Local cohomology with respect to a point. (Hartshorne III Ex 2.5)

Question

I'm trying to do Hartshorne's exercises on local cohomology at the moment and seem to be stuck in Exercise III 2.5. The problem goes as follows:

$X$ is supposed to be a Zariski space (i.e a Noetherian and sober topological space) and $P\in X$ a closed point. Let $X_P$ denote the subset of $X$ consisting of all points specializing to $P$, and give it the induced topology. Let $j:X_P\rightarrow X$ be the inclusion map and write $\mathcal F_P=j^{-1}\mathcal F$ for any sheaf (of abelian groups) $\mathcal F$ on $X$. Then for all $i\in\mathbb Z$ and sheaves $\mathcal F$ we have \begin{equation} H_P^i(X,\mathcal F) = H^i_P(X_P,\mathcal F_P) \end{equation}

I succeeded in showing that the identity holds for $i=0$, so to prove the claim it would suffice to know that the $H^i_P(X_P,(~\cdot~)_P)$ define a universal $\delta$-functor. I tried to do this with the standard procedure by showing effaceability of these functors for $i>0$. So first I tried to show that pulling back along $j$ preserves flsaqueness, but meanwhile I'm quite convinced that this isn't true. I do not see why injectives would go to injectives, either. Then I thought maybe for special sheaves like sheaves of discontinuous sections flasqueness is preserved, but, as before, it didn't lead anywhere. Does anyone have some good advice?

Thanks!

Answer 1

By the excision isomorphism (Exercise III.2.3(f)), we have already $$ H^i_P(X,\mathscr{F}) \cong H^i_P(V,\mathscr{F}\rvert_V) \cong \varinjlim_{V \ni P} H^i_P(V,\mathscr{F}\rvert_V), $$ where the direct limit is over the restriction maps induced by the inclusions $P \in W \subset V$. So it remains to show $\varinjlim_{V \ni P} H^i_P(V,\mathscr{F}\rvert_V) \cong H^i_P(X_P,\mathscr{F}_P)$. Note that for any inclusion $P \in W \subset V$, we have the morphism of long exact sequences from Exercise III.2.3(e) $$\require{AMScd}\begin{CD} \cdots @>>> H^{i-1}(V,\mathscr{F}\rvert_V) @>>> H^{i-1}(V^P,\mathscr{F}\rvert_{V^P}) @>>> H^{i}_P(V,\mathscr{F}\rvert_V) @>>> H^{i}(V,\mathscr{F}\rvert_V) @>>> H^{i}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \cdots\\ @. @VVV @VVV @VVV @VVV @VVV\\ \cdots @>>> H^{i-1}(W,\mathscr{F}\rvert_W) @>>> H^{i-1}(W^P,\mathscr{F}\rvert_{W^P}) @>>> H^{i}_P(W,\mathscr{F}\rvert_W) @>>> H^{i}(W,\mathscr{F}\rvert_W) @>>> H^{i}(W^P,\mathscr{F}\rvert_{W^P}) @>>> \cdots \end{CD} $$ where $V^P$ denotes the punctured open set $V \setminus \{P\}$, and similarly for $W$. Note that every open set $V$ containing $P$ contains $X_P$ as well by the fact that it is closed under generization by Exercise II.3.17(e), and so the direct limit taken over all $V \ni P$ is the same as the direct limit taken over all $V \supset X_P$. Using the universal property of the direct limit, and the fact that direct limits are exact in the category of abelian groups [Atiyah–Macdonald, Exc. 1.19], we therefore get the corresponding morphism of long exact sequences $$\begin{CD} \cdots @>>> \varinjlim_{V \ni P}H^{i-1}(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i-1}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \varinjlim_{V \ni P}H^{i}_P(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i}(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \cdots\\ @. @VVV @VVV @VVV @VVV @VVV\\ \cdots @>>> H^{i-1}(X_P,\mathscr{F}_P) @>>> H^{i-1}(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) @>>> H^{i}_P(X_P,\mathscr{F}_P) @>>> H^{i}(X_P,\mathscr{F}_P) @>>> H^{i}(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) @>>> \cdots \end{CD} $$ Now, we see that if the vertical arrows above are isomorphisms except for the middle arrow, then by the five lemma [Weibel, Exc. 1.33], the middle one would be as well. Thus, it suffices to show that $$\tag{*}\label{eq:III.2.5} \varinjlim_{V \ni P} H^i(V,\mathscr{F}\rvert_V) \cong H^i(X_P,\mathscr{F}_P), \quad \varinjlim_{V \ni P} H^i(V^P,\mathscr{F}\rvert_{V^P}) \cong H^i(X_P^P,\mathscr{F}_P\rvert_{X_P^P}). $$ We first show the isomorphisms for $i = 0$ (using the Lemma below): $$ \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}\rvert_V) = \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}) = \varinjlim_{V \supset X_P} \Gamma(V,\mathscr{F}) = \Gamma(X_P,\mathscr{F}_P)\\ \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}\rvert_{V^P}) = \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}) = \varinjlim_{V \supset X_P} \Gamma(V^P,\mathscr{F}) = \varinjlim_{V^P \supset X_P^P} \Gamma(V^P,\mathscr{F}) = \Gamma(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) $$ and so it suffices to show them for $i \ge 1$.

As in the proof of Prop. III.2.9, since the isomorphisms hold for $i = 0$, to show the functors agree for $i \ge 1$, it suffices to show they are both effaceable as functors $\mathfrak{Ab}(X) \to \mathfrak{Ab}$, since in that case they are both universal, hence must be isomorphic by Thm. III.1.3A. So let $\mathscr{G}$ be the sheaf of discontinuous functions of $\mathscr{F}$ from Exercise II.1.16(e). Then, $\mathscr{G}$ is flasque and there is a natural inclusion $\mathscr{F} \hookrightarrow \mathscr{G}$. Now the sheaves $\mathscr{G}\rvert_V$, $\mathscr{G}\rvert_{V^P}$, $\mathscr{G}\rvert_{X_P}$, and $\mathscr{G}\rvert_{X_P^P}$ are flasque by the Lemmas below. Thus, every functor in \eqref{eq:III.2.5} is zero when applied to $\mathscr{G}$ by Prop. III.2.5, hence the functors are effaceable so we are done. $\blacksquare$

EDIT. Andreas pointed out a gap in the proof, and provided the following Lemma to fix it:

Lemma. If $\mathscr{F}$ is a sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $$\varinjlim_{\tilde{U} \supset U} \Gamma(\tilde{U},\mathscr{F}) \cong \Gamma(U,\mathscr{F}\rvert_Y)$$ where $\tilde{U}$ are open in $X$, and so there is no need to sheafify in the definition of restriction.

Proof. Because open subsets $U$ of $Y$ are closed under generization in $X$, it suffices to show the claim for $U = Y$. Consider the natural map $$\varinjlim_{Y\subseteq U}\Gamma(U,\mathscr{F}) \to \Gamma(Y,\mathscr{F}\rvert_Y)$$ obtained from the definition of sheafification.

Injectivity. This follows from the fact that a separated presheaf injects into its sheafification [Stacks, Tag 0082].

Surjectivity. We want to show every global section of $\mathscr{F}\rvert_Y$ extends to some open neighborhood of $Y$. So let $s \in \Gamma(Y,\mathscr{F}\rvert_Y)$. Since $Y$ is noetherian, there is a maximal open subset $U \subset Y$ such that $s\rvert_U$ lifts to a section $t\in\Gamma(\tilde{U},\mathscr{F})$ with $\tilde{U} \cap Y=U$. Note it is nonempty since choosing an arbitrary point $z \in Y$, the germ of $s_z$ lives in $\mathscr{F}_z = (\mathscr{F}\rvert_V)_z$, and $s_z$ extends to some open subset of $X$.

Suppose $U \ne Y$, and let $y \in Y \setminus U$. Then, by the argument above using stalks, there exists an open neighorhood $V$ of $y$ such that $s\rvert_V$ lifts to a section $t'\in\Gamma(\tilde{V},\mathscr{F})$ with $\tilde{V} \cap Y=V$.

Now because $t,t'$ are both local lifts of $s$, there exists an open subset $W$ in $X$ such that $U \cap V \subset W \subset \tilde{U} \cap \tilde{V}$ and $t\rvert_W = t'\rvert_W$ (since the support of $t-t'$ is closed in $\tilde{U} \cap \tilde{V}$). Now let $A = \tilde{U} \cap \tilde{V} \setminus W$, and let $A=\bigcup_{i=1}^k A_i$ be an irreducible decomposition. We want to show $\overline{A} \cap Y = \emptyset$, where $\overline{A}$ denotes the closure of $A$ in $X$. For suppose $\overline{A} \cap Y=\bigcup_i\overline{A_i}\cap Y$ is non-empty; then, since $Y$ is closed under generization, we can assume the generic point ($X$ is sober) $x_i$ of some $\overline{A_i}$ also lies in $\overline{A} \cap Y$. But this is a contradiction, since then, $x_i \in \tilde{U} \cap \tilde{V} \cap Y = U \cap V \subset W$ implies that $x_i\notin A_i$. Since intersecting an open with an irreducible component yields an irreducible component of the intersection, we can realise the $A_i$ are $\tilde{U}\cap\tilde{V}\cap B_i$ for the irreducible components $B_i$ of the closed set $X\setminus W$ which meet $\tilde{U}\cap\tilde{V}$; these are closed in $X$ so have (unique) generic points $y_i$ by hypothesis, and if an open set touches the closure of a point it must contain that point, so $y_i\in\tilde{U}\cap\tilde{V}$ for all $i$, $y_i\in A_i$. By irreducibility $\overline{A_i}=B_i$ so we may take $x_i=y_i\in A_i$.

Now the two sections $t\rvert_{\tilde{U} \setminus \overline{A}}$ and $t'\rvert_{\tilde{V}\setminus\overline{A}}$ match on the overlap of their domains, which contains $U \cap V$, and hence patch together to give a lift of $s\rvert_{U\cup V}$, contradicting the maximality of $U$. $\blacksquare$

This is used in the following way:

Lemma. If $\mathscr{F}$ is a flasque sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $\mathscr{F}\rvert_Y$ is flasque.

Proof. This follows since direct limits are exact, and using the previous Lemma. $\blacksquare$

Answer 2

Here is how I would prove this:

Let $X$ be a Zariski space. Let $ p \in X$ be a closed point and let $X_{p} \subset X$ be the set of all points $ q \in X$ such that $p \in (q)^{-}$. We observe that $X_{p}$ has the induced topology. Now let $j: X_{p}\hookrightarrow X$ be the inclusion map. For any sheaf $\mathcal F$, let $\mathcal F_{p}=j^{*}\mathcal F$. We then state our claim as: $\Gamma_{p}(X, \mathcal F)=\Gamma_{p}(X, \mathcal F_{p})$. Any open set $U$ containing $p$ also contains $X_{p}$ so gluing sheaves will not affect $\Gamma(X_{p}, \mathcal F_{p})= \lim\underset {\rightarrow p \in U} \Gamma (U, \mathcal F)$. Taking $\pi \in \Gamma_{p} (X, \mathcal F)$, we get a section $\pi^{\prime} \in \Gamma_{p}(X_{p}, \mathcal F_{p})$. We may represent $\pi^{\prime}$ by $\pi \in \Gamma(U, \mathcal F)$. Shrink $U$ such that we may assume $\mathrm{Supp}(\pi)=p$. Glue $\pi$ and $0\in \Gamma (X/p, \mathcal F)$ to obtain a global section. Then there exists a bijection $\Gamma_{p}(X, \mathcal F_{p})\leftrightarrow \Gamma_{p}(X_{p}, \mathcal F_{p})$.

Finally, if $0 \rightarrow \mathcal F \rightarrow I_{0} \rightarrow I_{1} \rightarrow \cdots$ is a flasque resolution of $\mathcal F$, then $0 \rightarrow \mathcal F_{p} \rightarrow I_{0,p} \rightarrow I_{1,p} \rightarrow \cdots$ is an injective resolution of $\mathcal F_{p}$. By repeating this argument we obtain $\Gamma_{p}(X, I_{i}) \cong \Gamma_{p}(X_{p}, I_{i,p})$ and we have that $H^i_p(X, \mathcal F)=H^{i}_{p}(X_{p}, \mathcal F_{p})$.

I hope this helps!