Algebraic closure of the field of two elements

I think I learned about this from a comment on MathOverflow.

Recall that the field of two elements \mathbf F_2 is the ring \mathbf Z/2\mathbf Z of integers modulo 2. In other words, it consists of the elements 0 and 1 with addition 1+1 = 0 and the obvious multiplication. Clearly every nonzero element is invertible, so \mathbf F_2 is a field.

Lemma. The field \mathbf F_2 is algebraically closed.

Proof. We need to show that every non-constant polynomial f \in \mathbf F_2[x] has a root. Suppose f does not have a root, so that f(0) \neq 0 and f(1) \neq 0. Then f(0) = f(1) = 1, so f is the constant polynomial 1. This contradicts the assumption that f is non-constant. \qedsymbol

Internal Hom

This is an introductory post about some easy examples of internal Hom.

Definition. Let (\mathscr C, \otimes) be a symmetric monoidal category, i.e. a category \mathscr C with a functor \otimes \colon \mathscr C \times \mathscr C \to \mathscr C that is associative, unital, and commutative up to natural isomorphism. Then an internal Hom in \mathscr C is a functor

    \[\mathbf{Hom}(-,-) \colon \mathscr C\op \times \mathscr C \to \mathscr C\]

such that -\otimes Y is a left adjoint to \mathbf{Hom}(Y,-) for any Y \in \mathscr C, i.e. there are functorial isomorphisms

    \[\operatorname{Hom}(X \otimes Y, Z) \stackrel\sim\to \operatorname{Hom}(X, \mathbf{Hom}(Y,Z)).\]

Remark. In the easiest examples, we typically think of \mathbf{Hom}(Y,Z) as ‘upgrading \operatorname{Hom}(Y,Z) to an object of \mathscr C‘:

Example. Let R be a commutative ring, and let \mathscr C = \mathbf{Mod}_R be the category of R-modules, with \otimes the tensor product. Then \mathbf{Hom}(M,N) = \operatorname{Hom}_R(M,N) with its natural R-module structure is an internal Hom, by the usual tensor-Hom adjunction:

    \[\operatorname{Hom}_R(M \otimes_R N, K) \cong \operatorname{Hom}_R(M, \mathbf{Hom}(N, K)).\]

The same is true when \mathscr C =\!\ _R\mathbf{Mod}_R is the category of (R,R)-bimodules for a not necessarily commutative ring R.

However, we cannot do this for left R-modules over a noncommutative ring, because there is no natural R-module structure on \operatorname{Hom}_R(M,N) for left R-modules M and N. In general, the tensor product takes an (A,B)-bimodule M and a (B,C)-bimodule N and produces an (A,C)-bimodule M \otimes_B N. Taking A = C = \mathbf Z gives a way to tensor a right R-module with a left R-module, but there is no standard way to tensor two left R-modules, let alone equip it with the structure of a left R-module.

Example. Let \mathscr C = \mathbf{Set}. Then \mathbf{Hom}(X,Y) = \operatorname{Hom}(X,Y) = Y^X is naturally a set, making it into an internal Hom for (\mathscr C, \times):

    \[\operatorname{Hom}(X \times Y, Z) \stackrel\sim\to \operatorname{Hom}(X, \mathbf{Hom}(Y,Z)).\]

When \otimes is the categorical product \times, the internal \mathbf{Hom}(X,Y) (if it exists) is usually called an exponential object, in analogy with the case \mathscr C = \mathbf{Set} above.

Example. Another example of exponential objects is from topology. Let \mathscr C = \mathbf{Haus} be the category of locally compact Hausdorff topological spaces. Then the compact-open topology makes \mathbf{Hom}(X,Y) := Y^X into an internal Hom of topological spaces. (There are mild generalisations of this beyond the compact Hausdorff case, but for an arbitrary topological space X the functor - \times X does not preserve colimits and hence cannot admit a right adjoint.)

Example. An example of a slightly different nature is chain complexes: let R be a commutative ring, and let \mathscr C = \mathbf{Ch}(\mathbf{Mod}_R) be the category of cochain complexes

    \[\ldots \to C^{i-1} \to C^i \to C^{i+1} \to \ldots\]

of R-modules (meaning each C^i is an R-module, and the d^i \colon C^i \to C^{i+1} are R-linear maps satisfying d \circ d = 0). Homomorphisms f \colon C \to D are commutative diagrams

    \[\begin{array}{ccccccc}\ldots & \to & C^i & \to & C^{i+1} & \to & \ldots \\ & & \!\!\!\!\! f^i\downarrow & & \downarrow f^{i+1}\!\!\!\!\!\!\! & & \\ \ldots & \to & D^i & \to & D^{i+1} & \to & \ldots,\!\!\end{array}\]

and the tensor product is given by the direct sum totalisation of the double complex of componentwise tensor products.

There isn’t a natural way to ‘endow \operatorname{Hom}(C, D) with the structure of a chain complex’, but there is an internal Hom given by

    \[\mathbf{Hom}(C, D)^i = \prod_{m \in \mathbf Z} \operatorname{Hom}(C_m, D_{m+i}),\]

with differentials given by

    \[d^if = d_D f - (-1)^i f d_C.\]

Then we get for example

    \[\operatorname{Hom}(R[0], \mathbf{Hom}(C, D)) \cong \operatorname{Hom}(C, D),\]

since a morphism R[0] \to \mathbf{Hom}(C, D) is given by an element f \in \mathbf{Hom}(C, D)^0 such that df = 0, i.e. d_Df = f d_C, meaning that f is a morphism of cochain complexes.

Example. The final example for today is presheaves and sheaves. If X is a topological space, then the category \mathbf{Ab}(X) of abelian sheaves on X has an internal Hom given by

    \[\mathbf{Hom}(\mathscr F, \mathscr G)(U) = \operatorname{Hom}(\mathscr F|_U, \mathscr G|_U),\]

with the obvious transition maps for inclusions V \subseteq U of open sets. This is usually called the sheaf Hom. A similar statement holds for presheaves.

Local structure of finite unramified morphisms

It is well known that a finite étale morphism f \colon X \to Y of schemes is étale locally given by a disjoint union of isomorphisms, i.e. there exists an étale cover Y' \to Y such that the pullback X' \to Y' is given by X' = \coprod_{i=1}^n Y' \to Y'. Something similar is true for finite unramified morphisms:

Lemma. Let f \colon X \to Y be a finite unramified¹ morphism of schemes. Then there exists an étale cover Y' \to Y such that the pullback X' \to Y' is given by X' = \coprod_i Z_i \to Y', where Z_i \hookrightarrow Y' are closed immersions of finite presentation.

Proof. Let y \in Y be a point, let A = \mathcal O_{Y,y}^{\operatorname{sh}} be the strict henselisation of Y at y, and let \Spec B \to \Spec A be the base change of X \to Y along \Spec A \to Y. Then A \to B is unramified, so by Tag 04GL it splits as

    \[B = A_1 \times \ldots \times A_r \times C\]

whereA \to A_i is surjective for each i and no prime of C lies above \mathfrak m_y \subseteq A. But A \to C is also finite, so by Tag 00GU the map \Spec C \to \Spec A hits the maximal ideal if \Spec C \neq \varnothing. Thus, we conclude that C = 0, hence B is a product of quotients of A.

But A is the colimit of \mathcal O_{Y',y'} for (Y',y') \to (Y,y) an étale neighbourhood inducing a separable extension \kappa(y) \to \kappa(y'). Since f is of finite presentation, each of the ideals \ker(A \to A_i) and the projections B \to A_i are defined over some étale neighbourhood (Y',y') \to (Y,y). Then the pullback X' \to Y' is given by a finite disjoint union of closed immersions in Y'.

Then Y' \to Y might not be a covering, but since y \in Y was arbitrary we can do this for each point separately and take a disjoint union. \qedsymbol

Remark. The number of Z_i needed is locally bounded, but if Y is not quasi-compact it might be infinite. For example, we can take X \cong Y = \coprod_{i \in \N} \Spec k an infinite disjoint union of points, and f \colon X \to Y such that the fibre over y_i \in Y for i \in \N has i points.

Remark. In the étale case, we may actually take Y' \to Y finite étale, by taking Y' to be the Galois closure of X \to Y, which exists in reasonable cases². For example, if Y is normal, we may take Y' to be the integral closure of Y in the field extension corresponding to the Galois closure of k(Y) \to k(X). In general, if Y is connected it follows from Tag 0BN2 that a suitable component of the \deg(f)-fold fibre product of X over Y is a Galois closure Y' \to Y of X \to Y. If the connected components of Y are open, apply this construction to each component.

In the unramified case, this is too much to hope for. For example, if Y = \mathbf P^2_{\mathbf C}, then we may take X to be a nontrivial finite étale cover of an elliptic curve E \subseteq Y. This is finite and unramified, but does not split over any finite étale cover of \mathbf P^2 since there aren’t any. In fact, it cannot split over any connected étale cover Y' \to \mathbf P^2 whose image contains E, since that implies the image only misses finitely many points (as E is ample), which is again impossible since \pi_1(\mathbf P^2 \setminus \{p_1,\ldots,p_r\}) = 0.

¹For the purposes of this post, unramified means in the sense of Grothendieck, i.e. including the finite presentation hypothesis. In Raynaud’s work on henselisations, this was weakened to finite type. See Tag 00US for definitions.

²I’m not sure what happens in general.

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


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.

Rings are abelian

In this post, we prove the following well-known lemma:

Lemma. Let (R,+,\times,0,1) satisfy all axioms of a ring, except possibly the commutativity a + b = b + a. Then (R,+) is abelian.

That is, additive commutativity of a ring is implied by the other axioms.

Proof. By distributivity, we have 2(a+b) = 2a + 2b, so multiplication by 2 is a homomorphism. By our previous post, this implies R is abelian. \qedsymbol

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