# Wednesday April 22nd

Remarks:

1. In the category of $G\dash$sets, we have $\aut_{G\dash\text{set}}(G) = G$ where $G$ acts on itself by translation.

2. For $k$ a field, $G/k$ a smooth algebraic group, $X/k$ a variety, $\mu: G\cross X \to X$ is a torsor iff the map
  \begin{align*}
  \phi: G \cross X &\to X\cross X \\
  (g, x) &\mapsto (\mu(g, x), x)
  \end{align*}
    is an isomorphism.
  Letting $G_X \definedas G\cross X, X_X \definedas X\cross X$ be the base changes, this asks for a commuting diagram
  
  \begin{tikzcd}
  G_X \cross_X G_X \ar[r, "\mu_{G_X}"] & G_X \ar[d, "1\cross \phi"] \\
  G_X \cross_X X_X \ar[r, "\mu_{X_X}"] & X_X
  \end{tikzcd}
  
  thus the base change to $X$ is the trivial $G\dash$torsor on $X$.


Suppose $G$ is commutative and recall we have a Weil-Chevalley group $WC(k, G)$.
Question:
What is the difference between a twisted form $X/k$ of $G/k$ and a torsor under $fG/k$?

Better question:
Suppose $(X, \mu) \in WC(k, G)$.
How many other elements of $WC(k, G) \ni (X', \mu)$ have $X' \cong_k X$?

Recall that $\wait [(X, \mu)] = (X, \mu([-1]\circ \wait, \wait)  )$.
Letting $G = E$ an elliptic curve, we can consider the subtraction map
\begin{align*}
v: X\cross X &\to E \\
(p, g) &\mapsto g
\end{align*}

with $p= g+q$ iff


\begin{tikzcd}
\underline{\Pic}^1 X \cross \underline{\Pic}^1 X  \ar[rr, "v"]\ar[rd, "v"]
& & E \\ 
&  \underline{\Pic}^0 X  \ar[ru, dotted] &
\end{tikzcd}


For $X/k$ a nice genus one curve, $E/k$ an elliptic curve, $\mu: E\cross X \to X$ is a torsor iff the map $\underline{\Pic}^0 X \to E$ is an isomorphism.
Therefore two elements $X_1, X_2 \in WC(k, E)$ are isomorphic iff $X_1, X_2$ lie in the same $\Aut(E)\dash$orbit of $WC(k, E)$.

Remarks:
Thus the torsors over $E$ aren't much more interesting than $E$ itself.
E.g. characteristic zero, $j\neq 0, 1728$, you just mod out by $\pm 1$.
There is a version of this for abelian varieties.

- $WC(k, E) = (0) \iff$ every genus 1 curve with Jacobian $E$ has a $k\dash$rational point.
- $\forall E/k WC(k, E) = (0) \iff$ every genus 1 $C/k$ has $C(k) \neq\emptyset$.

Example
: For $k = \bar k$, all nonempty $V/k$ have $V(k) \neq \emptyset$.

Example
:   Say a $k$ is *pseudo algebraically closed* iff every geometrically integral $V/k$ has a $k\dash$rational point, i.e. $V(k) \neq \emptyset$.
    E.g., if $k$ is *separably closed* it is pseudo algebraically closed.

Example
:   For $k = \FF_q$ a finite field, if $X/k$ is nice of genus $g$, then $$\abs{\# X(\FF_q) - (q + 1)} \leq 2g \sqrt q.$$
    Thus for $g=1$, for elliptic curves, we get $$q+1 - 2\sqrt q \leq \# X(\FF_q) \leq q+1 + 2\sqrt{q}$$ and since $q>2$, the number of points is strictly positive.

Example (Non-Example)
:   Take $C: y^2 = P_4(x) \in k[x]$ for $P_4$ a separable degree 4 polynomial.
    Look at $C \mapsvia{x} \PP^1$, and define the *index* $I(C)$ of a genus 1 curve $C/k$ to be the least positive degree of a $k\dash$rational divisor on $C$, equivalently the gcd of degrees of closed points on $C$.

Exercise
: If $C$ is a genus 1 curve, then $C$ is given by $y^2 = P_4(x)$ iff $C$ has a $k\dash$rational divisor of degree 2 iff $I(C) \in \theset{1, 2}$.

Exercise
: If $C: y^2 = ax^4 + bx^3 + cx^2 + dx + e$, then $C(k)$ is ? iff there exists $x, y\in k$ such that $y^2 = P_4(x)$ or $a\in k\units$ is a square.

Example
:   Take $k=\RR$ and $C: y^2 = - \qty{x^4 + 1}$.
    The leading term is negative, and not a square, and the point at $\infty$ doesn't need to be check (this would yield exactly 2 real points, thus not a 1-dimensional real manifold).
    Thus $C$ is a nontrivial element of $WC(\PP, \underline{\Pic}^0 C)$.

Exercise
:   Let $p$ be a prime number and find $P_4(x) \in \QQ_p[x]$ such that $y^2 = P_4(x)$ has no $\QQ_p\dash$points.

    > Try not choosing $p=2$, and try polynomials in $\ZZ[x]$ and apply Hensel's lemma.

Exercise
:   If $C: F(x, y,z) = 0$ is a nice plane cubic curve over $k$
    
    a. Show that $C/k$ admits such a defining equation iff it has a rational divisor of degree 3 iff $I(c) \in \theset{1, 3}$.

    b. Take $k= \QQ_p$ and find $C/k$ with no $k\dash$rational points.

For $G/k$ a smooth (commutative, but not necessary) group, $X/k$ a $G\dash$torso, choose $p\in k\sep$.
Then defining $g\definedas \aut(k\sep / k)$, then $X(k\sep)$ has two actions: a galois action $g\actson$ the left, and a $G(k\sep)$ action on the right.
For all $\sigma \in g$, there exists a unique $a_\sigma \in G(k\sep)$ such that $\sigma p = p a_{\sigma}$.

This defines a map $a_\bullet: g\to G(k\sep)$ -- however, this is not a group morphism, it is a "twisted" version.
For $\sigma, \tau \in g$, by definition we have $p_{a_{\sigma \tau}} = \sigma \tau p = \sigma(\tau p) = \sigma (P a_\tau) = \cdots$ and we can conclude $$ a_{\sigma \tau} = a_\sigma^\sigma a_\tau, $$ which is in fact a 1-cocycle.