Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 126 / 3

A group scheme over $k$ is a $k$-scheme $G$ with multiplication $m_G:G{\times }_{k}G\to G$, identity $e_G:\mathrm{Spec}k\to G$, and inversion $i_G:G\to G$ satisfying the group axioms as identities of morphisms. A homomorphism of group schemes $f:H\to G$ is a $k$-morphism satisfyingand it then preserves the identity and inversion.

Assume $G$ and $H$ are commutative. The group $G(H)={\mathrm{Mor}}_k(H,G)$ has pointwise additionFor every $k$-algebra $R$ and $x,y\in H(R)$,where commutativity permits the middle terms to be reordered. Hence $f+g$ is a homomorphism. The zero morphism and pointwise inverse are also homomorphisms, so ${\mathrm{Hom}}_k(H,G)$ is a subgroup of $G(H)$. The definition immediately gives

Repeated pointwise addition gives $(nf{)}_R(x)=n{f}_R(x)$. Since $f_R$ is a group homomorphism,The Yoneda lemma turns equality on all $R$-valued points into equality of morphisms, proving

One form of the Mumford rigidity lemma says that if $X$ is a complete variety, $S$ is connected, and a morphism $F:X\times S\to Z$ maps $X\times \{s_0\}$ to one point, then $F$ factors through the projection to $S$. In particular, if $F$ also maps $\{x_0\}\times S$ to that point, then $F$ is constant.

Choose $s_0\in S(k)$ and put $g=f{|}_{X\times \{s_0\}}$. To see that the pointed morphism $g:X\to Y$ is a homomorphism, apply rigidity toIt vanishes on $X\times \{e\}$, so it factors through the second projection; it also vanishes on $\{e\}\times X$, so it is identically zero. Now defineThen $F(x,s_0)=e$ for every $x$ and $F(e,s)=e$ for every $s$. Rigidity forces $F$ to be identically $e$, so