Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 3

A useful form of the Mumford rigidity lemma says that if $X$ is complete and connected and $h:X\times Y\to Z$ is constant on one fiber $X\times \{y_0\}$, then under the pointed separated hypotheses it factors through $Y$.

For $f:X\to G$ with $f(e_X)=e_G$, defineAt $x=e_X$ this is constantly $e_G$. Applying rigidity with the complete factor $X$ in the $y$ variable shows that $h(x,y)$ is independent of $y$. At $y=e_X$ its value is $f(x)$, soThus $f$ is a homomorphism of group varieties.

Completeness is essential. On the additive group variety ${\mathbb{G}}_a$ over a field of characteristic different from two, the morphism $f(t)=t^2$ fixes zero but is not additive.

Solved by gpt-5.6-sol high.

For $x:S\to G$, defineThen $T_{x/S}=(T_x,{\mathrm{pr}}_2)$ is an $S$-morphism, and translation by $x^{-1}$ is its inverse.

Because an isomorphism preserves relative differentials and $T_{x/S}$ lies over $S$,Using base change, ${\Omega }_{G\times S/S}\cong {\mathrm{pr}}_1^{\ast }{\Omega }_{G/k}$, while the left side is $T_x^{\ast }{\Omega }_{G/k}$. Pulling this isomorphism back along the identity section $e\times 1_S$ givesFinally take $S=G$ and $x=1_G$. This yields the invariant differential on a group scheme trivializationso ${\Omega }_{G/k}$ is a free ${\mathcal{O}}_G$-module.

Solved by gpt-5.6-sol high.