Finiteness is not a local property

In this post, we consider the following question:

Question. Let be a Noetherian ring, and and -module. If is a finite -module for all primes , is 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 , and let . Then , because localisation commutes with direct sums and if is prime. Thus, is finitely generated for all primes . Finally, , because is torsion. But is obviously not finitely generated.

Example 2. Again, let , and let be the subgroup of fractions with such that is squarefree. This is a subgroup because can be written with denominator , and that number is squarefree if and are. Clearly is not finitely generated, because the denominators can be arbitrarily large. But , which is finitely generated over . If is a prime, then is the submodule , which is finitely generated over .

Another way to write is .

Remark. The second example shows that over a PID, the property that is free of rank can not be checked at the stalks. Of course it can be if is finitely generated, for then is finite projective [Tag 00NX] of rank , hence free since 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 is a finite morphism of schemes, is the pushforward 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 be the spectrum of a DVR , let be a finite extension of domains such that has exactly two primes above , and let . For example, and , or and if you prefer a more geometric example.

By my previous post, the global sections functor is exact. If the same were true for , then the global sections functor on would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection of sheaves on such that the map on global sections is not surjective.

The topological space of consists of closed points and a generic point . Let and ; then is open and is closed. Hence, for any sheaf on , we have a short exact sequence (see e.g. Tag 02UT)

where and are the inclusions. Let be the constant sheaf ; then the same goes for and . Then the map

is given by the diagonal map , since is connected by has two connected components. This is visibly not surjective.

Cohomology of a local scheme

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

Lemma. Let be a local scheme, and let be any abelian sheaf on . Then for all .

Proof. It suffices to show that the global sections functor is exact. Let be a surjection of abelian sheaves on , and let be a global section. Then can be lifted to a section of in an open neighbourhood of . But the only open neighbourhood of is . Thus, can be lifted to a section of .

What’s going on is that the functors and are naturally isomorphic, due to the absence of open neighbourhoods of .

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 is local if is contained in every nonempty closed subset of .

Example. If is a local ring, then is a local scheme. Indeed, is contained in every nonempty closed subset , because every strict ideal is contained in .

We prove that this is actually the only example.

Lemma. Let be a local scheme. Then is affine, and is a local ring whose maximal ideal corresponds to the point .

Proof. Let be an affine open neighbourhood of . Then the complement is a closed set not containing , hence . Thus, is affine. Let . Let be a maximal ideal of ; then . Since this contains , we must have , i.e. corresponds to the (necessarily unique) maximal ideal .

A fun example of a representable functor

This post is about representable functors:

Definition. Let be a functor. Then is representable if it is isomorphic to for some . In this case, we say that represents .

Exercise. If such exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation is uniquely determined by the element of corresponding to the identity of .

When is a natural isomorphism, the corresponding element is called the universal object of . It has the property that for every and any , there exists a unique morphism such that .

Example. The forgetful functor is represented by . Indeed, the natural map

is an isomorphism. The universal element is .

Example. Similarly, the forgetful functor is represented by . The universal element is .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor that associates to a topological space its topology is representable.

Proof. Consider the topological space with topology . Then there is a natural map

Conversely, given an open set , we can associate the characteristic function . This gives an inverse of the map above.

The space we constructed is called the Sierpiński space. The universal open set is .

Remark. The space represents the data of open sets for : for any continuous map , we have , where . If denotes the complementary open, then the form a cover of if and only if . This corresponds to the statement that lands in .

Thus, the open cover is the universal open cover, i.e. for every open covering there exists a unique continuous map such that .

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 that has infinitely many points whose residue field has characteristic 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 as a localisation of . Recall that prime ideals of come in four types:

• The generic point , of height 0;
• Height 1 primes for every prime ;
• Height 1 primes for every irreducible polynomial ;
• Height 2 closed points for a prime and a polynomial whose reduction is irreducible.

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

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

The primes of are with residue field ; the prime with residue field ; and the prime ideals with a polynomial whose constant coefficient is divisible by , whose residue field is a finite extension of .

Remark. The ring we constructed is essentially of finite type over (a localisation of a finite type -algebra). There are no examples of finite type over , because by Chevalley’s theorem the image of would be constructible. However, no set of the form for finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type -algebra has residue characteristic .)

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.

Cardinality of fraction field

This is a ridiculous lemma that I came up with.

Lemma. Let be a (commutative) ring, and let be its total ring of fractions. Then and have the same cardinality.

Proof. If is finite, my previous post shows that . If is infinite, then is a subquotient of , hence . But injects into , so .

Corollary. If is a domain, then .

Proof. This is a special case of the lemma.

Finite domains are fields

This is one of the classics.

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

Proof. If is not a zero-divisor, then the map is injective. Since is finite, it is also surjective, so there exists with .

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

Proof. The total ring of fractions is the ring , where is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change .

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

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

Étale and universally injective

This is the second in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism of schemes is étale if is flat and locally of finite presentation, and .

Lemma. Suppose is étale and universally injective. Then is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that is universally bijective; then we have to prove that is an isomorphism. Since étale morphisms are open, is in fact a universal homeomorphism. By Tag 04DE, is affine.

The question is local on , so we may assume is affine, and hence so is . Say is induced by . Now is proper and affine, hence finite. Moreover, since is finitely presented and finite as -algebra, and is a finitely presented -module, it is also a finitely presented -module (Tag 0564).

Now is flat of finite presentation over , hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume is free of rank .

Now let be a geometric point; that is, let be a map to an algebraically closed field. Then the tensor product is étale of dimension over . Hence, is a union of points. Since is universally bijective, we have . Then the result follows from my previous post.

Locally free algebras

This is the first in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains a preliminary lemma; the second post contains the result; and the third one contains the application that I was interested in.

Remark. If is a (locally) free -module of rank 1, then . Multiplication by is injective on if and only if is not a zero-divisor, and it is surjective if and only if . In particular, if it is surjective, it is also injective.

Lemma. Suppose is a ring homomorphism, such that is locally free of rank 1 over . Then is an isomorphism.

Proof. The question is local on , so (after replacing with a suitable localisation) we may assume that is free of rank 1. Let be a basis element.

Then we can write for some , hence . Also, we can write for some , hence . Therefore, is surjective, so by the remark above, it is an isomorphism.

Using a different argument, we can also prove:

Lemma. Suppose is a ring homomorphism, such that is locally free of rank over . Then is injective.

Proof. Since is locally free of rank , it is faithfully flat over . Thus it suffices to prove that is injective. But this map admits a contraction given by .