Lemma 5.2 (0P29) · ScalingStacks

0P29

Lemma 5.2. Let $s\in S$ . Assume $s{\not\in}Z(W)$ . Let $L=(V^{S\setminus s})^{\perp}\cap(V^{*})^{s}$ , a hyperplane of $(V^{S\setminus s})^{\perp}=(V_{S\setminus s})^{*}$ . We have a commutative diagram of $P_{S\setminus s}^{\mathrm{en}}$ -modules where the diagonal map is an isomorphism

Here, $Ps$ denotes the left $P_{S\setminus s}$ -module $P$ endowed with a right action of $a\in P_{S\setminus s}$ by multiplication by $s(a)$ .

0P2A

Proof. We will show that $\phi:P_{S\setminus s}\otimes_{S(L)}P_{S\setminus s}\to P,\ a\otimes b\mapsto as(b)$ is an isomorphism. It is a morphism of algebras, and a morphism of graded left $P_{S\setminus s}$ -modules.

By assumption, there is $t\in S\setminus s$ such that $m_{st}\not=2$ , so that $s(\alpha_{t})-\alpha_{t}$ is a non-zero multiple of $\alpha_{s}$ . It follows that $\phi(\alpha_{t}\otimes 1-1\otimes\alpha_{t})\in k^{\times}\alpha_{s}$ . Since $V^{*}=(V^{S\setminus s})^{\perp}\oplus k\alpha_{s}$ , we have $P=P_{S\setminus s}\otimes k[\alpha_{s}]$ and we deduce that $\phi$ is surjective.

Since $\phi$ is a morphism of graded free left $P_{S\setminus s}$ -modules with the same graded ranks $1+t+t^{2}+\cdots$ , we deduce that $\phi$ is an isomorphism. ∎