# Monday April 20th

Today: torsors.

Let $k$ be a (perfect, separable) field and $G/k$ a commutative algebraic group (a finite type reduced group ?).

Definition
:   A variety $X/k$ is a *torsor under $G$* is $\mu: G\cross X \to X$ a group action such that the map

    \begin{align*}
    G\cross X \to X \cross X \\
    (g, x) \mapsto (\mu(gx), x)
    \end{align*}
    is an isomorphism.

For $\ell/k$ any field extension, the base change to $X/\ell$ induces $\mu_\ell$ making $X/\ell$ a $G/\ell$ torsor.
$X$ is trivial iff it is isomorphic to 
\begin{align*}
\mu: G\cross G &\to G \\
(g, hg) &\mapsto gh
.\end{align*}

Claim
: $X$ is trivial iff $X(k) \neq \emptyset$.

First "proof": for $p\in X(k)$, define $\mu(\wait, p): G \to X$.
Want to get a map $G(k\sep) \to X(k\sep)$, when does this happen?
In characteristic zero, we have some map $G\to X$ (???) which is surjective with trivial kernel and thus an isogeny but has not $k\sep$ points.
But this doesn't work in positive characteristic.

Second proof: the map $G\cross X \to X \cross X$ being an isomorphism says that upon base change on $X \to \spec k$, $X$ becomes isomorphic to $G$.
But then it also becomes isomorphic over base change for which $X$ is intermediate.
So if we have

\begin{tikzcd}
T \ar[rr] \ar[rd] &           & \spec k \\
                  & X \ar[ur] &
\end{tikzcd}

which factors through $Y$, if $p\in X(k)$ then $\spec k\to X$ and thus $X/k \cong G/k$.

The form of the assumed isomorphic implies that the base change of the $G\dash$torsor $X$ from $\spec k$ to $X$ is trivial as a $G\cross X$ torsor over $X$.

$\qed$

For $k$ a field, $G/k$, an equivalent definition would be that a $G$ torsor is $X/k$ with a $G$ action that becomes trivial over $k\sep$.
Therefore A $G$ torsor $X$ is a $k\sep /k$ twisted form of $X$ where $X/k\sep \cong G/k\sep$.

Example:
Let $G = E$ an elliptic curve, and $X/k$ is a nice curve of genus 1, but $X(k)$ is likely empty.
Conversely, given such a curve of genus 1, we can take the Picard variety $\underline{\Pic}^0 X$, i.e. the Jacobian.
Then there is an isomorphism
\begin{align*}
X &\mapsvia{\cong} \underline{\Pic}^1 X \\
p &\mapsto [p]
.\end{align*}

So every nice curve is a torsor for its Jacobian (?).
Note that in higher dimensions, we'd need to take the albanese, and the same statement would work: every abelian variety is a torsor over its albanese.

For $G/k$ commutative, we can make the set of torsors $X$ for $G/k$ modulo equivalence into a commutative group.
We define the Weil-Chatelet group of $G/k$ as $WC(k, G)$. 
For two torsors, we can define the *Baer sum* $X_1 \oplus X_2$ by first defining a map 
\begin{align*}
\mu_{\pm}: G\cross \qty{X_1 \cross X_2} &\to X_1 \cross X_2 \\
(g, x_1, x_2) &\mapsto (\mu_1(g, x_1), \mu_2([-1]g, x_2)  )
\end{align*}
and defining $X_1 \oplus X_2 = \qty{X_1 \cross X_2} / \mu_{\pm}$.
Then the action $\mu_{\pm}$ on $X_1 \oplus X_2$ is a $G$ torsor.

This makes $WC(k , G)$ into a commutative group where $\mu: G\cross G\to G$ defines $[-1](X, \mu) \definedas (X, \mu([-1]\wait  ))$.

Exercise
:   For $C/k$ a nice genus one curve, $G = E = \underline{\Pic}^0 C$ and $C = \underline{\Pic}^1 C$.
    Show that $n[C] \underline{\Pic}^n C$.

> Note that by adding divisor classes, there is a map $\underline{\Pic}^1 C \cross \underline{\Pic}^1 C \to \underline{\Pic}^2 E$.

Corollary
:   For $E/k$ an elliptic curve, $WC(k, E)$ is a torsion abelian group iff for all genus 1 curves $C$, there exists an $n\in \ZZ^{\geq 0}$ such that $\qty{\underline{\Pic}^n C}(k) \neq \emptyset$.

We can define the *period* of an elliptic curve as the least $n$ for which the torsor becomes trivial, this is an interesting numerical invariant.

Next up: cocycles and descent.