# Tor-amplitude [0, 1] in the setting of intersection theory on a regular surface?

The question is coming from Definition 1.5 in Deligne's Expose X in SGA 7 on intersection theory.

Let $X$ be a connected regular scheme of dimension $2$ and $Y \subset X$ a reduced divisor that admits a proper morphism $Y \rightarrow \mathrm{Spec}(k)$ for some field $k$. Suppose also that $D$ and $E$ are two effective divisors on $X$ with the reduced closed scheme associated to $D$ being a subscheme of $Y$. The definition mentioned above defines the intersection number of $D$ and $E$ (with respect to $k$), and the definition involves $\mathcal{O}_D \otimes^{\mathbb{L}} \mathcal{O}_E$ (in the derived category of $\mathcal{O}_X$-modules). The claim is that this complex has vanishing cohomology in degrees other than $0$ and $1$, i.e., that $\mathrm{Tor}^n_{\mathcal{O}_X}(\mathcal{O}_D, \mathcal{O}_E) = 0$ for $n \ge 2$. Why is this so?

Do we somehow know that $\mathcal{O}_D$ is of tor-amplitude $[0, 1]$ because it is supported on a $1$-dimensional closed subscheme? Is there some relation of this sort between tor-amplitude and the dimension of the support?

• On a regular scheme, effective Weil divisors are Cartier divisors, i.e., the ideal sheaf $\mathcal{I}_D$ is an invertible $\mathcal{O}_X$-module. Thus the complex supported in degrees $[-1,0]$, $\mathcal{I}_D\hookrightarrow \mathcal{O}_X$, is quasi-isomorphic to $\mathcal{O}_D$ (in degree $0$) as a complex of $\mathcal{O}_X$-modules. Thus the tor-amplitude is $[0,1]$. Aug 1, 2015 at 18:47
• Regarding your second question about the relationship between tor-amplitude and dimension of the support of a sheaf, one such result is the New Intersection Theorem. Aug 1, 2015 at 19:17
• @JasonStarr: Thank you! I had a feeling that I was missing something basic. Aug 1, 2015 at 19:58

## 1 Answer

You did not explicitly ask the following, but it is related to your second question, and it comes up in practice quite often. So I would like to state this at any rate. Let $X$ be a locally Noetherian scheme. Let $D\subset X$ be a closed subscheme such that there exists a complex $E^\bullet$ of (finite rank) locally free $\mathcal{O}_X$-modules concentrated in degrees $[-c,0]$, and there exists a chain homomorphism, $$\phi:E^\bullet \to \mathcal{O}_D[0],$$ such that the induced map $$h^0(\phi):h^0(E^\bullet) \to \mathcal{O}_D$$ is an isomorphism. Moreover, assume that the restriction of $E^\bullet$ to $X\setminus D$ is acyclic. Then every irreducible component of $D$ has "codimension $\leq c$ in $X$".

Precisely, let $\eta$ be any generic point of $D$. Form the Noetherian local ring $R=\mathcal{O}_{X,\eta}$. Then the stalk $E^\bullet_\eta$ satisfies the hypotheses of the New Intersection Theorem. Therefore, the New Intersection Theorem says that the Krull dimension of $R$ is $\leq c$.

How does this come up? Here is a typical application. Let $Y$ be a regular locally Noetherian scheme, and let $C\subset Y$ be a closed subscheme that is Cohen-Macaulay and everywhere has codimension $c$. Then (at least locally), there exists a locally free resolution $$\psi:F^\bullet \xrightarrow{\text{qism}} \mathcal{O}_C[0],$$ with $F^\bullet$ concentrated in degrees $[-c,0]$. Now let $f:X\to Y$ be a morphism, let $D$ be $X\times_Y C$, let $E^\bullet$ be $f^*F^\bullet$, and let $\phi$ be $f^*\psi$. Then $E^\bullet$ and $\phi$ satisfy the hypotheses from above. Therefore every irreducible component of $D$ has codimension $\leq c$ in $X$. Note that this setup is more general than local complete intersection morphisms, the class of morphisms that arises most often in intersection theory.

• This is very nice. Thank you for sharing this. What reference would you recommend for the New Intersection Theorem (I've looked it up from some slides online)? Also, why does there exist a $\psi$ locally on $Y$? Is this some generalization of the fact that a regular local ring of dimension $n$ has a free resolution of length $\le n$? Aug 2, 2015 at 15:19
• I am not sure what is the best reference for the New Intersection Theorem. I think I learned of it from notes of Roberts. The existence of $\psi$ follows from the Auslander-Buchsbaum formula. Aug 2, 2015 at 20:09