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\title{A short proof of Klyachko's theorem about rational algebraic tori}
\author{Mathieu Florence}
\date{}
\begin{document}
\maketitle
\begin{abstract}
In this paper, we give another proof of a theorem by Klyachko (\cite{Kly}), which asserts that Zariski's conjecture holds for a special class of tori over an arbitrary ground field.


\end{abstract}
\section{Introduction}
\noindent The main purpose of this paper is to give a much simpler
proof of a theorem due to Klyachko (\cite{Kly}; see also \cite{Vosk}, chap. 2, 6.3), which is here theorem
\ref{klyachko}. To achieve this, we first prove a generalization
of a theorem due to Voskresenskii (\cite{Vosk}, chap. 2, 5.1, corollary). To be more precise, we
show stable rationality for a certain class of algebraic tori over
a given field $k$, strictly containing the cyclotomic ones. What
is more, we give an effective way of presenting the character
module of these tori as the kernel of a surjection between
permutation modules (that is, lattices that contain a basis which
is permuted by the action of the absolute Galois group of $k$).
Recall that, according to \textit{loc. cit.}, chap. 2, 4.7, theorem 2 , the existence of such a
surjection is a necessary and sufficient condition for a torus to
be stably rational. All the basic material concerning algebraic
tori and rationality questions related to these is contained in
\textit{loc. cit.}, chap.2; we shall assume that the reader is familiar with this
reference.

 \noindent In the following section, the symbol $\otimes$ alone
means $\otimes_\ZZ$. If $k$ is a field with separable closure
$k_s$, we denote by $\Gamma_k$ the profinite group ${\rm{Gal}}
(k_s/k)$. Let $\Gamma$ be a profinite group. By a
$\Gamma$-lattice, we mean a free $\ZZ$-module of finite rank,
endowed with a continuous action of $\Gamma$. We will say simply
'exact sequence' instead of 'exact sequence of $\Gamma$-lattices'.

\section{Stably rational and rational algebraic tori}
\noindent To begin this section, we prove an elementary but
crucial lemma.

\begin{lemme}
Let $\Gamma$ be a profinite group. Let $A_i,B_i,C_i$, $i=1,2$ be
$\Gamma$-lattices, fitting into two exact sequences
$$ 0 \longrightarrow A_i \stackrel {j_i} \longrightarrow B_i \stackrel {\pi_i}\longrightarrow C_i
\longrightarrow 0.$$ \noindent Assume we are given $s_i : C_i
\longrightarrow B_i$, and $d_1,d_2$ two coprime integers, such
that $\pi_i \circ s_i = d_i Id$, $i=1,2$. Let $$A_3=A_1 \otimes
A_2,$$
$$B_3=(B_1 \otimes B_2) \oplus (C_1 \otimes C_2),$$ and $$C_3=(C_1
\otimes B_2) \oplus (B_1 \otimes C_2).$$ \noindent Then there is
an exact sequence $$ 0 \longrightarrow A_3 \stackrel {j_3}
\longrightarrow B_3 \stackrel {\pi_3}\longrightarrow C_3
\longrightarrow 0,$$ together with a morphism $s_3: C_3
\longrightarrow B_3$, satisfying $\pi_3 \circ s_3 = d_1d_2 Id.$

\end{lemme}
\noindent \textit{Proof}. We have an exact sequence
$$ 0 \longrightarrow A_1 \otimes A_2 \longrightarrow B_1 \otimes B_2 \stackrel {(\pi_1 \otimes Id)\oplus (Id \otimes \pi_2)} \longrightarrow
(C_1 \otimes B_2) \oplus (B_1 \otimes C_2) \stackrel \pi
\longrightarrow C_1 \otimes C_2 \longrightarrow 0,$$ where $\pi=
Id \otimes \pi_2 - \pi_1 \otimes Id.$

 \noindent Select integers $u,v$ such that $vd_2-ud_1=1.$ Then the
 map
 $$ s: C_1 \otimes C_2 \longrightarrow (C_1 \otimes B_2) \oplus (B_1 \otimes
 C_2),$$ $$ c_1 \otimes c_2 \mapsto (vc_1 \otimes s_2(c_2),us_1(c_1)
 \otimes c_2)$$
 is a splitting of $\pi$. Hence we have an exact sequence
 $$ 0 \longrightarrow A_3 \stackrel {j_3} \longrightarrow B_3
\stackrel {\pi_3}\longrightarrow C_3 \longrightarrow 0$$ as
stated, where
$$ \pi_3:   (B_1 \otimes
B_2) \oplus (C_1 \otimes C_2) \stackrel  {((\pi_1 \otimes
Id)\oplus (Id \otimes \pi_2),s )}\longrightarrow (C_1 \otimes B_2)
\oplus (B_1 \otimes C_2).$$ \noindent The last assertion is
obvious: if $r_i : B_i \longrightarrow A_i$ ($i=1,2$) are such
that $ r_i \circ j_i = d_i Id$, then $$ r_3:=(r_1 \otimes r_2,0):
B_3 \longrightarrow A_3$$ satisfies $r_3 \circ j_3 = d_1 d_2 Id.
\square$\\

\noindent From this we can derive the following

\begin{theoreme} \label{presentation}
Let $k$ be a field, and $X_1,...,X_r$ be finite $\Gamma_k$-sets.
For $i=1,...,r$, denote by $J_i$ the kernel of the canonical
surjection $\ZZ ^{X_i} \stackrel {\pi_i} \longrightarrow \ZZ$. Let $J= \otimes_{i}
J_i$. If the orders of the $X_i$ are two by two coprime, then we
have an exact sequence

$$0 \longrightarrow J \longrightarrow \bigoplus_{I \in \mathcal I_0} \ZZ^ {\Pi_{i \in I} X_i}
  \stackrel \pi \longrightarrow \bigoplus_{I \in \mathcal I_1} \ZZ^ {\Pi_{i \in I} X_i} \longrightarrow 0,$$ where $\mathcal I_i$ is
the set of subsets of $ \{ 1,...,r \}$ whose cardinality is
congruent to $r-i$ mod $2$. In particular, a $k$-torus with
character module isomorphic to $J$ is stably rational over $k$.
What is more, let $d$ denote the product of the orders of the
$X_i$, $i=1,...,r$. Then there exists
$$s: \bigoplus_{I \in \mathcal I_1} \ZZ^ {\Pi_{i \in I} X_i}
\longrightarrow \bigoplus_{I \in \mathcal I_0} \ZZ^ {\Pi_{i \in I}
X_i}$$ such that $\pi \circ s=d Id$.
\end{theoreme}
\noindent \textit{Proof}. For $i=1,...,r$, we have a canonical map 
$$s_i: \ZZ \longrightarrow \ZZ^{X_i},$$
$$ 1 \mapsto \sum_{x \in X_i} x,$$
\noindent which satisfies $\pi_i \circ s_i=d_i Id $, where $d_i$ is the order of $X_i$. The proof is then an easy induction using the
previous lemma and the obvious isomorphism $\ZZ ^X \otimes \ZZ ^Y
\simeq \ZZ^{X \times Y}$, for any two finite sets $X$ and $Y$.
$\square$\\

\noindent As a particular case of this theorem, we recover a
result due to Voskresenskii (\cite{Vosk}, chap. 2, 5.1 corollary).

\begin{corollaire}
Let $k$ be a field, and $l/k$ a Galois extension with cyclic
Galois group $G$ of ordrer $ n=p_1...p_r$, where the $p_i$ are
prime numbers. Let $\sigma$ be a generator of this Galois group,
and $T/k$ the $n^{th}$ cyclotomic torus, i.e. the torus with
character group isomorphic to $\ZZ [X]/ \phi_n(X)$, where
$\phi_n(X)$ is the $n^{th}$ cyclotomic polynomial, the action of
$\sigma$ being given by multiplication by $X$ (in other words, the
character group of $T$ is isomorphic to the ring of integers of
the $n^{th}$ cyclotomic extension of $\QQ$, with the action of
$\sigma$ being given by multiplication by a primitive $n^{th}$
root of unity). Then $T$ is stably rational over $k$.
\end{corollaire}
\noindent \textit{Proof}. For $i=1,...,r$, let $X_i$ be the unique
quotient of $G$ isomorphic to $\ZZ/p_i$. With the notations of the
preceding theorem, the $\Gamma_k$-module $J$ is isomorphic to the
character module of $T$ (this is just the fact that the ring of
integers of the $n^{th}$ cyclotomic extension of $\QQ$ is
naturally isomorphic to the tensor product of the rings of
integers of the $p_i^{th}$ cyclotomic extensions of $\QQ$), whence
the
claim.$\square$\\

\noindent We are now able to give a simple proof of the following
theorem.

\begin{theoreme} \label{klyachko}(Klyachko) Let $k$ be a field, and $X$, $Y$ two
finite $\Gamma_k$-sets, of coprime orders $p$ and $q$,
respectively. Consider the two basic exact sequences
$$ 0 \longrightarrow J_X \longrightarrow \ZZ^X \longrightarrow \ZZ
\longrightarrow 0,$$
$$ 0 \longrightarrow J_Y \longrightarrow \ZZ^Y \longrightarrow \ZZ
\longrightarrow 0.$$ Then, a $k$-torus $T$ with character module
isomorphic to $J:=J_X \otimes J_Y$ is rational over $k$.
\end{theoreme}
\noindent \textit{Proof}. Select integers $u,v$ such that
$up-vq=1$. Theorem \ref{presentation} gives the following
presentation of $J$:
$$ 0 \longrightarrow J \longrightarrow \ZZ^{X \times Y} \oplus \ZZ
\stackrel \pi \longrightarrow \ZZ^X \oplus \ZZ^Y \longrightarrow
0,$$ where $\pi(x \otimes y,0)=(x,y)$ and $\pi(0,1)=(u\sum_{x \in
X} x, v\sum_{y \in Y} y)$. Let $E/k$ (resp. $F/k$) be the etale
extension of $k$ corresponding to $X$ (resp. to $Y$). Then, in
terms of tori, this exact sequence reads as
$$ 1  \longrightarrow {\rm{R}}_{E/k}(\mathbb{G}_m) \times {\rm{R}}_{F/k}(\mathbb{G}_m) \stackrel i \longrightarrow  {\rm{R}}_{E \otimes_k F/k}(\mathbb{G}_m) \times \mathbb{G}_m  \longrightarrow T
\longrightarrow 1,$$ where R denotes Weil scalar restriction. The
map $i$ is given on the $k$-points of the considered tori by the
following formula:
$$i(x,y)=(x \otimes y, N_{E/k}(x)^u N_{F/k}(y)^v), x \in E^*, y
\in F^*.$$ \noindent Thus, we have a generically free action of
the algebraic $k$-group $H:={\rm{R}}_{E/k}(\mathbb{G}_m) \times
{\rm{R}}_{F/k}(\mathbb{G}_m)$ on the $k$-vector space $V:= (E
\otimes_k F) \oplus k$, such that $T$ is birational to the
quotient $V/H$ (of course, such a quotient is defined up
to birational equivalence only).\\
 \noindent Assume that $p<q$. Let $G/k$ be the algebraic
 $k$-group ${\rm{GL}}_k(E) \times {\rm{R}}_{F/k}(\mathbb{G}_m)$ ($E$ being viewed as a $k$-vector space). I
 claim that the action of $H$ on $V$ can be naturally extended to
 an action of $G$ on $V$ ($H$ being viewed as a subgroup of $G$
 the obvious way). Indeed, this new action is given on the $k$-points by the
 formula, for $g=(\phi,y) \in G(k), v=(e\otimes f,\lambda) \in V$:

 $$ g.v= (\phi(e) \otimes yf, {\rm{det}}(\phi)^u N_{F/k}(y)^v \lambda).$$ This action is generically free. Indeed, this is an easy
consequence of the equality $up-vq=1$ and of the following lemma.

\begin {lemme} \label{genlibre}
Let $G$ act on $E \otimes_k F$ the obvious way. Then the
stabilizer of a generic element is the subgroup $\mathbb{G}_m$ of
$G$ given, on the level of $k$-points, by elements of the form
$(x,x^{-1}) \in {\rm{GL}}_k(E)\times F^*$, for $x \in k^*$.
\end{lemme}
\noindent We postpone the proof until the end of this section.
Assuming this lemma, we have a birational $G$-equivariant
isomorphism $V \simeq( V/G) \times G$, where the action of $G$ on
the right is given by translation. Indeed, this is a direct
consequence of Hilbert's theorem 90, asserting that $H^1(l,G)=1$
for any field extension $l$ of $k$. Hence we have birational
isomorphisms
$$ T \simeq V/H \simeq V/G \times G/H.$$
It is clear that the $k$-variety
$G/H={\rm{GL}}_k(E)/{\rm{R}}_{E/k}(\mathbb{G}_m)$ is $k$-rational.
As in Klyachko's original proof, the key point is here that the
$k$-variety (defined up to birational equivalence) V/G is
independent of $E$ (seen as an etale $k$-algebra). Hence, the
birational equivalence class of $T$ is independent of $E$; we may
therefore assume that $E$ is split, i.e. that the action of
$\Gamma_k$ on $X$ is trivial. But then $J$ is isomorphic to
${J_Y}^{p-1}$, hence $T$ is birational to
$({\rm{R}}_{F/k}(\mathbb{G}_m)/\mathbb{G}_m)^{p-1}$, which is a
rational variety (it is an open subvariety of
$(\mathbb{P}_k^{q-1})^{p-1}$). $\square$\\

\noindent \textit{Proof of lemma \ref{genlibre}}. We may assume
that $F$ is split, i.e. $F=k^q$ as an etale $k$-algebra. Let
$f_i$, $i=1,...,q$ denote the canonical $k$-basis of $F$. Consider
an element $w=\sum_i e_i \otimes f_i \in E \otimes_k F$ in general
position. Let $g=(\phi,(\lambda_1,...,\lambda_q)) \in
{\rm{GL}}_k{E} \times F^*$ be such that $g.w=w.$ This amounts to
saying that $\phi(e_i)=\lambda_i ^{-1} e_i$ for all $i$. Since
$p<q$ and since $w$ is in general position, $e_1,...,e_p$ form a
basis of $E$ with respect to which the $i$'th component of
$e_{p+1}$ is non zero for all $i=1,...,p$. This readily implies that
the $\lambda_i$ are all equal to some scalar $\lambda$ and that
$\phi=\lambda ^{-1} Id$, thus proving the claim. $\square$


\begin{thebibliography}{2}
\bibitem[Vo]{Vosk} \textsc{V. E. Voskresenskii}. ---   \textit{Algebraic Groups and their Birational Invariants}, Translations of Mathematical Monographs \textbf{179} (1991), Amer. Math. Soc.

\bibitem[Kl]{Kly} \textsc{A. A. Klyachko}. ---   \textit{On rationality of tori with a cyclic splitting field}, Arithmetic and Geometry of Varieties, Kuibyshev Univ. Press (1988), 73-78 (russian).
\end{thebibliography}
\end{document}

 