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I'm trying to understand forms of algebraic groups. There is a theorem that says that if $L/K$ is a finite Galois extension, then the $K$-isomorphic $L/K$-forms of an algebraic group $X$ over $k$, are in bijection with the first cohomology set $H^1(Gal(L/K),Aut_L(X))$, where Aut$_L(X)$ denotes the group of $L$-defined automorphisms of $X$. I'm trying to work this out for some examples. For example, if $X=\mathbb{G}_m$, the multiplicative group, then for any field $L$, Aut$_L(\mathbb{G}_m) \cong \{\pm 1\}$. According to Springer's book Linear Algebraic Groups, this implies that the classes of $\mathbb{G}_m$ correspond to quadratic extensions of $K$ contained in $L$. Could someone explain how this works?


Here's what I understand from how the forms are constructed from classes of cocycles. Suppose we are given a cocycle $a \in Z^1(Gal(L/K),Aut_L(X))$. This is by definition a map $a:Gal(L/K) \rightarrow Aut_L(X)$ such that for all $s,t \in Gal(L/K)$, $a(st)=a(s)s\cdot a(t)$. Denote the cohomology class by [a]. If $\sigma \in Gal(L/K)$, denote the image of $\sigma$ by $a_\sigma$. Then another notation for $a$ is $\{a_\sigma\}$. We want to construct a form from $X$, in other words, some algebraic group $Y$ over $k$ such that there exists an $L$-defined isomorphism $X \rightarrow Y$.

We can identify Aut$_L(X)$ with the automorphism group of the Hopf algebra $\mathcal{O}[X_L]=L \otimes_K \mathcal{O}[X]$. Then we define an action of Gal$(L/K)$ on $\mathcal{O}[X_L]$ by $^\sigma a = a_\sigma \circ \sigma a$. Then consider the set $B=\{a \in \mathcal{O}[X_L]:^\sigma a = a\} \subset \mathcal{O}[X_L]$. Then $L \otimes_K B \cong \mathcal{O}[X_L]$ and $B$ is a Hopf algebra. $B$ is the coordinate ring of $L$-defined functions on some algebraic group, denoted $_a X$, which is said to be obtained from $X$ by twisting using $a$. It depends only on the class of $a$.

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For the multiplicative group, here is how it works (at least if $K$ is not of characteristic 2, I don't want to take any risk). There are the so called non split tori of rank 1. They are defined as follow : choose $a\in K^*$ and define $T_a\subset\operatorname{SL}_2(k)$ to be the set of matrices of the form $\left( \begin{array}{cc} x & ay \\ y & x \end{array} \right)$ such that $x^2-ay^2=1$.

For example, if $K=\mathbb{R}$ and $a=-1$, you get the circle.

You can prove that, if $a$ is a square in $K$, then $T_a$ is isomorphic to $\mathbb{G}_m$, however if $a$ is not a square, you get a new group.

More generally, $T_a\simeq T_b$ if and only if $a/b$ is a square in $K$. So the set of groups of the form $T_a$ up to isomorphism is isomorphic to $K^*/{K^*}^2$. And this set is isomorphic to the set of quadratic extensions of $K$.

Of course, on $K[\sqrt{a}]$, $a$ becomes a square and so $T_a\otimes_K K[\sqrt{a}]\simeq\mathbb{G}_m$. In other words, $T_a$ is a form of $\mathbb{G}_m$ and becomes isomorphic to $\mathbb{G}_m$ in an extension $L$ if and only if $a$ is a square in $L$.

In fact, every form of $\mathbb{G}_m$ is isomorphic to $T_a$ for some $a\in K$. So the forms of $\mathbb{G}_m$ is in bijection with $K^*/{K^*}^2$ and with the set of quadratic extension of $K$. Those that are split in $L$ correspond to the classes $a$ such that $a$ is a square in $L$ and to the quadratic extensions contains in $L$.

Now, as you said, $\operatorname{Aut}_L(\mathbb{G}_m)=\{\pm1\}$ (with trivial $\operatorname{Gal}(L/K)$-action). So $$H^1(\operatorname{Gal}(L/K),\operatorname{Aut}_L(\mathbb{G}_m))=H^1(\operatorname{Gal}(L/K),\{\pm1\})=\operatorname{Hom}(\operatorname{Gal}(L/K),\{\pm1\}).$$ Thus the class of a form of $\mathbb{G}_m$ corresponds to a morphism $\varphi\in\operatorname{Hom}(\operatorname{Gal}(L/K),\{\pm1\})$. Now such a $\varphi$ is completely determined by $\ker\varphi$ which is a subgroup of $\operatorname{Gal}(L/K)$ and thus corresponds to an extension of $K$ in $L$ by the main theorem of Galois theory. It can be two different things :

  • either $\ker\varphi=\operatorname{Gal}(L/K)$ ($\varphi$ is trivial). The corresponding extension is of course $K$ itself. And the form of $\mathbb{G}_m$ is $\mathbb{G}_m$ itself.

  • or $\ker\varphi$ is a subgroup of index two in $\operatorname{Gal}(L/K)$. The corresponding extension is a quadratic extension of $K$ contain in $L$. It is of the form $K[\sqrt{a}]$. And the form of $\mathbb{G}_m$ is isomorphic to $T_a$.

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    $\begingroup$ I'd say with $\varphi({\scriptstyle\begin{pmatrix}u & av \\ u & v \end{pmatrix}}) = u+\sqrt{a}v$ we get an isomorphism from the matrix field $K({\scriptstyle\begin{pmatrix}x & ay \\ y & x \end{pmatrix}}), x \in K,y \in K^* $ to $K(\sqrt{a})$. So clearly $\{T_a, a \in K^*/ {K^*}^2\}$ is isomorphic to the set of quadratic extensions of $K$. $\endgroup$ – reuns Dec 29 '16 at 12:59
  • $\begingroup$ @user1952009 Yes good point. $\endgroup$ – Roland Dec 29 '16 at 15:43

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