Finiteness is not a local property

In this post, we consider the following question:

Question. Let A be a Noetherian ring, and M and A-module. If M_\mathfrak p is a finite A_\mathfrak p-module for all primes \mathfrak p \subseteq A, is M finite?

That is, is finiteness a local property?

For the statement where local means the property is true on a cover by Zariski opens, see Tag 01XZ. Some properties (e.g. flatness) can also be checked at the level of local rings; however, we show that this is not true for finiteness.

Example 1. Let A = \mathbb Z, and let M = \bigoplus_{p \text{ prime}} \mathbb Z/p\mathbb Z. Then M_{(p)} = \mathbb Z/p\mathbb Z, because localisation commutes with direct sums and (\mathbb Z/q\mathbb Z)_{(p)} = 0 if q \neq p is prime. Thus, M_{(p)} is finitely generated for all primes p. Finally, M_{(0)} = 0, because M is torsion. But M is obviously not finitely generated.

Example 2. Again, let A = \mathbb Z, and let M \subseteq \mathbb Q be the subgroup of fractions \frac{a}{b} with \gcd(a,b) = 1 such that b is squarefree. This is a subgroup because \frac{a}{b} + \frac{c}{d} can be written with denominator \lcm(b,d), and that number is squarefree if b and d are. Clearly M is not finitely generated, because the denominators can be arbitrarily large. But M_{(0)} = \mathbb Q, which is finitely generated over \mathbb Q. If p is a prime, then M_{(p)} \subseteq \mathbb Z_{(p)} is the submodule \frac{1}{p}\mathbb Z_{(p)}, which is finitely generated over \mathbb Z_{(p)}.

Another way to write M is \sum_{p \text{ prime}} \frac{1}{p}\mathbb Z \subseteq \mathbb Q.

Remark. The second example shows that over a PID, the property that M is free of rank 1 can not be checked at the stalks. Of course it can be if M is finitely generated, for then M is finite projective [Tag 00NX] of rank 1, hence free since A is a PID.

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

Cohomology of a local scheme

This is a continuation of my previous post on local schemes. Here is a ridiculous lemma.

Lemma. Let (X,x) be a local scheme, and let \mathscr F be any abelian sheaf on X. Then H^i(X,\mathscr F) = 0 for all i > 0.

Proof. It suffices to show that the global sections functor \Gamma \colon \Sh(X) \to \Ab is exact. Let \mathscr F \to \mathscr G be a surjection of abelian sheaves on X, and let s \in \mathscr G(X) be a global section. Then s can be lifted to a section of \mathscr F in an open neighbourhood U of x. But the only open neighbourhood of x is X. Thus, s can be lifted to a section of \mathscr F(X). \qedsymbol

What’s going on is that the functors \mathscr F \mapsto \Gamma(X,\mathscr F) and \mathscr F \mapsto \mathscr F_x are naturally isomorphic, due to the absence of open neighbourhoods of x.

Remark. It seems believable that there are suitable site-theoretic versions of this lemma as well. For example, a strictly Henselian local ring has no higher cohomology in the étale topology. The argument is essentially the same: every open neighbourhood of the closed point has a section; see e.g. the proof of Tag 03QO.

Local schemes

Consider the following definition. It seems to be standard, although I have not found a place where it is actually spelled out in this way.

Definition. A pointed scheme (X,x) is local if x is contained in every nonempty closed subset of X.

Example. If (A,\mathfrak m) is a local ring, then (\Spec A,\mathfrak m) is a local scheme. Indeed, \mathfrak m is contained in every nonempty closed subset V(I) \subseteq X, because every strict ideal I \subsetneq A is contained in \mathfrak m.

We prove that this is actually the only example.

Lemma. Let (X,x) be a local scheme. Then X is affine, and A = \Gamma(X,\mathcal O_X) is a local ring whose maximal ideal corresponds to the point x \in X = \Spec A.

Proof. Let U be an affine open neighbourhood of x. Then the complement V is a closed set not containing x, hence V = \varnothing. Thus, X = U is affine. Let A = \Gamma(X,\mathcal O_X). Let \mathfrak m be a maximal ideal of A; then V(\mathfrak m) = \{\mathfrak m\}. Since this contains x, we must have x = \mathfrak m, i.e. x corresponds to the (necessarily unique) maximal ideal \mathfrak m \subseteq A. \qedsymbol

A weird Dedekind domain

Here is a random question that I was wondering about at some point (just out of curiosity):

Question. Does there exist a Dedekind domain R that has infinitely many points whose residue field has characteristic 0 and finitely many points (at least one) whose residue field has positive characteristic?

I don’t fully recall why this question came up, but it had something to do with a similar property that was satisfied by an object involved in the definition of the Fargues–Fontaine curve. However, we don’t need such deep theory to discuss this elementary commutative algebra question.

Lemma. Such a Dedekind domain exists.

Proof. We will construct R as a localisation of \mathbb Z[x]. Recall that prime ideals of \mathbb Z[x] come in four types:

  • The generic point (0), of height 0;
  • Height 1 primes (p) for every prime p \in \mathbb Z;
  • Height 1 primes (f) for every irreducible polynomial f \in \mathbb Z[x];
  • Height 2 closed points (p,f) for p \in \mathbb Z a prime and f \in \mathbb Z[x] a polynomial whose reduction \bar f \in \mathbb F_p[x] is irreducible.

We first localise at (p,x); then the only primes we have left are the ones contained in (p,x). This is the generic point, the height 1 prime (p), the height 1 primes (f) where f \in \mathbb Z[x] is an irreducible polynomial whose constant coefficient is divisible by p (e.g. (x), (x-p), \ldots), and exactly one height 2 prime (p,x).

Next, we invert x; denoting the resulting ring by R. This gets rid of all prime ideals containing x, which are (p,x) and (x). In particular, there are no more height 2 primes, so R is 1-dimensional. It is a normal Noetherian domain because it a localisation of a normal Noetherian domain. Therefore, R is a Dedekind domain.

The primes of R are (0) with residue field \mathbb Q(x); the prime (p) with residue field \mathbb F_p(x); and the prime ideals (f) with f \neq x a polynomial whose constant coefficient is divisible by p, whose residue field is a finite extension of \mathbb Q. \qedsymbol

Remark. The ring R we constructed is essentially of finite type over \mathbb Z (a localisation of a finite type \mathbb Z-algebra). There are no examples of finite type over \mathbb Z, because by Chevalley’s theorem the image of \operatorname{Spec} R \to \operatorname{Spec} \mathbb Z would be constructible. However, no set of the form \{(0)\} \cup S for S \subseteq \operatorname{Spec} \mathbb Z finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type \mathbb Z-algebra has residue characteristic p > 0.)

I was a little surprised that we can make examples when we drop the finite type assumption. I don’t know if this type of ring has ever been used for anything.

Finite domains are fields

This is one of the classics.

Lemma. Let R be a finite commutative ring. Then every element is either a unit or a zero-divisor.

Proof. If x \in R is not a zero-divisor, then the map x \colon R \to R is injective. Since R is finite, it is also surjective, so there exists y \in R with xy = 1. \qedsymbol

Corollary 1. Let R be a finite commutative ring. Then R is its own total ring of fractions.

Proof. The total ring of fractions is the ring R[S^{-1}], where S is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change R. \qedsymbol

Corollary 2. Let R be a finite domain. Then R is a field.

Proof. In this case, the total ring of fractions is the fraction field. Therefore, R is its own fraction field by Corollary 1. \qedsymbol

Sites without a terminal object

Let \mathcal C be a site with a terminal object X. Then the cohomology on the site is defined as the derived functors of the global sections functor \Gamma(X,-). But what do we do if the site does not have a terminal object?

The solution is to define H^i(\mathcal C,-) as \Ext{\mathcal O}{i}(\mathcal O,-), where \mathcal O denotes the structure sheaf if \mathcal C is a ringed site. If \mathcal C is not equipped with a ring structure, we take \mathcal O to be the constant sheaf \underline{\mathbb Z}; this makes \mathcal C into a ringed site.

Lemma. Let \mathcal C be a site with a terminal object X. Then the above definitions agree, i.e.

    \[H^i(X,-) = \Ext{\mathcal O}{i}(\mathcal O,-).\]

Proof. Note that \Hom_{\mathcal O}(\mathcal O, \mathscr F) = \Gamma(X, \mathscr F), since any map \mathcal O(X) \to \mathscr F(X) can be uniquely extended to a morphism of (pre)sheaves \mathcal O \to \mathscr F, and conversely every such morphism is determined by its map on global sections. The result now follows since \Ext{\mathcal O}{i}(\mathcal O, -) and H^i(X,-) are defined as the derived functors of \Hom_{\mathcal O}(\mathcal O,-) and \Gamma(X,-) respectively. \qedsymbol

Remark. From this perspective, it seems quite magical that for a sheaf \mathscr F of \mathcal O_X-modules on a ringed space (X,\mathcal O_X), the cohomology groups \Ext{\mathcal O_X}{i}(\mathcal O_X,\mathscr F) and \Ext{\underline{\Z}}{i}(\underline{\Z},\mathscr F) agree. It turns out that this is true in the setting of ringed sites as well; see Tag 03FD.

So why is this useful? Let’s give some examples of sites that do not have a terminal object.

Example. Let G be a group scheme over k. Then we have a stack BG of G-torsors. The objects of BG are pairs (U,P), where U is a k-scheme and P is a G-torsor over U. Morphisms (U,P) \to (U',P') are pairs (f,g) \colon (U,P) \to (U',P') making the diagram

Rendered by

commutative. This forces the diagram to be a pullback, since all maps between G-torsors are isomorphisms.

The (large) Zariski site on BG is defined by declaring coverings \{(U_i, P_i) \to (U,P)\} to be families such that \{U_i \to U\} is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category BG have a terminal object? This would be a G-torsor P_0 \to U_0 such that every other G-torsor P \to U admits a unique map to it, realising P as the pullback of P_0 along U \to U_0. But this object would exactly be the classifying stack U_0 = BG, which does not exist as a scheme (or algebraic space). The fact that a terminal object does not exist is the whole reason we need to define it as a stack in the first place!

Example. Let X/k be a variety in characteristic p > 0; for simplicity, let’s say k = \mathbb F_p. Then consider the crystalline site of X/\Spec(\Z/p^n\Z). Roughly speaking, its objects are triples (U,T,\delta), where U \to X is an open immersion, U \to T is a thickening with a map to \Spec{\F_p} \to \Spec{\Z/p^n\Z}, and \delta is a divided power structure on the ideal sheaf \mathcal I_U \subseteq \mathcal O_T (with a compatibility condition w.r.t. \Spec{\F_p} \to \Spec{\Z/p^n\Z}). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on U = X with the respective compatibilities. (I am admittedly no expert, and it could very well be true that this is not 100% correct. However, I am certain that the crystalline site in general does not have a terminal object.)