For $x:S\to G$, define
$$
T_x:G\times_kS\to G,
\qquad(g,s)\mapsto x(s)g.
$$
Then $T_{x/S}=(T_x,\operatorname{pr}_2)$ is an $S$-morphism, and translation by $x^{-1}$ is its inverse.

Because an isomorphism preserves relative differentials and $T_{x/S}$ lies over $S$,
$$
T_{x/S}^*\Omega_{G\times S/S}\cong\Omega_{G\times S/S}.
$$
Using base change, $\Omega_{G\times S/S}\cong\operatorname{pr}_1^*\Omega_{G/k}$, while the left side is $T_x^*\Omega_{G/k}$. Pulling this isomorphism back along the identity section $e\times1_S$ gives
$$
x^*\Omega_{G/k}\cong\mathcal O_S\otimes_k\Omega_{G/k}(e).
$$
Finally take $S=G$ and $x=1_G$. This yields the <invariant differential on a group scheme> trivialization
$$
\Omega_{G/k}\cong\mathcal O_G\otimes_k\Omega_{G/k}(e),
$$
so $\Omega_{G/k}$ is a free $\mathcal O_G$-module.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-126/3/ii/solution