Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 126 / 2

A group scheme over $k$ is a $k$-scheme $G$ equipped with multiplication $m:G{\times }_{k}G\to G$, identity $e:\mathrm{Spec}k\to G$, and inversion $j:G\to G$ satisfying the associativity, identity, and inverse diagrams.

ConsiderThe diagonal morphism is ${\Delta }_{G/k}=q^{-1}(e)$. For a finite-type $k$-scheme the rational identity point is closed, so its inverse image is closed. Thus the diagonal is a closed immersion and every such group scheme over a field is a separated scheme.

In characteristic $p>0$, the infinitesimal additive groupis a nonreduced group scheme. Its comultiplication is $t\mapsto t\otimes 1+1\otimes t$, which is well defined because $(t\otimes 1+1\otimes t{)}^p=t^p\otimes 1+1\otimes t^p$.

The Mumford rigidity lemma says that if $A$ is complete, $B$ is connected, and a morphism $F:A\times B\to Z$ maps the fiber over some $b_0\in B$ to a point, then $F$ is constant on every $A$-fiber and factors through $B$.

Put $y=f(e_X)$ and normalizeso $g(e_X)=e_G$. DefineWhen the first coordinate is $e_X$, this is constantly $e_G$. Apply rigidity with the second copy of the complete variety $X$ as the complete factor. It follows that $F(x,z)$ is independent of $z$, and evaluation at $z=e_X$ gives $F(x,z)=g(x)$. Thereforeso $g$ is a homomorphism of group varieties and $f=T_y\circ g$.

Completeness is essential. Take $X=G={\mathbb{G}}_a$. In characteristic different from two, the morphism $f(t)=t^2$ satisfies $f(0)=0$ but is not additive. If it were $T_y\circ g$ with $g$ a group homomorphism, evaluation at zero would give $y=0$ and hence $f=g$, a contradiction. In characteristic two the same argument works with $f(t)=t^3$.

For an abelian variety $X$, consider the commutator morphismIt is the identity whenever either coordinate is the identity. The Mumford rigidity lemma applied successively to the two complete connected factors makes $c$ constant everywhere; its value at $(e,e)$ is $e$. Therefore every pair of points commutes, so the group law is commutative.

Because $p:X_1\times X_2\to X$ is an isomorphism and $X$ is connected, each $X_i$ is connected. They are complete because they are closed in the complete variety $X$. Letbe the two component morphisms.

For $x,x^{\prime }\in X_1$, the morphismfrom $X_1\times X_1$ to $X_2$ is constantly $e$ on either coordinate axis. Rigidity therefore makes it constantly $e$, so $X_1$ is closed under addition. The same argument applies to $X_2$. If $-x=u+v$ is the unique decomposition with $u\in X_1$ and $v\in X_2$, then $e=(x+u)+v$; uniqueness of the decomposition of $e$ gives $v=e$ and $u=-x$. Thus each $X_i$ is also closed under inversion.

The restrictions of the multiplication and inversion morphisms of $X$ now make each $X_i$ a complete connected group variety, hence an abelian variety. Since the group law on $X$ is commutative,Thus $p$ is a homomorphism. It is already an isomorphism of varieties, and its inverse consequently respects the group operations as well, so $p$ is an isomorphism of group schemes.