# Does $H^i(X,F)\cong H^i(Y,f_*F)$ hold for $X\to Y$ finite but $F$ not necessarily quasi-coherent?

Let $X\to Y$ be a finite morphism between schemes,$F$ be a sheaf of abelian groups but not necesarily quasi-coherent. Does $H^i(X,F)\cong H^i(Y,f_*F)$ still hold for sheaf cohomology with Zariski topology? (It holds for etale cohomology)

• @Nefertiti How do we know the relative sheaf is zero? I can think the stalks as colimit of $H^k(f^{-1}(U),F)$, but I am not sure why the colimit are zero..Shall we use cohomology and base change to describe the stalk? But we need quasi-coherence of $F$ to apply $R^if_*F\otimes k(y)\cong H^i(X_y,F_y)$. – Qixiao Nov 16 '15 at 13:33
• Ah, I see your point. Let me think a bit more. – Nefertiti Nov 16 '15 at 13:53

Great question! Here is a counterexample:

Let $(A,\mathfrak m)$ be a DVR, and let $A \subseteq B$ be a finite extension of domains such that $B$ has exactly two primes above $\mathfrak m$. For example, $A = \mathbb Z_{(5)}$, and $B = \mathbb Z_{(5)}[i]$, with the primes $(1+2i), (1-2i)$ lying above $(5)$.

Then $Y = \operatorname{Spec} A$ is the space $\{x,\eta\}$ in which $x$ is closed and $\eta$ is open, and $X = \operatorname{Spec} B$ is the space $\{x, y,\eta\}$ in which $x$ and $y$ are closed, but $\eta$ is not. The map sends both $x$ and $y$ to $x$, and $\eta$ to $\eta$.

Lemma 1. Let $\mathcal F$ be a sheaf on $Y$. Then $H^i(Y, \mathcal F) = 0$ for all $i > 0$.

Proof. We will show that taking global sections is exact. Given a surjection $\mathcal F \to \mathcal G$ on $Y$, and a global section $s \in \mathcal G(Y)$, there exists a covering on which $s$ comes from $\mathcal F$. But every covering of $Y$ has to include a copy of $Y$ itself, so $s$ comes from $\mathcal F(Y)$. $\square$

Remark. The lemma actually generalises (with the exact same proof!) to arbitrary local rings. Indeed, the only open that contains the closed point of $Y$ is $Y$ itself.

It now suffices to construct a sheaf $\mathcal F$ on $X$ whose first cohomology is nontrivial. To do this, recall from Hartshorne Exercise II.1.19 the exact sequence $$0 \to j_! (\mathcal F|_U) \to \mathcal F \to i_* (\mathcal F|_Z) \to 0,$$ where $j \colon U \to X$ and $i \colon Z \to X$ are a complementary open and closed set in a topological space $X$.

Lemma 2. Let $U = \{\eta\}$, $Z = \{x,y\}$, and $\mathcal F = \mathbb Z$ (the constant sheaf). Then $\mathcal F(X) \to i_*(\mathcal F|_Z) (X)$ is not surjective. Consequently (since the constant sheaf $\mathbb Z$ on the irreducible space $X$ is flasque), $$H^1(X, j_!(\mathcal F|_U)) \neq 0.$$

Proof. Since $X$ is irreducible, it is connected, so $\mathcal F(X) = \mathbb Z$. On the other hand, $Z$ is a discrete space of two points, so $\underline{\operatorname{Sh}}(Z) \cong \underline{\operatorname{Ab}} \times \underline{\operatorname{Ab}}$, given by $$\mathcal G \mapsto (\mathcal G_x, \mathcal G_y).$$ We use that $\mathcal F|_Z = i^{-1} \mathcal F$ is the sheafification of the presheaf $$V \mapsto \operatorname*{colim}_{W \supseteq i(V)} \mathcal F(W).$$ Since sheafification doesn't alter stalks, and $\{x,\eta\}$ is the unique minimal open set containing $x$, we see that $$(\mathcal F|_Z)_x = \mathcal F|_Z(\{x\}) = \mathcal F(\{x,\eta\}) = \mathbb Z,$$ and similarly $(\mathcal F|_Z)_y = \mathbb Z$. Thus, we see that $$i_*(\mathcal F|_Z)(X) = \mathcal F|_Z(U) = \mathbb Z \oplus \mathbb Z.$$ Thus, the map $\mathcal F(X) \to (i_*\mathcal F|_Z)(X)$ can never be surjective, as $\mathbb Z$ does not surject onto $\mathbb Z \oplus \mathbb Z$. $\square$

Remark. The proof of the vanishing of higher direct images for étale cohomology for a finite morphism relies on the fact that strictly Henselian local rings (the local rings for the étale topology) have no higher cohomology, and that a finite covering of a strictly Henselian local ring is again a finite product of strictly Henselian local rings. However, as we saw above, finite coverings of local rings for the Zariski topology do have higher cohomology. Maybe we should see this as a hint that the Zariski topology does not have a good local theory in the same way that the étale topology does.

• Fantastic answer, Remy ! Lemma 1 is both unexpected and very amusing: I had never thought of such a spectacular result for the spectrum of a local ring. Did you invent it on the spot or did you know it beforehand? Anyway, you made my day! I have also upvoted a few of your other answers today, and brace yourself for more :-) – Georges Elencwajg Nov 5 '16 at 19:42
• @GeorgesElencwajg: I don't remember if I knew this before (the post is already a year old). I think I arrived at it from thinking about the cohomology of a DVR (the space is very very simple). The general case is perhaps more surprising. – Remy Nov 5 '16 at 23:30
• Maybe this type of philosophy should work on any reasonable site. For example, strict henselisations also don't have any higher étale cohomology. The difference is that these have stronger permanence properties: a finite extension of a strictly henselian local ring is a finite product of strictly henselian local rings. – Remy Nov 5 '16 at 23:35
• Thanks for your explanations, Remy. Tussen haakjes, bent U Nederlander of Belg? – Georges Elencwajg Nov 7 '16 at 12:59