The Stacks project

Lemma 48.15.5. Let $Y$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals. Set $\mathcal{O}_ X = \mathcal{O}_ Y/\mathcal{I}$ and $\mathcal{N} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_ X)$. If $\mathcal{I}$ is Koszul-regular (Divisors, Definition 31.20.2) then

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) = \wedge ^ r \mathcal{N}[-r] \]

where $r : Y \to \{ 1, 2, 3, \ldots \} $ sends $y$ to the minimal number of generators of $\mathcal{I}$ needed in a neighbourhood of $y$.

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

\[ \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ r_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ r_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \]

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

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ Y) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \]

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

\[ \mathcal{O}_ Y\langle \delta _1, \ldots , \delta _ r\rangle \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \]

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

\[ \text{d}(\delta _{j_1} \ldots \delta _{j_ i}) = (- \sum f_ j \delta _ j )\delta _{j_1} \ldots \delta _{j_ i} \]

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

\[ 1 \circ \delta _{j_1} \ldots \delta _{j_ i} \longmapsto (\delta _{j_1} \circ \ldots \circ \delta _{j_ i})(\xi _1 \ldots \xi _ r) \]

will define an isomorphism of complexes

\[ \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \longrightarrow \mathcal{A}^\bullet [-r] \]

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

\[ \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ Y) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ X) \]

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 $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{O}_ Y) = \wedge ^ r \mathcal{N}[-r]$.) $\square$

Comments (3)

Comment #8644 by nkym on

is defined as an element of , but I guess it is just an element of .

Comment #8645 by nkym on

I was wondering if in the statement and at the end of the proof, the shift by was meant instead of that by .

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