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\begin{document}
\title{Very free R-equivalence on toric models}
\author{David A. Madore\footnote{D\'epartement de math\'ematiques
et applications, \'Ecole normale sup\'erieure, 45~rue d'Ulm,
F75230 Paris cedex 05, France.  Email address:
\texttt{david.madore@ens.fr}}}
\maketitle
\begin{abstract}
Using the theory of the universal torsor, we prove that two rational
points on a smooth projective toric variety over an infinite field
that are rationally equivalent can in fact be connected by a very free
rational curve.  We also show a similar result over del
Pezzo surfaces of degree~$5$.
\end{abstract}
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\section*{Introduction}

Let $X$ be a smooth projective variety over an infinite field $k$, and
assume that $X$ is (geometrically) \emph{separably rationally
connected}, meaning that, over the algebraic closure $\bar k$, there
exists an $f\colon\mathbb{P}^1_{\bar k}\to X_{\bar k}$ which is ``very
free'' in the sense that $f^*T_X$ is ample (in other words,
$H^1(\mathbb{P}^1, (f^*T_X)(-2)) = 0$).  If $x$ and $y$ are two
$k$-rational points of $X$ (assuming there are any) which are
``R-equivalent'', that is, which can be joined by a chain of
rational curves on $X$ (each defined over $k$), we can ask ourselves
whether there exists $f\colon\mathbb{P}^1_k \to X$ defined over $k$
such that $f(0)=x$, and $f(\infty)=y$ and $H^1(\mathbb{P}^1,
(f^*T_X)(-2)) = 0$: if such is the case, we say that $x$ and $y$ are
R-equivalent by a single very free rational curve.

It is true over $k$ algebraically closed that any two points on a
smooth projective separably rationally connected variety $X$ are in
fact joined by a single very free rational curve
(see~\cite{KollarBook} for a proof of this fact as well as all
introductory material on rationally connected varieties).

In the case where $k$ is no longer algebraically closed but ``large'',
meaning that every irreducible variety that has a smooth $k$-point has
a Zariski dense subset of them, for example when $k$ is a local field,
then the answer is again affirmative: this is the result of a recent
work by Koll\'{a}r~(\cite{KollarSpec}, theorem~23)---any two points
(on a smooth projective rationally connected variety) which are
R-equivalent are so by a single very free rational curve.

For other fields $k$, however, the answer to the question is unknown,
even in some simple cases.

When $X$ is a smooth del Pezzo surface of degree~$4$ over $k$, for
example, it is known that every universal torsor over $X$ (a term
which we will define below) having a $k$-point is $k$-rational
(see~\cite{DP4}), so two rationally equivalent points on $X$ are
R-equivalent, but it is not known whether they can be joined by a
single very free rational curve.  Perhaps more to the point, it can be
shown, using the technique of the present paper, that, for every
R-equivalence class $\alpha$ of $X(k)$, there is a nonempty Zariski
open set $U_\alpha$ of $X$ such that if $P$ and $Q$ are in $\alpha$
and in $U_\alpha$ then they are joined by a single very free rational
curve---but it remains unknown whether, in fact, $U_\alpha$ can be
taken to be $X$.

A positive answer to the question in full generality (for any infinite
field $k$ and any separably rationally connected variety $X$) is
conceivable, but seems out of reach with present techniques.

In this article we prove a positive result when $X$ is a toric model
(i.e., a smooth equivariant compactification of a torus) over an infinite
field $k$: this is possible because a universal torsor can be
explicitly constructed, and because rational curves can be moved thanks to
the action of the torus.  In the next section, we also prove a
positive result in the case where $X$ is a del Pezzo surface of
degree~$5$ (another case in which the universal torsor is well
controlled).  We start with some general remarks on very free
R-equivalence.

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\section{General framework}

We introduce a notation: if $k$ is a field and $X$ an irreducible
variety over $k$, and if $x,y\in X(k)$, let $x\related{X}y$ stand for
the following statement: there exists an irreducible variety $M$ over
$k$ such that $M(k)$ is Zariski-dense in $M$, and a dominant and
separable rational map $F\colon M\times\mathbb{P}^1 \dasharrow X$ such
that $F$ restricted to $M\times\{0\}$ is constant equal to $x$ and $F$
restricted to $M\times\{\infty\}$ is constant equal to $y$.

The following proposition summarizes some general facts about this
relation:
\begin{prop}
\label{GeneralFacts}
Assume $k$ is a field and $X$ is an irreducible variety over $k$.
Then:
\begin{enumerate}
\item If $U\subseteq X$ is a Zariski open set and $x,y \in U(k)$ then
  $x\related{X}y$ if and only if $x\related{U}y$.
\item If $X=\mathbb{P}^n_k$ then $x\related{X}y$ for any two $x,y$.
\item Suppose $p\colon Z\dasharrow X$ is a dominant and separable
  rational map with $Z$ an irreducible variety over $k$: then, for any
  $x,y \in Z(k)$ at which $p$ is defined, if
  $x\related{Z}y$ then $p(x)\related{X}p(y)$.
\item Suppose $X$ is smooth projective: then, for any $x,y \in X(k)$,
  if $x\related{X}y$ then $x$ and $y$ are R-equivalent by a single
  very free rational curve.
\end{enumerate}
\end{prop}

The first fact is trivial (restrict $F$ to $U$ on the range).

To prove the second, consider $x, y \in \mathbb{P}^n_k(k)$ and take
the family of all smooth conics passing through $x$ and $y$ and
parameterize them rationally: obviously we can find an open set $M$ in
some affine space over $k$ (so certainly $M(k)$ is dense) and a
dominant and separable morphism $F \colon M\times \mathbb{P}^1 \to
\mathbb{P}^n$ which takes $M\times\{0\}$ to $x$ and
$M\times\{\infty\}$ to $y$.

The third statement is trivial: merely compose $F$ with $p$.

To get the fourth, first notice that when $X$ is projective we can by
restricting $M$ assume that $F$ is a morphism; now apply the following
geometric result (see, e.g.,~\cite{KollarBook}, II.3.10):
\begin{prop}
\label{Kollarerie}
Let $\bar k$ be an algebraically closed field, $M$ an irreducible
variety over $\bar k$, and $X$ a smooth projective variety over $\bar
k$.  Let $x\in X$.  Finally, let $F\colon M\times\mathbb{P}^1\to X$ be
a separable and dominant morphism such that $F(M\times\{0\})=\{x\}$.
Then there exists a \emph{nonempty} Zariski open set $M^0$ of $M$
such that for all $p\in M^0$ the morphism $F_p\colon\mathbb{P}^1\to X$
satisfies the condition that $F_p^* T_X$ be ample.
\end{prop}
---and make use of the fact that $M^0$ has a point over $k$ by
assumption.

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\section{R-equivalence on universal torsors}

The goal of this section is to prove the following result:
\begin{prop}
Let $T$ be an algebraic torus over an infinite field $k$, and $X$ a
smooth equivariant compactification of $T$; then given two
$k$-rational points $x,y$ of $X$, if $x$ and $y$ are rationally
equivalent, they are R-equivalent by a single very free rational curve.
\label{MainGoal}
\end{prop}

To do this, we use the following result, whose proof will be given in
the appendix:
\begin{prop}
Let $T$ be an algebraic torus on a field $k$, and $X$ a smooth
equivariant compactification of $T$; then there exists a torus $S$
over $k$, a ``universal'' $S$-torsor $p\colon\mathscr{T}\to X$, and an
$S$-equivariant open embedding of $\mathscr{T}$ in an affine space on
which $S$ acts linearly.
\label{MPStatement}
\end{prop}

``Universal'' is to be taken in the sense of \cite{DescenteUn}, II.C
(or \cite{DescenteDeux}, example~2.3.3), which we presently recall.
Call $H^1(X,S)$ the \'{e}tale cohomology group classifying $S$-torsors on
$X$, and $[\mathscr{T}]$ the class of $p$ in it.  Define a map
$\chi\colon H^1(X,S) \to\Hom_{\Gal(\bar k/k)} (S^*, \Pic \bar X)$
which sends the class of an $S$-torsor on $X$, say $\mathscr{S}$, and
a character $\lambda \in S^* = \Hom(\bar S, \bar{\mathbb{G}}_m)$ to
the class of the $\bar{\mathbb{G}}_m$-torsor on $\bar X$ deduced from
$\bar{\mathscr{S}}$ by $\lambda$.  To say that $\mathscr{T}$ is
universal means that $S^* = \Pic \bar X$ and that
$\chi([\mathscr{T}])$ is the identity on $\Pic \bar X$.

We will need the following fact:
\begin{lem}
Let $T$ and $X$ be as in proposition~\ref{MainGoal}, and let
$p\colon\mathscr{T}\to X$ be a universal torsor on $X$.  Then there
exists a point $z\in T(k)$ such that the class $[\mathscr{T}\times_X
\Spec k_z] \in H^1(k,S)$ of the fiber of $\mathscr{T}$ over $z$ is
trivial, i.e.~$\mathscr{T}$ has a $k$-point over $z$.
\label{HasPointLemma}
\end{lem}
\begin{proof}
Let $\alpha=[\mathscr{T}\times_X \Spec k_o] \in H^1(k,S)$ be the class
of the fiber of $\mathscr{T}$ over the origin $o\in T(k)$.  Let
$\mathscr{T}^o$ be the torsor defined by $[\mathscr{T}^o] =
[\mathscr{T}] - \alpha$: then $\mathscr{T}^o$ is the universal torsor
that is trivial\footnote{In fact, if the torsor $\mathscr{T}$ is that
which we shall construct in the appendix, it is easy
to see that it is already the universal torsor trivial over $o$;
however, we shall not use this fact, which only very slightly
simplifies the proof.} above $o$, and, from the discussion in
\cite{DescenteUn},~III (see also~\cite{DescenteDeux},~2.4.4), the map
$T(k) \to H^1(k,S),\;\penalty-100 z\mapsto [\mathscr{T}^o\times_X
\Spec k_z]$ is surjective.  In particular, there exists $z$ such that
$[\mathscr{T}^o\times_X \Spec k_z] = -\alpha$, so
$[\mathscr{T}\times_X \Spec k_z] - \alpha = -\alpha$, which proves
that $[\mathscr{T}\times_X \Spec k_z]$ is nil, what we wanted.
\end{proof}

Now apply lemma~\ref{HasPointLemma} to the universal torsor
$\mathscr{T}$ given by proposition~\ref{MPStatement}: we see that
there exists $z'\in T(k)$ such that the fiber of $\mathscr{T}$ over
$z'$ is trivial.  Apply now the same lemma to the universal torsor
$\mathscr{T}^x$ with trivial fiber over $x$ (in other words the torsor
given by $[\mathscr{T}^x] = [\mathscr{T}] - [\mathscr{T} \times_X
\Spec k_x]$): so there exists $z\in T(k)$ such that the fiber of
$\mathscr{T}^x$ over $z$ is trivial.  Let $\tau_{z'-z} \colon X\to X$
be the translation by $z'-z$: the torsor $\tau_{z'-z}^* \mathscr{T}$
is still universal (since $\tau_{z'-z}$ acts trivially on $\Pic \bar
X$) and it is trivial over $z$---therefore it is isomorphic to
$\mathscr{T}^x$ (which has the same property).

Let $x' = \tau_{z'-z}(x)$ and $y' = \tau_{z'-z}(y)$.  Since
$\mathscr{T}^x \cong \tau_{z'-z}^*\mathscr{T}$ is trivial over $x$, it
follows that $\mathscr{T}$ is trivial over $x'$.  But, since $y$ is
rationally equivalent to $x$ by \cite{DescenteUn},~II.B,
proposition~1, $\mathscr{T}^x$ is also trivial over $y$, and therefore
so is $\mathscr{T}$ over $y'$.  So there exist points $P$ and $Q$ of
$\mathscr{T}(k)$ over $x'$ and $y'$ respectively, and
proposition~\ref{MPStatement} shows that $P$ and $Q$ live inside an
open set of an affine space $\mathbb{A}$ over $k$.

Finally, using the general facts laid out in
proposition~\ref{GeneralFacts}~(1--4), we have
$P\related{\mathbb{A}}Q$ (use facts 1--2) so $P\related{\mathscr{T}}Q$
(fact~1 again) and therefore $x'\related{X}y'$ (fact~3: compose with
$p$) so $x\related{X}y$ (compose with $\tau_{z-z'}$) which gives the
desired conclusion (from fact~4).\qed

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\section{Del Pezzo surfaces of degree~$5$}

We now turn to the case where $X$ is a del Pezzo surface of degree~$5$
over $k$.  Then it is known that there is a unique universal torsor
$p\colon \mathscr{T}\to X$ on $X$ (``unique'' up to non-unique
isomorphism), trivial over every point, and that it is an open set of
the Grassmanian variety $\Gr(2,5)$ of lines in $\mathbb{P}^4$
(Skorobogatov, \cite{Skorobogatov}, theorem~3.1.4).

If now $x$ and $y$ are two arbitrary $k$-rational points on $X$, pick
$k$-rational points in $p^{-1}(x)$ and $p^{-1}(y)$ (which exist
because $\mathscr{T}$ is trivial over $x$ and $y$), corresponding to
two lines $\Delta$ and $\Lambda$ in $\mathbb{P}^4$.  Now let $\Pi$ and
$\Pi'$ be two hyperplanes in $\mathbb{P}^4$ neither of which contains
either $\Delta$ or $\Lambda$ and such that the intersection points
$P,P'$ of $\Pi,\Pi'$ with $\Delta$ are distinct and similarly for the
intersection points $Q,Q'$ of $\Pi,\Pi'$ with $\Lambda$.  Then we have
a rational map $\Pi\times\Pi' \dasharrow \mathscr{T} \to X$ taking a
point on $\Pi$ and one on $\Pi'$ to the line they define (in general)
and then to the image point by $p$ in $X$.  Again by the general facts
laid out in proposition~\ref{GeneralFacts}, since
$(P,P')\related{\Pi\times\Pi'} (Q,Q')$, we get $x\related{X}y$ and
consequently $x$ and $y$ are R-equivalent by a single very free
rational curve.

Thus, we have shown:
\begin{prop}
Let $X$ be a del Pezzo surface of degree~$5$ over an infinite field
$k$; then given any two $k$-rational points $x,y$ of $X$, there exists
$f\colon\mathbb{P}^1_k\to X$ such that $f(0)=x$ and $f(\infty)=y$
with, further, $f^* T_X$ ample.
\end{prop}

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\section*{Appendix:
Explicit construction of a universal torsor over a toric variety}

Proposition~\ref{MPStatement} remains to be settled.  A proof
can be found in \cite{Salberger} (proposition~8.5), but the
one we give below, for the reader's convenience, seems much more
straightforward.

{\medbreak\footnotesize \textbf{Historical remark:} The construction
described here was introduced in \cite{Delzant} and \cite{Cox}.  Here
we give a presentation similar to the one contained in
\cite{Merkurjev}, although universality of the torsor is not shown
there.\medbreak}

Let $T^*=\Hom_{\bar k}(\bar T, \bar{\mathbb{G}}_m)$ be the lattice of
characters of the torus $T$, and $T_*=\Hom_{\bar k}
(\bar{\mathbb{G}}_m, \bar T)$ the lattice, dual to the former, of
cocharacters.  One and the other are endowed with an action of the
Galois group $\Gamma = \Gal(\bar k/k)$.  We write $T^*_{\mathbb{R}} =
T^*\otimes_{\mathbb{Z}} \mathbb{R}$ for the real vector space in which
$T^*$ lives, and $T_{*\mathbb{R}} = T_*\otimes_{\mathbb{Z}} \mathbb{R}$ for
the real vector space, dual to the former, in which $T_*$ lives.  The
general theory of toric varieties (cf.~\cite{Fulton}, in particular
\S2.3) allows us to describe $X$ by means of a fan $\Sigma$ of strongly
convex rational polyhedral cones in $T_{*\mathbb{R}}$.  The fact that $X$ is
smooth means (cf.~\cite{Fulton},~\S2.1) that every cone
$\sigma\in\Sigma$ is spanned by part of a basis of $T_*$, determined
uniquely by $\sigma$: call $B_\sigma$ the part in question, and let
$P=\bigcup_{\sigma\in\Sigma}B_\sigma$ be the union of the $B_\sigma$ for
all $\sigma\in\Sigma$.  Then $P$ is a finite part of $T_*$ which spans
the latter and is stable under the action of $\Gamma$.  For every
$\sigma\in\Sigma$, we have $B_\sigma=\sigma\cap P$, and $\sigma$ is
spanned by $\sigma\cap P$.

Now let $V_*$ be the (free) lattice with basis $P$ (with the obvious
action of $\Gamma$ making it a permutation lattice), and $V^*$ the
dual lattice, and $V_{*\mathbb{R}}$ and $V^*_{\mathbb{R}}$ the real
vector spaces in which they respectively live.  We call $V$ the dual
torus to $V^*$ (i.e.~the torus of which $V^*$ is the character
lattice), so $\bar V = \Spec \bar k[z^u: u\in V^*]$: since $V^*$ is a
permutation lattice, $V$ is a quasi-trivial torus.  And let
$\mathbf{A}$ be the affine space defined by the cone of
$V_{*\mathbb{R}}$ spanned by the elements of $P$.  Since $P$ spans
$T_*$, we have a surjective morphism $V_*\to T_*$ and thus an injection
$T^*\to V^*$.

From the description in \cite{Fulton},~\S3.3, the lattice $V^*$ is
precisely the group $\Divisors_{\bar X\setminus\bar T} \bar X$ of
$\bar T$-invariant divisors of $\bar X$, by the arrow which sends a
$u\in V^*$ to $\sum_{p\in P} u(p) D_p$ (where $D_p$ is the closure of
the orbit of $\bar T$ acting on $\bar X$ associated to the ray spanned
by $p$ in $V_{*\mathbb{R}}$).  With this identification, $T^*\to V^*$
sends a $u\in T^*$ to the principal divisor $\divisor(t^u)$, and its
cokernel (\cite{Fulton},~\S3.4) is the Picard group of $\bar X$, which
is itself a lattice, say $S^*$, dual to a torus $S$.  We therefore
have the short exact sequence of lattices $0\to T^*\to V^*\to S^*\to
0$, equal to $0\to \bar k[\bar T]^\times/\bar k^\times \to
\Divisors_{\bar X \setminus \bar T} \bar X\to \Pic\bar X\to 0$, and
the dual short exact sequence of tori $1\to S\to V\to T\to 1$.

For every cone $\sigma\in\Sigma$, let $\sigma^\vee = \{u\in T^*_{\mathbb{R}}
: (\forall v\in\sigma)(\langle u,v\rangle\geq 0)\}$ denote the dual cone, and
let $\bar X(\sigma) = \Spec \bar k[t^u: u\in T^* \cap \sigma^\vee]$
be the spectrum of the semigroup algebra of $T^*\cap\sigma^\vee$: thus,
$\bar X$ is obtained precisely by gluing the $\bar X(\sigma)$ for
$\sigma\in\Sigma$ (identifying the open set $\bar X(\sigma\cap\sigma')$
in $\bar X(\sigma)$ and $\bar X(\sigma')$).  Similarly, given a cone
$\sigma\in\Sigma$, which is, therefore, spanned by a finite set (called
$B_\sigma$) of elements of $P$, we can consider the cone
$\tilde\sigma$ in $V_{*\mathbb{R}}$ spanned by the same elements of $P$, and
its dual $\tilde\sigma^\vee$, a cone in $V^*_{\mathbb{R}}$: let us call
$\bar{\mathbf{A}}(\sigma) = \Spec \bar k[z^u: u\in V^* \cap
\tilde\sigma^\vee]$ the spectrum of the corresponding semigroup
algebra.  Thus $\bar{\mathbf{A}}(\sigma)$ is an open set in
$\bar{\mathbf{A}}$, containing $\bar V$.  Furthermore, the inclusion
$T^* \to V^*$, which manifestly sends $T^*\cap\sigma^\vee$ inside $V^*
\cap \tilde\sigma^\vee$, defines a morphism $\bar{\mathbf{A}}(\sigma)
\to \bar X(\sigma)$.

To make the situation clearer, let us presently prove the following
lemma (lemma~5.1 of \cite{Merkurjev}):
\begin{lem}
Let $\delta\in S^*$ and let $\sigma\in\Sigma$.  Then there exists a
$u_\delta\in V^*$ (not necessarily unique) which maps to $\delta\in
S^*$ (by the arrow $V^*\to S^*$ defined above) and such that $\langle
u_\delta, p\rangle=0$ for all $p\in B_\sigma$ (in other words
$u_\delta\in V^*\cap \tilde\sigma^\vee \cap (-\tilde\sigma^\vee)$).
\label{DeltaLemma}
\end{lem}
\begin{proof}
The morphism $V^*\to S^*$ being surjective, there exists $v\in V^*$
which maps to $\delta\in S^*$.  Since $B_\sigma$ is a subset of a
basis of $T_*$, there exists $\tilde v\in T^*$ such that $\langle
\tilde v, p\rangle = \langle v, p\rangle$ for all $p\in B_\sigma$.  We
then take $u_\delta = v-\tilde v$.  \qed

A $u_\delta$ as given by the previous lemma defines a $z^{u_\delta}
\in \bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]$ which is invertible
in this algebra, since $-u_\delta$ manifestly also belongs to
$\tilde\sigma^\vee$.  We deduce the following description:
\end{proof}
\begin{fact}
$\bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]$, seen as a module over
$\bar k[t^u: u\in T^* \cap \sigma^\vee]$, is free and a basis is
formed by invertible elements $z^{u_\delta}$, one for each $\delta$ in
$S^*$; the free sub-module of rank $1$ corresponding to a $\delta$ in
$S^*$ is precisely the set of linear combinations of the $z^u$ for
those $u\in V^* \cap \tilde\sigma^\vee$ for which $u|_{S_*}$ (that is,
the image of $u$ by $V^*\to S^*$) is $\delta$.  This can also be
expressed by saying that $\bar k[z^u: u\in V^* \cap
\tilde\sigma^\vee]$ is graded by $S^*$ as an algebra over $\bar k[t^u:
u\in T^* \cap \sigma^\vee]$, each graded component containing an
invertible element.
\label{AlgebraDescription}
\end{fact}

In particular, we see that if $\sigma'\subseteq\sigma$ in $\Sigma$,
the tensor product of $k[z^u: u\in V^* \cap \tilde\sigma^\vee]$ with
$\bar k[t^u: u\in T^* \cap \sigma^{\prime\vee}]$ over $\bar k[t^u:
u\in T^* \cap \sigma^\vee]$ is $k[z^u: u\in V^* \cap
\tilde\sigma^{\prime\vee}]$, which means that the inverse image by
$\bar{\mathbf{A}}(\sigma) \to \bar X(\sigma)$ of $\bar X(\sigma')$ is
$\bar{\mathbf{A}}(\sigma')$, and, more precisely, that the morphism
$\bar{\mathbf{A}}(\sigma') \to \bar X(\sigma')$ is exactly the
restriction of $\bar{\mathbf{A}}(\sigma) \to \bar X(\sigma)$ to $\bar
X(\sigma')$.  The union of the $\bar{\mathbf{A}}(\sigma)$ for
$\sigma\in\Sigma$, which we call $\bar{\mathscr{T}}$, comes from a
variety $\mathscr{T}$ defined over $k$ and open in $\mathbf{A}$, and
by gluing we have a morphism $\mathscr{T}\to X$.

We also see that $\bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]$ is
faithfully flat over $\bar k[t^u: u\in T^* \cap \sigma^\vee]$.  Thus,
the morphism $\mathscr{T}\to X$ is faithfully flat.  We get an action
of $V$ on $\mathscr{T}$ because $\mathscr{T}$ has been constructed as
a toric variety (with cones $\tilde\sigma\subseteq V_{*\mathbb{R}}$);
therefore, by restriction, we get an action of $S$ on $\mathscr{T}$,
which by construction leaves $X$ invariant.  To see that this gives us
a torsor under $S$, it is enough to see that each
$\bar{\mathbf{A}}(\sigma) \to \bar X(\sigma)$ is a torsor under $\bar
S$.  In other words, we must show that the morphism
\[
\theta\colon \bar S\times \bar{\mathbf{A}}(\sigma)
\to \bar{\mathbf{A}}(\sigma) \times_{\bar X(\sigma)}
\bar{\mathbf{A}}(\sigma)\;,\;\;
(s,a)\mapsto (s\cdot a, a)
\]
is an isomorphism.  But the (co)morphism of the associated algebras
from which it comes is given by
\[
\begin{array}{l}
\theta^*\colon \bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]
\otimes_{\bar k[t^u: u\in T^* \cap \sigma^\vee]}
\bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]\\
\phantom{\theta^*\colon \bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]}
\to \bar k[\chi^\lambda: \lambda\in S^*]
\otimes_{\bar k} \bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]\\
\strut\phantom{theta^*\colon}z^u\otimes z^{u'}\mapsto
\chi^{u|_{S_*}}\otimes z^{u+u'}\\
\end{array}
\]
To see that this is indeed an isomorphism, notice that according to
fact~\ref{AlgebraDescription}, the left-hand side has a basis over
$\bar k[t^u: u\in T^* \cap \sigma^\vee]$ formed by the
$z^{u_\delta}\otimes z^{u_{\delta'}}$ with $u_\delta$ as given in
lemma~\ref{DeltaLemma}, and the right-hand side has a basis formed by the
$\chi^\lambda\otimes z^{u_{\delta''}}$.  And on these two bases, the
homomorphism in question is represented by a diagonal matrix whose
coefficients are $t^{u_\delta+u_{\delta'}-u_{\delta''}}$ (for
$\delta''=\delta+\delta'$ and $\lambda=\delta$), which are invertible
in $\bar k[t^u: u\in T^* \cap \sigma^\vee]$.

It remains to see that this torsor $p\colon \mathscr{T}\to X$ is
indeed universal.

If $\sigma\in\Sigma$, since $\bar X(\sigma)$ is smooth, it is abstractly
isomorphic to $\bar{\mathbb{A}}^d\times \bar{\mathbb{G}}_m^{n-d}$
(where $d$, say, is the dimension of $\sigma$ and $n$ that of $T$).
In particular we have $\Pic \bar X(\sigma) = 0$; and furthermore $\bar
k[t^u: u\in T^* \cap \sigma^\vee]^\times = \{t^u\colon u\in T^*\cap
\sigma^\vee \cap (-\sigma^\vee)\}$.  The general exact sequence $0\to
\bar k[\bar U]^\times/\bar k^\times \to \Divisors_{\bar X \setminus
\bar U} \bar X\to \Pic\bar X\to 0$ (cf.~\cite{DescenteDeux}, (2.3.10))
when $\Pic\bar U=0$ becomes, for $\bar U = \bar X(\sigma)$,
\[
0\to T^* \cap \sigma^\vee \cap (-\sigma^\vee) \to
V^* \cap \tilde\sigma^\vee \cap (-\tilde\sigma^\vee) \to S^*\to 0
\]
The dual short exact sequence of tori is $1\to \bar S\to \bar
M_\sigma\to \bar R_\sigma\to 1$, where $\bar R_\sigma$ and $\bar
M_\sigma$ are quotients of $\bar T$ and $\bar V$ respectively.
Furthermore, the quotient morphism $\bar T\to \bar R_\sigma$ extends
to $\bar X(\sigma)$ (of which $\bar T$ is an open set): precisely, the
morphisms $\bar T \to \bar X(\sigma) \to \bar R_\sigma$ give, on the
associated algebras,
\[
\bar k[t^u: u\in T^* \cap \sigma^\vee \cap (-\sigma^\vee)]
\to \bar k[t^u: u\in T^* \cap \sigma^\vee]
\to \bar k[t^u: u\in T^*]
\]
By corollary~2.3.4 of \cite{DescenteDeux}, it is now sufficient to
prove that $\bar{\mathbf{A}}(\sigma)\to \bar X(\sigma)$ is obtained as
the pull\-back of $\bar M_\sigma\to \bar R_\sigma$ by the arrow $\bar
X(\sigma)\to \bar R_\sigma$, moreover in a way compatible with the
restrictions when $\sigma'\subseteq\sigma$.  In other words, we are to
determine (in a natural way) the fiber product $\bar M_\sigma
\times_{\bar R_\sigma} \bar X(\sigma)$; this is the affine scheme
whose algebra is the tensor product
\[
\bar k[z^u: u\in V^* \cap \tilde\sigma^\vee \cap (-\tilde\sigma^\vee)]
\otimes_{\bar k[t^u: u\in T^* \cap \sigma^\vee \cap (-\sigma^\vee)]}
\bar k[t^u: u\in T^* \cap \sigma^\vee]
\]
But (from fact~\ref{AlgebraDescription}) $\bar k[z^u: u\in V^* \cap
\tilde\sigma^\vee]$ is free over $\bar k[t^u: u\in T^* \cap
\sigma^\vee]$ with basis $\{z^{u_\delta}\}$ for $\delta\in S^*$; and for
precisely the same reasons, $\bar k[z^u: u\in V^* \cap
\tilde\sigma^\vee \cap (-\tilde\sigma^\vee)]$ is free over $\bar
k[t^u: u\in T^* \cap \sigma^\vee \cap (-\sigma^\vee)]$ with the same
basis.  That is to say that the above tensor product is (by the
natural map) $\bar k[z^u: u\in V^* \cap \tilde\sigma^\vee]$, in other
words that $\bar M_\sigma \times_{\bar R_\sigma} \bar X(\sigma) =
\bar{\mathbf{A}}(\sigma)$ (naturally).

This shows that the torsor $p\colon \mathscr{T}\to X$, obtained by
gluing these different $\bar{\mathbf{A}}(\sigma)\to \bar X(\sigma)$,
is indeed universal.

\bigbreak
{\footnotesize \textbf{Acknowledgements:} The author wishes to thank Jean-Louis
Colliot-Th\'{e}l\`{e}ne for his illuminating explanations on the
universal torsor and its use, and Emmanuel Peyre for providing some of
the references below and for inviting me to Grenoble to give a talk on
this construction.  I am also indebted to Laurent Moret-Bailly for
showing me how to state clearly (and gather in a single place) the
facts listed in proposition~\ref{GeneralFacts}.\par}
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\end{document}
