Higher pushforwards along finite morphisms

This post is about one of my favourite answers I have given on MathOverflow, although it seems to have gone by mostly unnoticed. In the post, Qixiao asks (essentially) the following:

Question. If f \colon X \to Y is a finite morphism of schemes, is the pushforward f_* \colon \Sh(X) \to \Sh(Y) exact?

Note that this is true on the subcategory of quasicoherent sheaves because affine morphisms have no quasicoherent higher pushforwards. Also, in the étale topology the pushforward along a finite morphism is exact on the category of all abelian sheaves; see e.g. Tag 03QP.

However, we show that the answer to the question above is negative.

Example. Let Y be the spectrum of a DVR (R,\mathfrak m), let R \to S be a finite extension of domains such that S has exactly two primes \mathfrak p, \mathfrak q above \mathfrak m, and let X = \Spec S. For example, R = \Z_{(5)} and S = \Z_{(5)}[i], or R = k[x]_{(x)} and S = k[x]_{(x)}[\sqrt{x+1}] if you prefer a more geometric example.

By my previous post, the global sections functor \Gamma \colon \Sh(Y) \to \Ab is exact. If the same were true for f_* \colon \Sh(X) \to \Sh(Y), then the global sections functor on X would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection \mathscr F \to \mathscr G of sheaves on X such that the map on global sections is not surjective.

The topological space of X consists of closed points x,y and a generic point \eta. Let U = \{\eta\} and Z = U^{\operatorname{c}} = \{x,y\}; then U is open and Z is closed. Hence, for any sheaf \mathscr F on X, we have a short exact sequence (see e.g. Tag 02UT)

    \[0 \to j_! (\mathscr F|_U) \to \mathscr F \to i_* (\mathscr F|_Z) \to 0,\]

where j \colon U \to X and i \colon Z \to X are the inclusions. Let \mathscr F be the constant sheaf \Z; then the same goes for \mathscr F|_U and \mathscr F|_Z. Then the map

    \[H^0(X,\mathscr F) \to H^0(X,i_*(\mathscr F|_Z)) = H^0(Z,\mathscr F|_Z)\]

is given by the diagonal map \Z \to \Z \oplus \Z, since X is connected by Z has two connected components. This is visibly not surjective. \qedsymbol

Flat and projective

See the previous post for the notion of k-finitely presented modules.

Lemma. Let M be a 2-finitely presented flat module. Then M is projective.

Proof. For every prime \fr{p} \sbq R, the module M_{\fr{p}} is finitely presented and flat, hence free (use Nakayama). In particular, it is projective over R_{\fr{p}}, hence

    \[ \Ext{R_{\fr{p}}}{i}(M_{\fr{p}},-)=0 \]

for all i > 0. By our previous lemma, we conclude that

    \[ \Ext{R}{1}(M,N)_{\fr{p}} = \Ext{R_{\fr{p}}}{1}(M_{\fr{p}},N_{\fr{p}}) = 0 \]

for any R-module N, as M is 2-finitely presented. Since \fr{p} is arbitrary, this forces

    \[ \Ext{R}{1}(M,N) = 0 \]

for any R-module N. Hence M is projective. \qedsymbol

Remark. Using the equational criterion for flatness, one can in fact prove that any finitely presented flat module is projective. However, I thought the above proof was nice enough to make up for this slight loss of generality.

Remark. The Stacks project gives an example of a finitely generated (but not finitely presented) flat module that is not projective.

Ext and localisation

This post and the next are related, but I found this result interesting enough for a post of its own.

Lemma. Let M be a finitely presented R-module, and let S \sbq R be a multiplicative subset. Then

    \[ \Hom_R(M,N)[S^{-1}] = \Hom_{R[S^{-1}]}(M[S^{-1}],N[S^{-1}]). \]

Proof. The result is true when M is finite free, since

    \[ \Hom_R(R^n, N)[S^{-1}] = (N^n)[S^{-1}] = N[S^{-1}]^n, \]


    \[ \Hom_{R[S^{-1}]}(R[S^{-1}]^n, N[S^{-1}]) = N[S^{-1}]^n. \]

Now consider a finite presentation F_1 \ra F_0 \ra M \ra 0 of M. Since \Hom is left exact and localisation is exact, we get a commutative diagram

Rendered by QuickLaTeX.com

with exact rows (where the \Hom in the bottom row is over R[S^{-1}]). The right two vertical maps are isomorphisms, hence so is the one on the left. \qedsymbol

Definition. Let M be an R-module. Then M is k-finitely presented if there exists finite free modules F_0, \ldots, F_k and an exact sequence

    \[ F_k \ra F_{k-1} \ra \ldots \ra F_0 \ra M \ra 0. \]

For example, M is finitely generated if and only if it is 0-finitely presented, and finitely presented if and only if it is 1-finitely presented. Over a Noetherian ring, any finitely generated module is k-finitely presented for any k \in \Z_{\geq 0}.

[I do not know if this is standard terminology, but it should be.]

Corollary. Let k \geq 1, let M be a k-finitely presented module, and let S \sbq R be a multiplicative subset. Then

    \[ \Ext{R}{k-1}(M,N)[S^{-1}] = \Ext{R[S^{-1}]}{k-1}(M[S^{-1}],N[S^{-1}]). \]

Proof. Given an exact sequence F_k \ra \ldots \ra F_0 \ra M \ra 0 with F_i finite free, let M' be the kernel of F_0 \ra M. Then M' is (k-1)-finitely presented, and we have a short exact sequence

    \[ 0 \ra M' \ra F_0 \ra M \ra 0. \]

Now the result follows by induction, using the long exact sequence for \Ext{}{}. \qedsymbol

Remark. As Sebastian pointed out to me, we never used any specific properties of localisation, and the same result (with the same proof) works for any flat R-algebra.