# Highest derived inverse image

Suppose $$i_Z \hookrightarrow X$$ be a closed immersion, with $$Z$$ and $$X$$ being smooth varieties over $$\mathbb{C}$$, and $$c, d$$ are the dimensions of $$Z$$ and $$X$$ respectively.

$$\textbf{Question}:$$ Is it true that $$L_n i^*_{Z} \mathcal{F} = 0$$ for any $$n>d-c$$ and $$\mathcal{F}\in Coh(X)$$ ?

The reason I ask is the following inconsistency I encounter (suppose $$d=2$$ and $$c=1$$ for simplicity), $$$$E_2^{p,q}:= Ext_Z^p(L_q i^*_{Z}\mathcal{F},\mathcal{O}_Z) \Longrightarrow E^{p+q}_{\infty} = Ext_Z^{p+q}(L i^*_{Z}\mathcal{F},\mathcal{O}_Z)=Ext_X^{p+q}(\mathcal{F},i_{Z *} \mathcal{O}_Z).$$$$

Note that $$E_2^{1,2}$$ survives to infinity, so $$Ext_X^{3}(\mathcal{F},i_{Z *} \mathcal{O}_Z) \ne 0$$, this is absurd because $$X$$ is a surface and $$Ext_X^3 = 0$$. But this must be true for any curve $$Z\subset X$$, and any sheaf over that (not just the structure sheaf $$\mathcal{O}_Z$$). So I got the question above.

This is puzzling for me because in general there is a resolution by locally free sheaves (again I just assume $$d=2$$),

$$$$0 \longrightarrow V^{2}\longrightarrow V^1 \longrightarrow V^0 \longrightarrow \mathcal{F}\longrightarrow 0,$$$$ and $$L_n i_Z^* \mathcal{F} = \mathcal{H}^{-n} i_Z^* (V^{\bullet})$$, so why if $$n=2$$ the corresponding cohomology should be zero?

Since $$i_*$$ is exact, we get isomorphisms $$i_*L_ji^* \mathscr F \cong L_j(i_*i^*) \mathscr F.$$ Moreover, $$i_*$$ is faithful (in fact, it has a left inverse $$i^*$$), thus to show that $$L_ji^* \mathscr F$$ vanishes for $$j > \dim(X) - \dim(Z)$$, it suffices to show the same statement for $$L_j(i_*i^*)\mathscr F$$. But the functor $$i_*i^*$$ can also be described as $$-\otimes_{\mathcal O_X} \mathcal O_Z$$, so its derived functor is $$\mathscr Tor_j^{\mathcal O_X}(-,\mathcal O_Z) = \underline H^{-j}\left(-\overset{\mathbb L}{\underset{\mathcal O_X}\otimes}\mathcal O_Z\right).$$ Now $$\mathscr Tor(-,\mathcal O_Z)$$ (equivalently, $$-\otimes^{\mathbb L} \mathcal O_Z$$) can be computed using a resolution for $$\mathcal O_Z$$ instead.
Lemma. Let $$i \colon Z \to X$$ be a regular closed immersion of Noetherian schemes that is locally cut out by $$\leq n$$ equations. Then $$\mathcal O_Z \in D(\mathcal O_X)$$ has tor amplitude $$[-n,0]$$.
For example, a smooth subvariety $$Z$$ of a smooth variety $$X$$ is a regular immersion that is locally cut out by $$\dim X - \dim Z$$ equations [Tag 0E9J]. See also Hartshorne, Theorem II.8.17.
Proof. The statement is local on $$X$$ [Tag 08BS], so we may assume that $$X = \operatorname{Spec} A$$ is affine and $$Z$$ is cut out by an ideal $$I = (f_1,\ldots,f_n)$$. Then we may take the Koszul resolution $$K_\bullet(A,f_1,\ldots,f_n) \stackrel\simeq\to \mathcal O_Z[0] = A/I[0].$$ This is a resolution of length $$n$$ by free $$A$$-modules, showing that $$A/I$$ has tor amplitude $$[-n,0]$$ in $$D(A)$$ by [Tag 0654]. $$\square$$
If you prefer a less derived argument, what we're really doing is computing $$\mathscr Tor_j(-,\mathcal O_Z)$$ by locally using the Koszul resolution of $$\mathcal O_Z$$ associated with a regular sequence cutting out $$Z$$; this can be done in a none-fancy setting.
• Thanks a lot. I totally agree with using Koszul (+ projection formula), and indeed it seems easier in practice, but could you explain to me (at least intuitively), how can we see this directly from the definition? i.e. in above notation, why $\mathcal{H}^{-n}(i^* V^2) =0$ – Mohsen Karkheiran Nov 17 '18 at 3:08