Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 167 / 3

Let $I=I(H)\subseteq k[G]$. Choose finitely many generators of $I$ and a finite-dimensional $G$-submodule $U\subseteq k[G]$ containing them, using local finiteness of the right regular action. Set $W=U\cap I$. Right translation by $H$ preserves $I$, so $H$ stabilizes $W$. Conversely, if $gW=W$, every chosen generator $f$ has $R_{g}f\in I$, and evaluation at the identity gives $f(g)=0$. Thus $g\in H$, and ${\mathrm{Stab}}_G(W)=H$.

Put $d=\dim W$ and take $V={\bigwedge }^{d}U$. The linedetermines $W$ uniquely, so its stabilizer is the stabilizer of $W$, namely $H$.

Solved by gpt-5.6-sol high.

A flat algebraic-group torsor $X\to Y$ is a faithfully flat morphism with a right $H$-action for whichis an isomorphism.

The orbit $G\ell$ is a locally closed subvariety of $\mathbb{P}(V)$ by the orbit theorem for algebraic-group actions. The fibers of the orbit map $G\to G\ell$ are precisely the right cosets of $H={\mathrm{Stab}}_G(\ell )$, and the displayed action map is therefore an isomorphism. The theorem on quotients of affine algebraic groups by closed subgroups says that $G/H$ exists and $G\to G/H$ is faithfully flat; the induced map $G/H\to G\ell$ is an isomorphism. Hence the orbit map is a flat $H$-torsor.

Solved by gpt-5.6-sol high.

For $G={\mathrm{SL}}_2$ and the upper triangular Borel subgroup $B$, the quotient is $G/B\cong {\mathbb{P}}^1$: a matrix is sent to the line spanned by its first column. Over $U_0=\{[1:t]\}$ use the sectionand over $U_1=\{[s:1]\}$ useEvery matrix above $U_i$ is uniquely $s_i(u)b$ with $b\in B$. Thus each inverse image is $U_i\times B$, proving that $G\to G/B$ is a Zariski $B$-torsor.

Solved by gpt-5.6-sol high.

For $G={\mathbb{G}}_m$ and $H={\mu }_2$, the quotient isIt is a faithfully flat ${\mu }_2$-torsor, including in characteristic two where ${\mu }_2$ is nonreduced. It is not a Zariski torsor: a local section over any nonempty open set would put a square root of the coordinate $t$ in the function field $k(t)$, but $t$ is not a square there.

Take $V=k\oplus k$ with $z\in {\mathbb{G}}_m$ acting by $z\cdot (x,y)=(x,z^{2}y)$, and let $\ell =k(1,1)$. The equality $z\ell =\ell$ holds exactly when $z^2=1$, so the scheme-theoretic stabilizer is ${\mu }_2$.

Solved by gpt-5.6-sol high.