Rings that are localisations of each other

This is a post about an answer I gave on MathOverflow in 2016. Most people who have ever clicked on my profile will probably have seen it.

Question. If A and B are rings that are localisations of each other, are they necessarily isomorphic?

In other words, does the category of rings whose morphisms are localisations form a partial order?

In my previous post, I explained why k[x] and k[x,x^{-1}] are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let k be a field, and let K = k(x_1, x_2, \ldots). Let

    \[A = K[x_0,x_{-1},\ldots]\]

be an infinite-dimensional polynomial ring over K, and let

    \[B = A\left[\frac{1}{x_0}\right].\]

Then B is a localisation of A, and we can localise B further to obtain the ring

    \[k(x_0,x_1,\ldots)[x_{-1},x_{-2},\ldots]\]

isomorphic to A by shifting all the indices by 1. To see that A and B are not isomorphic as rings, note that A^\times \cup \{0\} is closed under addition, and the same is not true in B. \qed


Is there a moral to this story? Not sure. Maybe the lesson is to do mathematics your own stupid way, because the weird arguments you come up with yourself may help you solve other problems in the future. The process is more important than the outcome.

Is the affine line isomorphic to the punctured affine line?

This is the story of Johan Commelin and myself working through the first sections of Hartshorne almost 10 years ago (nothing creates a bond like reading Hartshorne together…). This post is about problem I.1.1(b), which is essentially the following:

Exercise. Let k be a field. Show that k[x] and k[x,x^{-1}] are not isomorphic.

In my next post, I will explain why I’m coming back to exactly this problem. There are many ways to solve it, for example:

Solution 1. The k-algebra k[x] represents the forgetful functor \mathbf{Alg}_k \to \mathbf{Set}, whereas k[x,x^{-1}] represents the unit group functor R \mapsto R^\times. These functors are not isomorphic, for example because the inclusion k \to k[x] induces an isomorphism on unit groups, but not on additive groups. \qed

A less fancy way to say the same thing is that all k-algebra maps k[x,x^{-1}] \to k[x] factor through k, while the same evidently does not hold for k-algebra maps k[x] \to k[x].

However, we didn’t like this because it only shows that k[x] and k[x,x^{-1}] are not isomorphic as k-algebras (rather than as rings). Literal as we were (because we’re undergraduates? Lenstra’s influence?), we thought that this does not answer the question. After finishing all unstarred problems from section I.1 and a few days of being unhappy about this particular problem, we finally came up with:

Solution 2. The set k[x]^\times \cup \{0\} is closed under addition, whereas k[x,x^{-1}]^\times \cup \{0\} is not. \qed

This shows more generally that k[x] and \ell[x,x^{-1}] are never isomorphic as rings for any fields k and \ell.

P¹ is simply connected

This is a cute proof that I ran into of the simple connectedness of \mathbb P^1. It does not use Riemann–Hurwitz or differentials, and instead relies on a purely geometric argument.

Lemma. Let k be an algebraically closed field. Then \mathbb P^1_k is simply connected.

Proof. Let f \colon C \to \mathbb P^1 be a finite étale Galois cover with Galois group G. We have to show that f is an isomorphism. The diagonal \Delta_{\mathbb P^1} \subseteq \mathbb P^1 \times \mathbb P^1 is ample, so the same goes for the pullback D = (f \times f)^* \Delta_{\mathbb P^1} to C \times C [Hart, Exc. III.5.7(d)]. In particular, D is connected [Hart, Cor. III.7.9].

But D \cong C \times_{\mathbb P^1} C is isomorphic to |G| copies of C because the action

    \begin{align*} G \times C &\to C \times_{\mathbb P^1} C\\ (g,c) &\mapsto (gc,c) \end{align*}

is an isomorphism. If D is connected, this forces |G| = 1, so f is an isomorphism. \qed

The proof actually shows that if X is a smooth projective variety such that \Delta_X is a set-theoretic complete intersection of ample divisors, then X is simply connected.

Example. For a smooth projective curve C of genus g \geq 1, the diagonal cannot be ample, as \pi_1(C) \neq 0. We already knew this by computing the self-intersection \Delta_C^2 = 2-2g \leq 0, but the argument above is more elementary.

References.

[Hart] Hartshorne, Algebraic geometry. GTM 52, Springer, 1977.

Number theory is heavy metal

As some of you may be aware, I am a musician as well as a mathematician. I often like to compare my experiences between the two. For example, I found that my approach to the creative process is not dissimilar (in both, I work on the more technical side, with a particular interest in the larger structure), and I face the same problems of excessive perfectionism (never able to finish anything).

This post is about some observations about the material itself, as opposed to my interaction with it.

Universal donor.

A look at the arXiv listing for algebraic geometry shows the breadth of the subject. Ideas from algebraic geometry find their way into many different areas of mathematics; from representation theory and abstract algebra to combinatorics and from number theory to mathematical physics. But when I say algebraic geometry is a universal donor (of ideas, techniques, etc.), I also mean that its applications to other fields far outnumber the applications of other fields to algebraic geometry, and that the field of algebraic geometry is largely self-contained.

Much the same role is played by classical music inside musical composition. Common practise theory is used throughout Western music, whether you’re listening to hip hop, trance, blues, ambient electronic, bluegrass, or hard rock. Conversely, the influence of popular genres music on [contemporary] classical is comparatively little. One could therefore argue that classical music is a universal donor in the field of musical composition.

Universal receptor.

The opposite is the case for number theory. Another vast area, number theory often uses ingenious arguments combining ideas from algebra, combinatorics, analysis, geometry, and many other areas. In practical terms, the amount of material that a number theorist needs to master is immense: whatever solves your particular problem.

So is there any musical genre that plays a similar role? I claim that metal fits the bill. With a vast list of subgenres including thrash metal, black metal, doom metal, progressive metal, death metal, symphonic metal, nu metal, grindcore, hair metal, power metal, and deathcore, metal writing often contains creative combinations of other genres, from classical music to ambient electronic, and from free jazz to hip hop. As Steven Wilson of Porcupine Tree said about his rediscovery of metal:

I said to myself, this is where all the interesting musicians are working! Because for a long time I couldn’t find where all these creative musicians were going… You know in the 70’s they had a lot of creative musicians like Carlos Santana, Jimmy Page, Frank Zappa, Neil Young, I was thinking “where are all these people now?” and I found them, they were working in extreme metal.

I sometimes think of metal as a small microcosmos reflecting the full range of Western (and some non-Western) music, tied together by distorted guitars and fast, technical drumming.

In summary, algebraic geometry is classical music, and number theory is heavy metal.

Not every open immersion is an open immersion

An immersion (or locally closed immersion) of schemes is a morphism f \colon X \to Y that can be factored as X \to U \to Y, where X \to U is a closed immersion and U \to Y is an open immersion. If it is moreover an open morphism, it need not be an open immersion:

Example. Let X be a nonreduced scheme, and let X_{\text{red}} \to X be the reduction. This is a closed immersion, whose underlying set is the entire space. Thus, it is a homeomorphism, hence an open morphism. It is not an open immersion, for that would force it to be an isomorphism. \qedsymbol

Remark. However, every closed immersion is a closed immersion; see Tag 01IQ.

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.