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 fun example of a representable functor

This post is about representable functors:

Definition. Let F \colon \mathscr C \to \Set be a functor. Then F is representable if it is isomorphic to \Hom(A,-) for some A \in \ob \mathscr C. In this case, we say that A represents F.

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

Really one should encode the isomorphism \Hom(A,-) \stackrel\sim\to F as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation \Hom(A,-) \to F is uniquely determined by the element of F(A) corresponding to the identity of A.

When \Hom(A,-) \to F is a natural isomorphism, the corresponding element a \in F(A) is called the universal object of F. It has the property that for every B \in \mathscr C and any b \in F(B), there exists a unique morphism f \colon A \to B such that f(a) = b.

Example. The forgetful functor \Ab \to \Set is represented by \Z. Indeed, the natural map

    \begin{align*} \Hom(\Z,M) &\to M\\ f &\mapsto f(1) \end{align*}

is an isomorphism. The universal element is 1 \in \Z.

Example. Similarly, the forgetful functor \Ring \to \Set is represented by \Z[x]. The universal element is x.

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 \Top\op \to \Set that associates to a topological space (X,\mathcal T_X) its topology \mathcal T_X is representable.

Proof. Consider the topological space Y = \{0,1\} with topology \{\varnothing, \{1\},\{0,1\}\}. Then there is a natural map

    \begin{align*} \Hom(X,Y) &\to \mathcal T_X\\ f &\mapsto f^{-1}(\{1\}). \end{align*}

Conversely, given an open set U, we can associate the characteristic function \mathbb I_U. This gives an inverse of the map above. \qedsymbol

The space Y we constructed is called the Sierpiński space. The universal open set is \{1\}.

Remark. The space Y^I represents the data of open sets U_i for i \in I: for any continuous map f \colon X \to Y^I, we have U_i = f^{-1}(Y_i), where Y_i = \pi_i^{-1}(\{1\}) \subseteq Y^I. If Z_i denotes the complementary open, then the U_i form a cover of X if and only if \bigcap_{i \in I} Z_i = \varnothing. This corresponds to the statement that f lands in Y^I\setminus\{(0,0,\ldots)\}.

Thus, the open cover Y^I\setminus\{0\} = \bigcup_{i \in I} Y_i is the universal open cover, i.e. for every open covering X = \bigcup U_i there exists a unique continuous map f \colon X \to Y^I\setminus\{0\} such that U_i = f^{-1}(Y_i).

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.

Cardinality of fraction field

This is a ridiculous lemma that I came up with.

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

Proof. If R is finite, my previous post shows that R = K. If R is infinite, then K is a subquotient of R^2, hence |K| \leq |R^2| = |R|. But R injects into K, so |R| \leq |K|. \qedsymbol

Corollary. If R is a domain, then |\operatorname{Frac}(R)| = |R|.

Proof. This is a special case of the lemma. \qedsymbol