The Stacks project

Proof. We can use Lemmas 48.15.3 and 48.15.4 to see that we have isomorphisms $\wedge ^ i\mathcal{N} \to \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X)$ for $i \geq 0$. Thus it suffices to show that the map $\mathcal{O}_ Y \to \mathcal{O}_ X$ induces an isomorphism

and that $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y)$ is zero for $i \not= r$. These statements are local on $Y$. Thus we may assume that we have global sections $f_1, \ldots , f_ r$ of $\mathcal{O}_ Y$ which generate $\mathcal{I}$ and which form a Koszul regular sequence. Let $\mathcal{A}^\bullet $ be the Koszul complex on $f_1, \ldots , f_ r$ as introduced in the proof of Lemma 48.15.4. Then

by Cohomology, Lemma 20.46.9. Denote $1 : \mathcal{A}^\bullet \to \mathcal{O}_ Y$ the map of differential graded $\mathcal{O}_ Y$-algebras given by the identity map of $\mathcal{A}^0 = \mathcal{O}_ Y \to \mathcal{O}_ Y$ in degree $0$. With $\delta _ j$ as in the proof of Lemma 48.15.4 we get an isomorphism of graded $\mathcal{O}_ Y$-modules

by mapping $\delta _{j_1} \ldots \delta _{j_ i}$ to $1 \circ \delta _{j_1} \circ \ldots \circ \delta _{j_ i}$ in degree $i$. Via this isomorphism the differential on the right hand side induces a differential $\text{d}$ on the left hand side. By our sign rules we have $\text{d}(1) = - \sum f_ j \delta _ j$. Since $\delta _ j : \mathcal{A}^\bullet \to \mathcal{A}^\bullet [1]$ is a morphism of complexes, it follows that

Observe that we have $\text{d} = \sum f_ j \delta _ j$ on the differential graded algebra $\mathcal{A}$. Therefore the map defined by the rule

will define an isomorphism of complexes

if $r$ is odd and commuting with differentials up to sign if $r$ is even. In any case these complexes have isomorphic cohomology, which shows the desired vanishing. The isomorphism on cohomology in degree $r$ under the map

also follows in a straightforward manner from this. (We observe that our choice of conventions regarding Koszul complexes does intervene in the definition of the isomorphism $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{O}_ Y) = \wedge ^ r \mathcal{N}[-r]$.) $\square$