $KL:=K\tens_F L$. The exact
sequence of $G$-modules
\begin{equation}\label{augg}
0\to I\to R\to \Z \to 0
\end{equation}
gives an exact sequence of the tori
\[
1\to \gm \to R_{L/F}(\gml) \to U \to 1
\]
and then an exact sequence
\[
0\to H^1(K,U^\Phi)\to H^2(K,\gm)\to H^2(KL,\gm).
\]
Hence
\begin{equation}\label{hu}
H^1(K,U^\Phi)\simeq\Br(KL/K).
\end{equation}

\begin{lemma}\label{compute}
The homomorphism $(K^\times)^r\to H^1(K,U^\Phi)\simeq\Br(KL/K)$
induced by the first row of the diagram {\rm (\ref{diag})} takes
$(x_1,\dots,x_r)$ to $\sum_{i=1}^r \((\chi_i)_K\cup(x_i)\)$.
\end{lemma}

\begin{proof}
Consider the composition
\begin{equation}\label{comp7}
h:\Hom_G(\Z^r,\Z)\to \Ext^1_G(I,\Z)\to
\Ext^2_G(\Z,\Z)=H^2(G,\Z)=\Ch(G),
\end{equation}
where the first homomorphism is induced by the bottom row of the
diagram (\ref{diagmod}) and the second one - by the exact sequence
(\ref{augg}).


We claim that for any $k$, the image of the $k$-th projection
$p_k:\Z^r\to\Z$ under the composition (\ref{comp7}) coincides with
$\chi_k$. Consider the $G$-homomorphism $R^r\to \Q$, taking $e_k$ to
$1/p$ and $e_i$ to $0$ for all $i\neq k$. By Lemma \ref{genforn},
this homomorphism vanishes on $M$ and hence it factors through a map
$Q\to\Q$. Thus, we have a commutative diagram
\begin{equation}\label{diag5}
\begin{CD}
0  @>>> \Z^r  @>>> Q @>>> I @>>> 0       \\
@.    @V{p_k}VV  @VVV  @V{f_k}VV @. \\
0  @>>> \Z  @>>> \Q @>>> \Q/\Z @>>> 0         \\
\end{CD}
\end{equation}
for the map $f_k$ defined by $f_k(\sigma_k-1)=1/p+\Z$ and
$f_k(\sigma_i-1)=0$ for all $i\neq k$.

Let $\alpha$ be the image of the class of the top row of
$(\ref{diag5})$ under the map $p_k^*:\Ext^1_G(I,\Z^r)\to
\Ext^1_G(I,\Z)$. Then $h(p_k)$ is the image of $\alpha$ under the
second map in the composition (\ref{comp7}). Hence $h(p_k)$ is also
the image of the class $\beta$ of the sequence (\ref{augg}) under
the connecting map $H^1(G,I)=\Ext^1_G(\Z,I)\to
\Ext^2_G(\Z,\Z)=H^2(G,\Z)$ induced by the exact sequence
representing the class $\alpha$.

The diagram (\ref{diag5}) yields a commutative diagram
\[
\begin{CD}
H^1(G,I)  @>\partial>> H^2(G,\Z^r)       \\
@V{f_k^*}VV    @V{p_k^*}VV      \\
H^1(G,\Q/\Z)  @= H^2(G,\Z)      \\
\end{CD}
\]
As we have shown, $p_k^*(\partial(\beta))=h(p_k)$. Therefore, it
suffices to prove that $f_k^*(\beta)=\chi_k$. The cocycle $\beta$
satisfies $\beta(\sigma_i)=\sigma_i-1$. It follows that
$f_k^*(\beta)(\sigma_k)=f_k(\sigma_k-1)=1/p+\Z$ and
$f_k^*(\beta)(\sigma_i)=0$ for all $i\neq k$. This proves the claim.

Consider the commutative diagram
\[
\xymatrix{ (K^\times)^r=\Hom_G(\Z^r,\Z)\tens K^\times \ar@{=}
[d]\ar@{->}[r] & \Ext^1_G(I,\Z)\tens
K^\times \ar@{->}[d] \ar@{->}[r]& \Ext^2_G(\Z,\Z)\tens K^\times  \ar@{->}[d]  \\
(K^\times)^r=\Hom_G(\Z^r,KL^\times)   \ar@{->}[r]  &  \Ext^1_G(I,KL^\times)
\ar@{->}[r] &  \Ext^2_G(\Z,KL^\times),\\
}
\]
\begin{comment}
\[
\begin{CD}
(K^\times)^r=\Hom_G(\Z^r,\Z)\tens K^\times  @>>> \Ext^1_G(I,\Z)\tens
K^\times
@>>> \Ext^2_G(\Z,\Z)\tens K^\times     \\
@|    @VVV  @VVV    \\
(K^\times)^r=\Hom_G(\Z^r,KL^\times)  @>>> \Ext^1_G(I,KL^\times)
@>>> \Ext^2_G(\Z,KL^\times),   \\
\end{CD}
\]
\end{comment}
where the vertical homomorphisms are given by the
cup-products. By the claim, the image of the tuple $(x_1,\dots,x_r)$
under the diagonal composition is equal to $\sum_{i=1}^r
\((\chi_i)_K\cup(x_i)\)$. On the other hand, the bottom composition
coincides with $(K^\times)^r\to H^1(K,U^\Phi)\simeq\Br(KL/K)$.
\end{proof}


\begin{corollary}\label{compute2}
The map $H^1(K,U^\Phi)\to H^1(K,S^\Phi)$ induces an isomorphism
$H^1(K,S^\Phi)\simeq \Br_{\ind}(KL/K)$.
\end{corollary}


It follows from Corollary \ref{compute2} the triviality of the group
$H^1(K,P^\Phi)$ that we have a commutative diagram
\begin{equation}\label{diag2}
\begin{CD}
V(K)  @>>> H^1(K,U^\Phi) @= \Br(KL/K)        \\
@VVV    @VVV  @VVV     \\
T(K)  @>>>  H^1(K,S^\Phi) @= \Br_{\ind}(KL/K)       \\
\end{CD}
\end{equation}
with surjective homomorphisms.

\subsection{The element $a$}\label{thea}

Let $a'$ be the image of the generic point of $V$ over $K=F(V)$ in
$\Br\(L(V)/F(V)\)$ in the diagram (\ref{diag2}). Choose also an
element $a\in \Br\(L(T)/F(T)\)$ corresponding to the generic point
of $T$ over $F(T)$. The field $F(T)$ is a subfield of $F(V)$ and the
classes $a_{F(V)}$ and $a'$ are equal in $\Br_{\ind}\(L(V)/F(V)\)$.
It follows that $pa_{F(V)}=pa'$ in $\Br F(V)$.

The exact sequence of $G$-modules
\[
0\to L^\times\oplus N\to L(V)^\times\to \Div(V_L)\to 0
\]
induces an exact sequence
\[
H^1\(G,\Div(V_L)\)\to H^2(G,L^\times)\oplus H^2(G,N)\to
H^2\(G,L(V)^\times\).
\]
As $\Div(V_L)$ is a permutation $G$-module, the first term in the
sequence is trivial. Therefore, we get an injective homomorphism
\[
\varphi: H^2(G,N)\to \Br F(V)/\Br(F).
\]
Then (\ref{diagmod}) and (\ref{augg}) yield
\[
H^2(G,N)\simeq H^1(G,I)\simeq \hat H^0(G,\Z)=\Z/p^r\Z,
\]
thus, $H^2(G,N)$ has a canonical generator $\xi$ of order $p^r$.

\begin{lemma} {\rm (cf., \cite[Lemma 2.4]{Merkurjev09a})}
We have $\varphi(\xi)=-a' +\Br(F)$.
\end{lemma}

\begin{proof}
Consider the following diagram

\[
\xymatrix{
&  &  \Hom_G(\Z,\Z)  \ar@{->} [d] \\
&  \Hom_G(I,I)  \ar@{->}[d]\ar@{->} [r]  &
\Ext^1_G(\Z,I)\ar@{->}[d] \\
\Hom_G(N,N)  \ar@{->}[d]\ar@{->} [r] & \Ext^1_G(I,N) \ar@{->}_l[d]
\ar@{->} [r] & \Ext^2_G(\Z,N) \ar@{->}[d] \\
\Hom_G\(N,L(V)^\times\)  \ar@{->} [r] & \Ext^1_G\(I,L(V)^\times\)
\ar@{->} [r] & \Ext^2_G\(\Z,L(V)^\times\)\\  \\
}
\]
By \cite[Ch. XIV]{Cartan}, the images of $1_{\Z}$ and $-1_I$ agree
in $\Ext^1_G(\Z,I)$ and the images of $1_N$ and $-1_I$ agree in
$\Ext^1_G(I,N)$. It follows from \cite[Ch. V, Prop. 4.1]{Cartan}
that the upper square is anticommutative. The image of $1_{\Z}$ is
equal to $\varphi(\xi)$ and the image of $1_{N}$ is equal to $a'
+\Br(F)$ in the right bottom corner.
\end{proof}

\begin{corollary}\label{notinimage}
If $r\geq 2$, then the class $p^{r-1}a$ in $\Br F(T)$ does not
belong to the image of $\Br(F)\to\Br F(T)$.
\end{corollary}
\begin{proof}
The image of $p^{r-1}a$ in $\Br F(V)$ coincides with $p^{r-1}a'$.
Modulo the image of the map $\Br(F)\to\Br F(V)$, the class
$p^{r-1}a'$ is equal to $-\varphi(p^{r-1}\xi)$ and therefore, is
nonzero as $\varphi$ is injective.
\end{proof}
\begin{comment}
\subsection{The torus $S^\Phi$}
Let $\Phi\subset_p\!\Ch(F)$ be a finite subgroup of rank $r$ and
$L=F(\Phi)$ and $G=\Gal(L/F)$. Let $X=Q$. As $N/M\simeq\Z^r$, by
???, we have an exact sequence of $G$-modules
\[
0\to \Z^r\to X\to I\to 0.
\]
Let $S^\Phi$ be the torus with the $G$-module of characters $X$.
Thus, we have an exact sequence
\[
1\to U\to S^\Phi\to (\gm)^r \to 1.
\]
It yields an exact sequence
\[
(K^\times)^r\to H^1(K,U)\to H^1(K,S^\Phi)\to 1
\]
for any field extension $K/F$. It follows from (\ref{hu}) and ???
that
\begin{equation}\label{su}
H^1(K,S^\Phi)\simeq \Br(KL/K)/\Br_{\dec}(KL/K).
\end{equation}
\end{comment}


\section{Essential dimension of algebraic tori}




Let $S$ be an algebraic torus over $F$ with the splitting group
$G$. We assume that $G$ is a $p$-group of order $p^r$. Let $X$ be
the $G$-module of characters of $S$. A \emph{$p$-presentation} of
$X$ is a $G$-homomorphism $f:P\to X$ with $P$ a permutation
$G$-module and finite cokernel of order prime to $p$. A
$p$-presentation with the smallest $\rank(P)$ is called
\emph{minimal}.

Essential $p$-dimension of algebraic tori was determined in
\cite[Th. 1.4]{LMMR}:

\begin{theorem}\label{reich}
Let $S$ be an algebraic torus over $F$ with the splitting $p$-group
$G$, $X$ the $G$-module of characters of $S$ and $f:P\to X$ a
minimal $p$-presentation of $X$. Then $\ed_p(S)=\rank\(\Ker(f)\)$.
\end{theorem}

\begin{corollary}\label{reich2}
Suppose that $X$ admits a surjective minimal $p$-presentation
$f:P\to X$. Then $\ed(S)=\ed_p(S)=\rank\(\Ker(f)\)$.
\end{corollary}

\begin{proof}
As explained in Example \ref{toritorsors2}, a surjective
$G$-homomorphism $f$ yields a generically free representation of $S$
of dimension $\rank(P)$. By \cite[\S 3]{Reichstein00},
\[
\ed_p(S)\leq\ed(S)\leq \rank(P)-\dim(S)=\rank\(\Ker(f)\).\qedhere
\]
\end{proof}

In this section we derive from \ref{reich} an explicit formula for
the essential $p$-dimension of algebraic tori.


Define the group $\overline X:=X/(pX+IX)$, where $I$ is the
augmentation ideal in $R=\Z[G]$. For any subgroup $H\subset G$,
consider the composition $X^H\hookrightarrow X\to \overline X$. For
every $k$, let $V_k$ denote the image of the homomorphism
\[
\coprod_{H\subset G} X^H\to\overline X,
\]
where the coproduct is taken over all subgroups $H$ with $[G:H]\leq
p^k$. We have the sequence of subgroups
\begin{equation}\label{vis}
0=V_{-1}\subset V_0\subset\dots\subset V_r=\overline X.
\end{equation}




\begin{theorem}\label{edtori}
We have the following explicit formula for the essential
$p$-dimension of $S$:
\[
\ed_p(S)=\sum_{k=0}^{r}(\rank V_k-\rank V_{k-1})p^k -\dim(S).
\]
\end{theorem}

\begin{proof}
Set$b_k=\rank(V_k)$. By Theorem \ref{reich}, it suffices to prove
that the smallest rank of the $G$-module $P$ is a $p$-presentation
of $X$ is equal to $\sum_{k=0}^{r}(b_k-b_{k-1})p^k$.

Let $f:P\to X$ be a $p$-presentation of $X$ and $A$ a $G$-invariant
basis of $P$. The set $A$ is the disjoint union of the $G$-orbits
$A_j$, so that $P$ is the direct sum of the permutation $G$-modules
$\Z[A_j]$.


The composition $\bar f:P\to X\to \overline X$ is surjective. As $G$
acts trivially on $\overline X$, the rank of the group $\bar
f(\Z[A_j])$ is at most $1$ for all $j$ and $\bar f(\Z[A_j])\subset
V_k$ if $|A_j|\leq p^k$. It follows that the group $\overline X/V_k$
is generated by the images under the composition $P\xra{\bar
f}\overline X\to\overline X/V_k$ of all $\Z[A_j]$ with $|A_j|> p^k$.
Denote by $c_k$ the number of such orbits $A_j$, so we have
\[
c_k\geq\rank(\overline X/V_k)=b_r-b_k.
\]
Set $c'_k=b_r-c_k$, so that $b_k\geq c'_k$ for all $k$ and
$b_r=c'_r$.


Since the number of orbits $A_j$ with $|A_j|= p^k$ is equal to
$c_{k-1}-c_k$, we have
\begin{multline*}
\rank (P)=\sum_{k=0}^{r}(c_{k-1}-c_k)p^k=
\sum_{k=0}^{r}(c'_k-c'_{k-1})p^k=\\
c'_rp^r+\sum_{k=0}^{r-1}c'_k(p^k-p^{k+1})\geq
b_rp^r+\sum_{k=0}^{r-1}b_k(p^k-p^{k+1})=\sum_{k=0}^{r}(b_k-b_{k-1})p^k.
\end{multline*}

It remains to construct a $p$-presentation with $P$ of rank
$\sum_{k=0}^{r}(b_k-b_{k-1})p^k$. For every $k\geq 0$ choose a
subset $X_k$ in $X$ of the pre-image of $V_k$ under the canonical
map $X\to \overline X$ with the property that for any $x\in X_k$
there is a subgroup $H_x\subset G$ with $x\in X^{H_x}$ and $[G:H_x]=
p^k$ such that the composition
\[
X_k\to V_k\to V_k/V_{k-1}
\]
yields a bijection between $X_k$ and a basis of $V_k/V_{k-1}$. In
particular, $|X_k|=b_k-b_{k-1}$. Consider the $G$-homomorphism
\[
f:P:=\coprod_{k=0}^{r}\coprod_{x\in X_k} \Z[G/H_x]\to X,
\]
taking $1$ in $\Z[G/H_x]$ to $x$ in $X$.

By construction, the composition of $f$ with the canonical map $X\to
\overline X$ is surjective. As $G$ is a $p$-group, the ideal
$pR_{(p)}+I$ of $R_{(p)}$ is the Jacobson radical of the ring
$R_{(p)}:=R\tens\Z_{(p)}$. By Nakayama Lemma, $f_{(p)}$ is
surjective. Hence the cokernel of $f$ is finite of order prime to
$p$. The rank of the permutation $G$-module $P$ is equal to
\[
\sum_{k=0}^{r} \sum_{b\in B_k} p^k=\sum_{k=0}^{r} |B_k|
p^k=\sum_{k=0}^{r}(b_k-b_{k-1})p^k.\qedhere
\]
\end{proof}

\subsection{Examples}\label{edtoric}



Let $F$ be a field, $\Phi$ a subgroup of $_p\Ch(F)$ of rank $r$,
$L=F(\Phi)$ and $G=\Gal(L/F)$. Consider the torus $U^\Phi$ with the
character group the augmentation ideal $I$ defined in \ref{toruss}.



The middle row of (\ref{diagmod}) yields an exact sequence
\[
\overline N\to (\overline R)^r\to\overline I\to 0.
\]
It follows from Lemma \ref{gene} that $N\subset pR^r+I^r$, hence the
first homomorphism in the sequence is trivial. The middle group is
isomorphic to $(\Z/p\Z)^r$, hence $\rank(\overline I)=r$.


For any subgroup $H\subset G$, the Tate cohomology group $\hat
H^0(H,I)\simeq\hat H^{-1}(H,\Z)$ is trivial. It follows that the
group $I^H$ is generated by $N_Hx$ for all $x\in I$, where
$N_H=\sum_{h\in H}h\in R$. Since $\overline I$ is of period $p$ with
the trivial $G$-action, the classes of the elements $N_Hx$ in
$\overline I$ are trivial if $H$ is a nontrivial subgroup of $G$. It
follows that the maps $I^H\to\overline I$ are trivial for all $H\neq
1$. In the notation of (\ref{vis}), $V_0=\dots =V_{r-1}=0$ and
$V_r=\overline I$. By Theorem \ref{edtori},
\[
\ed_p(U^\Phi)=rp^r-\dim(U^\Phi)=rp^r-p^r+1=(r-1)p^r+1
\]
and the rank of the permutation module in a minimal $p$-presentation
of $I$ is equal to $rp^r$. Therefore, $k:R^r\to I$ is a minimal
$p$-presentation of $I$ that appears to be surjective. Therefore, by
Corollary \ref{reich2},
\begin{equation}\label{kuku1}
\ed(U^\Phi)=\ed_p(U^\Phi)=(r-1)p^r+1.
\end{equation}



Let $S^\Phi$ be the torus with the character group $Q$ defined in
\ref{toruss}. As in (\ref{diagmod}), the homomorphism $k$ factors
through a surjective map $R^r\to Q$ that is then necessarily a
minimal $p$-presentation of $Q$. According to Theorem \ref{edtori}
and Corollary \ref{reich2},
\begin{equation}\label{kuku}
\ed(S^\Phi)=\ed_p(S^\Phi)=rp^r-\dim(S^\Phi)=(r-1)p^r-r+1.
\end{equation}


\section{Degeneration}

In this section we study the behavior of the essential
$p$-dimension under degeneration, i.e. we compare the essential
$p$-dimension of an object over a complete discrete valued field
and its specialization over the residue field (Proposition
\ref{f2f3}). The iterated degeneration (Corollary \ref{nera})
connects a class in the Brauer group degree $p^r$ over some
(large) field and the elements of the indecomposable relative
Brauer group that are torsors for a certain torus.

\subsection{A simple degeneration}
Let $F$ be a field, $p$ a prime integer different from $\ch(F)$ and
$\Phi\subset\ _p\!\Ch(F)$ a finite subgroup. For an integer $k\geq
0$ and a field extension $K/F$, let
\[
\cS_k^\Phi(K)=\{a\in \Br(K)\{p\}\quad\text{such that}\quad\ind
a_{K(\Phi)}\leq p^k\}.
\]
Two elements $a$ and $a'$ in $\cS_k^\Phi(K)$ are \emph{equivalent}
if $a-a'\in\Br_{\dec}\(K(\Phi)/K\)$. Write $\cF_k^\Phi(K)$ for the
set of equivalence classes in $\cS_k^\Phi(K)$. Abusing notation we
shall write $a$ for the equivalence class of an element
$a\in\cS_k^\Phi(K)$ in $\cF_k^\Phi(K)$.



We view $\cS_k^\Phi$ and $\cF_k^\Phi$ as functors from
$\cat{Fields}/F$ to $\cat{Sets}$.

\begin{example}\label{extreme}
$(1)$ If $\Phi$ is the zero subgroup, then
$\cF_k^\Phi=\cS_k^\Phi\simeq\Alg(p^r)\simeq
\gPGL(p^r)\text{-}\tors$.

\smallskip

$(2)$ The set $\cS_0^\Phi(K)$ is naturally bijective to
$\Br\(K(\Phi)/K\)$ and $\cF_0^\Phi(K)\simeq
\Br_{\ind}\(K(\Phi)/K\)$. By Corollary \ref{compute2}, the latter
group is naturally isomorphic to $H^1(K,S^\Phi)$, where $S^\Phi$ is
the torus defined in \ref{toruss}, thus, $\cF_0^\Phi\simeq
S^\Phi\text{-}\tors$.
\end{example}

Let $\Phi'\subset\Phi$ be a subgroup of index $p$ and
$\eta\in\Phi\setminus \Phi'$, hence $\Phi=\langle\Phi',\eta\rangle$.
Let $E/F$ be a field extension such that $\eta_E\notin\Phi'_E$ in
$\Ch(E)$. Choose an element $a\in\cS_k^\Phi(E)$, i.e.,
$a\in\Br(E)\{p\}$ and $\ind(a_{E(\Phi)})\leq p^k$.

Let $E'$ be a field extension of $F$ that is complete with respect
to a discrete valuation $v'$ over $F$ with residue field $E$ and set
\begin{equation}\label{bbb}
a'=\widehat a+\(\widehat\eta_E\cup (x)\)\in \Br(E'),
\end{equation}
for some $x\in {E'}^\times$ such that $v'(x)$ is not divisible by
$p$. By Proposition \ref{propbr}(2),
$\ind(a_{E'(\Phi')})=p\cdot\ind(a_{E(\Phi)})\leq p^{k+1}$, hence
$a'\in \cS_{k+1}^{\Phi'}(E')$.










\begin{proposition}\label{f2f3}
Suppose that for any finite field extension $N/E$ of degree prime to
$p$ and any character $\rho\in\Ch(N)$ of order $p^2$ such that
$p\cdot\rho\in \Phi_N\setminus\Phi'_N$, we have $\ind
a_{N(\Phi',\rho)}>p^{k-1}$. Then
\[
\ed_p^{\cF_{k+1}^{\Phi'}}(a')\geq \ed_p^{\cF_{k}^{\Phi}}(a)+1.
\]
\end{proposition}

\begin{proof}
Let $M/E'$ be a finite field extension of degree prime to $p$,
$M_0\subset M$ a subfield over $F$ and $a'_0\in
\cS_{k+1}^{\Phi'}(M_0)$ such that $(a'_0)_M=a'_M$ in
$\cF_{k}^{\Phi}$ and $\td_F(M_0)=\ed_p^{\cF_{k+1}^{\Phi'}}(a')$. We
have
\begin{equation}\label{1111}
a'_M-(a'_0)_M\in\Br_{\dec}\(M(\Phi')/M\).
\end{equation}

It follows from (\ref{bbb}) that
\begin{equation}\label{cde}
a'_M=\widehat a_N+\(\widehat\eta_N\cup (x)\)
\end{equation}
and $\partial_{v'}(a')=q\cdot\eta_E$, where $q=v'(x)$ is relatively
prime to $p$. We extend the discrete valuation $v'$ on $E'$ to a
(unique) discrete valuation $v$ on $M$. The ramification index $e'$
and inertia degree are both prime to $p$. Thus, the residue field
$N$ of $v$ is a finite extension of $E$ of degree prime to $p$. By
Proposition \ref{propbr}(3),
\begin{equation}\label{biabia}
\partial_v(a'_M)=e'\cdot\partial_{v'}(a')=e'q\cdot\eta_E.
\end{equation}






Let $v_0$ be the restriction of $v$ to $M_0$ and $N_0$ its residue
field. It follows from (\ref{1111}) that
\begin{equation}\label{mmn}
\partial_v(a'_M)-\partial_v((a'_0)_M)\in\Phi'_N.
\end{equation}
Recall that $\eta_E\notin\Phi'_E$. As $[N:E]$ is not divisible by
$p$, it follows that
\begin{equation}\label{netnet}
\eta_N\notin\Phi'_N.
\end{equation}
By (\ref{biabia}), (\ref{mmn}) and (\ref{netnet}),
$\partial_v\((a'_0)_M\)\neq 0$, i.e., $(a'_0)_M$ is ramified and
therefore $v_0$ is nontrivial, i.e., $v_0$ is a discrete valuation
on $M_0$.

Let $\eta_0:=\partial_{v_0}(a'_0)\in\Ch(N_0)\{p\}$. By Proposition
\ref{propbr}(3),
\begin{equation}\label{nhyju}
\partial_v((a'_0)_M)=e\cdot(\eta_{0})_N,
\end{equation}
where $e$ is the ramification index of $M/M_0$, hence
$(\eta_{0})_N\neq 0$. It follows from (\ref{biabia}),(\ref{mmn}) and
(\ref{nhyju}) that
\begin{equation}\label{14}
e'q\cdot\eta_N- e\cdot(\eta_{0})_N\in\Phi'_N.
\end{equation}





%The valuation $v'$ on $M$ extends to a discrete valuation on $R$
%such that $R/M$ is unramified.

As $e'q$ is relatively prime to $p$,
\begin{equation}\label{gen}
\eta_N\in\langle \Phi'_N,(\eta_0)_N\rangle\quad\text{in}\quad
\Ch(N).
\end{equation}
Let $p^t$ ($t\geq 1$) be the order of $(\eta_{0})_N$. It follows
from (\ref{netnet}) and (\ref{14}) that $v_p(e)=t-1$ and
\begin{equation}\label{innot}
p^{t-1}\cdot(\eta_0)_N\in\Phi_N\setminus\Phi'_N.
\end{equation}


Choose a prime element $\pi_0$ in $M_0$ and write
\begin{equation}\label{12}
(a'_0)_{\widehat M_0}=\widehat a_0+\(\widehat\eta_0\cup(\pi_0)\)
\end{equation}
in $\Br(\widehat M_0)$, where $a_0\in\Br(N_0)\{p\}$.

Applying the specialization homomorphism $s_\pi:\Br(M)\{p\}\to
\Br(N)\{p\}$ (for a prime element $\pi$ in $M$) to (\ref{1111}),
(\ref{cde}) and (\ref{12}), using (\ref{miau}) and (\ref{gen}), we
get
\begin{equation}\label{16}
a_N-(a_0)_N\in \Br_{\dec}\(N(\Phi',\eta_0)/N\).
\end{equation}
It follows from (\ref{16}) that
\begin{equation}\label{eqqq}
a_{N(\Phi',\eta_0)}=(a_0)_{N(\Phi',\eta_0)}
\end{equation}
in $\Br\(N(\Phi',\eta_0)\)$.

%It follows from (\ref{innot}) that the group $\langle
%\Phi_N,(\eta_0)_N\rangle$ has rank ${k+s}$.

By (\ref{12}),
\[
(a'_0)_{\widehat M_0(\Phi')}=\widehat {(a_0)}_{N_0(\Phi')}+
\(\widehat{(\eta_0)}_{N_0(\Phi')}\cup(\pi_0)\).
\]
As no nontrivial multiple of $(\eta_0)_N$ belongs to $\Phi'_N$ by
(\ref{innot}), the order of the character $(\eta_0)_{N_0(\Phi')}$ is
at least $p^t$. It follow from Proposition \ref{propbr}(2) that
\begin{equation}\label{firstin}
\ind (a_0)_{N_0(\Phi',\eta_0)}=\ind (a'_0)_{\widehat
M_0(\Phi')}/\ord (\eta_0)_{N_0(\Phi')}\leq p^{k+1}/p^t=p^{k-t+1}.
\end{equation}

By (\ref{eqqq}) and (\ref{firstin}),
\begin{equation}\label{indad}
\ind (a_{N(\Phi',\eta_0)})\leq p^{k-t+1}.
\end{equation}


%Since $\deg(A)=p^r$, the field $N(\Phi,\eta_0)$ is embedded into
%$A_N$.

Suppose that $t\geq 2$ and consider the character
$\rho=p^{t-2}\cdot(\eta_0)_N$ of order $p^2$ in $\Ch(N)$. We have
$p\cdot\rho=p^{t-1}(\eta_0)_N\in \Phi_N\setminus\Phi'_N$ by
(\ref{innot}). Moreover, the degree of the field extension
$N(\Phi',\eta_0)/N(\Phi',\rho)$ is equal to $p^{t-2}$. Hence by
(\ref{indad}),
\[
\ind (a_{N(\Phi',\rho)})\leq \ind (a_{N(\Phi',\eta_0)})\cdot
p^{t-2}\leq p^{k-t+1}\cdot p^{t-2}=p^{k-1}.
\]
This contradicts the assumption. Therefore, $t=1$, i.e.,
$\ord(\eta_{0})_N=p$. Then $(e,p)=1$ and it follows from (\ref{14})
that $(\eta_0)_N\in\langle\Phi'_N,\eta_N\rangle$. Moreover,
\begin{equation}\label{commm}
\langle\Phi',\eta_0\rangle_N=\langle\Phi',\eta\rangle_N=\Phi_N.
\end{equation}

By Lemma \ref{charup}, there is a finite subextension $N_1/N_0$ of
$N/N_0$ such that $\langle\Phi',\eta_0\rangle_{N_1}=\Phi_{N_1}$.
Replacing $N_0$ by $N_1$ and $a_0$ by $(a_0)_{N_1}$, we may assume
that $\langle\Phi',\eta_0\rangle_{N_0}=\Phi_{N_0}$. In particular,
$\eta_0$ is of order $p$ in $\Ch(N_0)$.

Since by (\ref{firstin}),
\[
\ind (a_0)_{N_0(\Phi)}=\ind (a_0)_{N_0(\Phi',\eta_0)}\leq p^{k},
\]
we have $a_0\in \cS_k^\Phi(N_0)$.



It follows from (\ref{16}) that
\[
a_N-(a_0)_N\in\Br_{\dec}\(N(\Phi)/N\).
\]
Hence the classes of $a_N$ and $(a_0)_N$ are equal in
$\cF_k^\Phi(N)$. The class of $a_N$ in $\cF_k^\Phi(N)$ is then
defined over $N_0$, therefore,
\[
\ed_p^{\cF_{k+1}^{\Phi'}}(a')=\td_F(M_0)\geq\td_F(N_0)+
1\geq\ed_p^{\cF_{k}^{\Phi}}(a)+1.\qedhere
\]
\end{proof}

\subsection{Multiple degeneration}\label{degen}

In this subsection we assume that the base field $F$ contains a
primitive $p^2$-th root of unity.



Let $\Phi$ be a subgroup in $_p\Ch(F)$ of rank $r$. Choose a basis
$\chi_1,\chi_2,\dots,\chi_r$ of $\Phi$. Let $E/F$ be a field
extension such that $\rank(\Phi_E)=r$ and let $a\in\Br(E)\{p\}$ be
an element that is split by $E(\Phi)$.

Let $E_0=E$, $E_1,\dots, E_r$ be field extensions of $F$ such that
for any $k=1,2,\dots, r$, the field $E_k$ is complete with respect
to a discrete valuation $v_k$ over $F$ and $E_{k-1}$ is its residue
field. For any $k=1,2,\dots, r$, choose elements $x_k\in E_k^\times$
such that $v_k(x_k)$ is not divisible by $p$ and define the elements
$a_k\in\Br(E_k)\{p\}$ inductively by $a_0=a$ and
$a_k=\widehat{a_{k-1}}+\(\widehat{(\chi_k)}_{E_{k-1}}\cup(x_k)\)$.

Let $\Phi_k$ be the subgroup of $\Phi$ generated by
$\chi_{k+1},\dots,\chi_r$. Thus, $\Phi_0=\Phi$, $\Phi_r=0$ and
$\rank(\Phi_k)=r-k$. Note that the character
$(\chi_k)_{E_{k-1}(\Phi_k)}$ is not trivial. It follows from
Proposition \ref{propbr}(2) that
\[
\ind (a_k)_{E_k(\Phi_k)}=p\cdot\ind (a_{k-1})_{E_{k-1}(\Phi_{k-1})}
\]
for any $k=1,\dots, r$. As $\ind a_{E(\Phi)}=1$, we have $\ind
(a_k)_{E_k(\Phi_k)}=p^k$ for all $k=0,1,\dots,r$. In particular,
$a_k\in\cS_k^{\Phi_k}(E_k)$.

The followings lemma assures that under a certain restriction on the
element $a$, the conditions of Proposition \ref{f2f3} are satisfied
for the fields $E_k$, the groups of characters $\Phi_k$ and the
elements $a_k$.

\begin{lemma}\label{cond}
Suppose that $p^{r-1}a\notin\Im\(\Br(F)\to\Br(E)\)$. Then for every
$k=0,1,\dots, r-1$, and any finite field extension $N/E_k$ of degree
prime to $p$ and any character $\rho\in\Ch(N)$ of order $p^2$ such
that $p\cdot\rho\in\(\Phi_k)_N\setminus (\Phi_{k+1}\)_N$, we have
\begin{equation}\label{iiii}
\ind (a_k)_{N(\Phi_{k+1},\rho)}> p^{k-1}.
\end{equation}
\end{lemma}

\begin{proof}
Induction on $r$. The case $r=1$ is obvious. Suppose that the
inequality (\ref{iiii}) does not hold for some $k=1,\dots, r-1$, a
finite field extension $N/E_k$ and a character $\rho\in\Ch(N)$.
Suppose first that $k<r-1$. Consider the fields $F'=F(\Phi_{k+1})$,
$E'=E(\Phi_{k+1})$, $E'_i=E_i(\Phi_{k+1})$, $N'=N(\Phi_{k+1})$, the
sequence of characters $(\chi_i)_{F'}$ and the sequence of elements
$a'_i:=(a_i)_{E_i'}\in\Br(E_i)$ for $i=0,1,\dots, k+1$. As
$(a'_k)_{N'(\rho)}=(a_k)_{N(\Phi_{k+1},\rho)}$, the inequality
(\ref{iiii}) does not hold for the term $a'_k$ of the new sequence,
the field extension $N'/E'_k$ and the character $\rho_{N'}$.


Note that $p^k a_{E'}\notin\Im\(\Br(F')\to\Br(E')\)$, because
otherwise, taking the norm map for the extension $F'/F$ of degree
$p^{r-k-1}$, we would get $p^{r-1}a\in\Im\(\Br(F)\to\Br(E)\)$. By
induction, the inequality (\ref{iiii}) holds for all the terms of
the new sequence, in particular for $a'_k$, a contradiction.

Thus we can assume that $k=r-1$. We construct a new sequence of
fields $\tilde E_0, \tilde E_1,\dots, \tilde E_r$ such that each
$\tilde E_i$ is a finite extension of $E_i$ of degree prime to $p$
as follows. We set $\tilde E_{r-1}=N$ and let $\tilde E_r$ be an
unramified extension of $E_r$ with the residue field $\tilde
E_{r-1}$. The fields $\tilde E_j$ with $j<r-1$ are constructed by
descending induction on $j$. If we have constructed $\tilde E_j$ as
a finite extension of $E_j$ of degree prime to $p$, then we extend
the valuation $v_j$ to $\tilde E_j$ and let $\tilde E_{j-1}$ to be
its residue field. Replacing $E_i$ by $\tilde E_i$ and $a_i$ by
$(a_i)_{\tilde E_i}$, we may assume that $N=E_{r-1}$.

\smallskip

{\it Case} $1$: The character $\rho$ is unramified with respect to
$v_{r-1}$, i.e., $\rho=\widehat\mu$ for a character
$\mu\in\Ch(E_{r-2})$ of order $p^2$. By Lemma \ref{base}(1),
\begin{equation}\label{swsw}
\ind(a_{r-2})_{E_{r-2}(\chi_{r-1},\mu)}=\ind(a_{r-1})_{E_{r-1}(\rho)}/p=
\ind(a_{r-1})_{E_{r-1}(\Phi_r,\rho)}/p\leq p^{r-3}.
\end{equation}
Consider the fields $F'=F(\chi_{r-1})$, $E'=E(\chi_{r-1})$,
$E'_i=E_i(\chi_{r-1})$, $N'=N(\chi_{r-1})$, the sequence of
characters $\chi_1,\dots,\chi_{r-2},\chi_r$ and the elements
$a'_i\in\Br(E'_i)$ for $i=0,1,\dots, r-1$ defined by
$a'_i=(a_i)_{E_i'}$ for $i\leq r-2$ and $a'_{r-1}=\widehat
a_{r-2}+\(\widehat\chi_r\cup(x_{r-1})\)$ over $E'_{r-1}$. As
$(a'_{r-2})_{N'(\mu)}=(a_{r-2})_{N(\chi_{r-1},\rho)}$, the
inequality (\ref{swsw}) shows that (\ref{iiii}) does not hold for
the term $a'_{r-2}$ of the new sequence, the field extension
$N'/E'_{r-2}$ and the character $\mu_{N'}$.


Note that $p^{r-2} a_{E'}\notin\Im\(\Br(F')\to\Br(E')\)$, as
otherwise, taking the norm map for the extension $F'/F$ of degree
$p$, we get $p^{r-1}a\in\Im\(\Br(F)\to\Br(E)\)$. By induction, the
inequality (\ref{iiii}) holds for all the terms of the new sequence,
in particular for $a'_{r-2}$, a contradiction.

\smallskip

{\it Case} $2$: The character $\rho$ is ramified. Note that
$p\cdot\rho$ is a nonzero multiple of $(\chi_r)_{E_{r-1}}$. As the
inequality (\ref{iiii}) fails for $a_{r-1}$, we have
\[
\ind (a_{r-1})_{E_{r-1}(\rho)}\leq p^{r-2}.
\]


By Lemma \ref{base}(2), there exists a unit $u\in E_{r-1}$ such
that $E_{r-2}(\chi_r)=E_{r-2}(\bar u^{1/p})$ and
\[
\ind\(a_{r-2}-(\chi_{r-1}\cup(\bar
u^{1/p}))\)_{E_{r-2}(\chi_r)}=\ind (a_{r-1})_{E_{r-1}(\rho)}\leq
p^{r-2}.
\]
By descending induction on $j=0,1,\dots, r-2$ we show that there
exist a unit $u_j$ in $E_{j+1}$ and a subgroup $\Theta_j\subset\Phi$
of rank $r-j-1$ such that
$\langle\chi_1,\dots,\chi_j,\chi_{r-1}\rangle\cap \Theta_j=0$,\
$E_j(\chi_r)=E_j(\bar u_j^{1/p})$ and
\begin{equation}\label{laastineq}
\ind\(a_j-(\chi_{r-1}\cup(\bar u_j^{1/p}))\)_{E_j(\Theta_j)}\leq
p^j.
\end{equation}
If $j=r-2$, we set $u_j=u$ and $\Theta_j=\{\chi_r\}$.

$(j\Rightarrow j-1)$: The field $E_j(\bar u_j^{1/p})=E_j(\chi_r)$ is
unramified over $E_j$, hence $v_j(\bar u_j)$ is divisible by $p$.
Modifying $u_j$ by a $p^2$-th power, we may assume that $\bar
u_j=u_{j-1}x_j^{mp}$ for a unit $u_{j-1}\in E_j$ and an integer $m$.
Then
\[
\(a_j-(\chi_{r-1}\cup(\bar u_j^{1/p}))\)_{E_j(\Theta_j)}=\widehat b
+\(\widehat\eta\cup(x_j)\)_{E_j(\Theta_j)},
\]
where $\eta=\chi_j-m\chi_{r-1}$ and
$b=\(a_{j-1}-(\chi_{r-1}\cup(\bar
u_{j-1}^{1/p}))\)_{E_{j-1}(\Theta_j)}$. As $\eta$ is not contained
in $\Theta_j$, the character $\eta_{E_{j-1}(\Theta_j)}$ is not
trivial. Set $\Theta_{j-1}=\langle\Theta_j,\eta\rangle$. It follows
from Proposition \ref{propbr}(2) that
\[
\ind(b_{E_{j-1}(\Theta_{j-1})})=\ind \(a_j-(\chi_{r-1}\cup(\bar
u_j^{1/p}))\)_{E_j(\Theta_j)}/p\leq p^{j-1}.
\]


Applying the inequality (\ref{laastineq}) in the case $j=0$, we get
\[
a_{E(\Theta_0)}=\(\chi_{r-1}\cup(w^{1/p})\)_{E(\Theta_0)}
\]
for an element $w\in E^\times$ such that $E(w^{1/p})=E(\chi_r)$. The
degree of the extension $E(\Theta_0)/E$ is equal to $p^{r-1}$ and
$E(w^{1/p})\subset E(\Theta_0)$. Taking the norm for the extension
$E(\Theta_0)/E$, we get that $p^{r-1}a$ is a multiple of
$\chi_{r-1}\cup(w)$. As the character $\chi_r$ is defined over $F$,
we may assume that $w\in F^\times$, hence
$p^{r-1}a\in\Im\(\Br(F)\to\Br(E)\)$, a contradiction. Thus, we have
shown that the inequality (\ref{iiii}) holds.
\end{proof}

By Example \ref{extreme}(2), we can view $a$ as an $S^\Phi$-torsor
over $E$.

\begin{corollary}\label{nera}
Suppose that $p^{r-1}a\notin\Im\(\Br(F)\to\Br(E)\)$. Then
\[
\ed_p^{\Alg(p^r)}(a_r)\geq \ed_p^{S^\Phi\text{-}\tors}(a)+r.
\]
\end{corollary}

\begin{proof}
By iterated application of Proposition \ref{f2f3} and Example
\ref{extreme},
\begin{multline*}
\ed_p^{\Alg(p^r)}(a_r)=\ed_p^{\cF_{r}^{\Phi_r}}(a_r)\geq
\ed_p^{\cF_{r-1}^{\Phi_{r-1}}}(a_{r-1})+1\geq\dots\\\geq
\ed_p^{\cF_1^{\Phi_1}}(a_1)+(r-1)\geq
\ed_p^{\cF_0^{\Phi_0}}(a_0)+r=\ed_p^{S^\Phi\text{-}\tors}(a)+r.
\end{multline*}
\end{proof}


\section{Proof of the main theorem}

\begin{theorem}\label{main}
Let $F$ be a field and $p$ an integer different from $\ch(F)$. Then
\[
\ed_p\(\Alg_F(p^r)\)\geq (r-1)p^r+1.
\]
\end{theorem}

\begin{proof}
As $\ed_p\(\Alg_F(p^r)\)\geq \ed_p\(\Alg_{F'}(p^r)\)$ for any field
extension $F'/F$ by \cite[Prop. 1.5]{Merkurjev09}, we can replace
$F$ by any field extension. In particular, we may assume that $F$
contains a primitive $p^2$-th root of unity and there is a subgroup
$\Phi$ of $_p\Ch(F)$ of rank $r$. Let $T^\Phi$ be the algebraic
torus constructed in Section \ref{brtori} for the field extension
$L=F(\Phi)$ of $F$. Set $E=F(T^\Phi)$ and let $a\in \Br(EL/E)$ be
the element defined in \ref{thea}. Let $a_r\in\Br(E_r)$ be the
element of index $p^r$ constructed in \ref{degen}. By Corollary
\ref{notinimage}, the class $p^{r-1}a$ in $\Br (E)$ does not belong
to the image of $\Br(F)\to\Br (E)$. It follows from Corollary
\ref{nera} that
\begin{equation}\label{1111a}
\ed_p^{\Alg(p^r)}(a_r)\geq \ed_p^{S^\Phi\text{-}\tors}(a)+r.
\end{equation}
The $S^\Phi$-torsor $a$ is the generic fiber of the versal
$S^\Phi$-torsor $P^\Phi\to S^\Phi$ (see Example \ref{toritorsors2}),
hence $a$ is a generic torsor. By \cite[\S 6]{Reichstein00a} or
\cite[Th. 2.9]{Merkurjev09}
\begin{equation}\label{2222}
\ed_p^{S^\Phi\text{-}\tors}(a)=\ed_p(S^\Phi).
\end{equation}
The essential $p$-dimension of $S^\Phi$ was calculated in
(\ref{kuku}):
\begin{equation}\label{3333}
\ed_p(S^\Phi)= (r-1)p^r-r+1.
\end{equation}
Finally, it follows from (\ref{1111a}), (\ref{2222}) and
(\ref{3333}) that
\[
\ed_p\(\Alg_F(p^r)\)\geq \ed_p^{\Alg(p^r)}(a_r)\geq
\ed_p^{S^\Phi\text{-}\tors}(a)+r=(r-1)p^r+1.\qedhere
\]
\end{proof}

\section{Remarks}

Let $K/F$ be a field extension and $G$ an elementary abelian group
of order $p^r$. Consider the subset $\cat{Alg}_K(G)$ of
$\Alg_K(p^r)$ consisting of all classes admitting a splitting Galois
$K$-algebra with the Galois group $G$. Equivalently,
$\cat{Alg}_K(G)$ consists of all classes represented by crossed
product algebras with the group $G$ (see \cite[\S 4.4]{Herstein94}).

Write $\cat{Pair}_K(G)$ for the set of pairs $(a,E)$, where $a\in
\cat{Alg}_K(G)$ and $E$ is a Galois $G$-algebra splitting $a$.

Finally, fix a Galois field extension $L/F$ with $\Gal(L/F)\simeq G$
and consider the subset $\Alg_K(L/F)$ of $\cat{Alg}_K(G)$ consisting
of all classes split by $KL$. Thus, $\Alg(L/F)$ is a subfunctor of
$\cat{Alg}(G)$ and there is the obvious surjective morphism of
functors $\cat{Pair}_K(G)\to \cat{Alg}_K(G)$.

\begin{theorem}
Let $F$ be a field, $p$ an integer different from $\ch(F)$, $G$ an
elementary abelian group of order $p^r$, $r\geq 2$, and $L/F$ a
Galois field extension with $\Gal(L/F)\simeq G$. Let $\cF$ be one of
the three functors $\Alg(L/F)$, $\cat{Alg}(G)$ and
$\cat{Pair}_K(G)$. Then
\[
\ed(\cF)=\ed_p(\cF)=(r-1)p^r+1.
\]
\end{theorem}

\begin{proof}
The functor $\Alg(L/F)$ is isomorphic to $U^\Phi\text{-}\tors$ by
(\ref{hu}). It follows from (\ref{kuku1}) that
\[
\ed\(\Alg(L/F)\)=\ed_p\(\Alg(L/F)\)=(r-1)p^r+1.
\]
Let $a_r$ be the element in $\Br(E_r)$ in the proof of Theorem
\ref{main}. It satisfies $\ed_p^{\Alg(p^r)}(a_r)\geq (r-1)p^r+1$. By
construction, $a_r\in \cat{Alg}_{E_r}(G)$. As $\cat{Alg}(G)$ is a
subfunctor of $\Alg(p^r)$, we have
\[
\ed_p\(\cat{Alg}(G)\)\geq \ed_p^{\cat{Alg}(G)}(a_r)\geq
\ed_p^{\Alg(p^r)}(a_r)\geq (r-1)p^r+1.
\]
The upper bound $\ed\(\cat{Alg}(G)\)\leq (r-1)p^r+1$ was proven in
\cite[Cor. 3 10]{LRR03}.

The split \'etale $F$-algebra $E:=\Map(G,F)$ has the natural
structure of a Galois $G$-algebra over $F$. The group $G$ acts on
the split torus $U:=R_{E/F}(\gme)/\gm$. Let $A$ be the split
$F$-algebra $\End_F(E)$. The semidirect product $H:=U\rtimes G$ acts
naturally on $A$ by $F$-algebra automorphisms. Moreover, by the
Skolem-Noether Theorem, $H$ is precisely the automorphism group of
the pair $(A,E)$. It follows that the functor $\cat{Pair}_K(G)$ is
isomorphic to $H\text{-}\tors$.

The character group of $U$ is $G$-isomorphic to the ideal $I$ in
$R=\Z[G]$. By \cite[\S 3]{MR09}, the $G$-homomorphism $k:R^r\to I$
constructed in \ref{toruss} yields a representation $W$ of the group
$H$ of dimension $rp^r$. As $r\geq 2$, by Lemma \ref{gene}, $G$ acts
faithfully on the kernel $N$ of $k$. By \cite[Lemma 3.3]{MR09}, the
action of $H$ on $W$ is generically free, hence
\[
\ed\(\cat{Pair}(G)\)=\ed(H)\leq \dim(W)-\dim(H)=(r-1)p^r+1.
\]
Since $\cat{Pair}(G)$ surjects onto $\cat{Alg}(G)$, we have
\[
\ed\(\cat{Pair}(G)\)\geq\ed_p\(\cat{Pair}_K(G)\)\geq
\ed_p\(\cat{Alg}(G)\)=(r-1)p^r+1.\qedhere
\]
\end{proof}

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\end{thebibliography}




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