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 f \colon X \ra Y of schemes is étale if f is flat and locally of finite presentation, and \Omega_{X/Y} = 0.

Lemma. Suppose f \colon X \ra S is étale and universally injective. Then f is an open immersion.

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

The question is local on S, so we may assume S is affine, and hence so is X. Say f is induced by g \colon A \ra B. Now f is proper and affine, hence finite. Moreover, since B is finitely presented and finite as A-algebra, and B is a finitely presented B-module, it is also a finitely presented A-module (Tag 0564).

Now B is flat of finite presentation over A, 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 B is free of rank r.

Now let \bar{s} \ra S be a geometric point; that is, let A \ra \bar{K} be a map to an algebraically closed field. Then the tensor product B \tens_A \bar{K} is étale of dimension r over \bar{K}. Hence, X_{\bar s} = X \times_{S} \bar{s} is a union of r points. Since f is universally bijective, we have r = 1. Then the result follows from my previous post. \qedsymbol

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 M is a (locally) free A-module of rank 1, then \End_A(M) = A. Multiplication by a \in A is injective on M if and only if a is not a zero-divisor, and it is surjective if and only if a \in A\x. In particular, if it is surjective, it is also injective.

Lemma. Suppose f \colon A \ra B is a ring homomorphism, such that B is locally free of rank 1 over A. Then f is an isomorphism.

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

Then we can write 1 = f(a)b for some a \in A, hence b \in B\x. Also, we can write b^2 = f(c)b for some c \in A, hence b = f(c). Therefore, f is surjective, so by the remark above, it is an isomorphism. \qedsymbol

Using a different argument, we can also prove:

Lemma. Suppose f \colon A \ra B is a ring homomorphism, such that B is locally free of rank r > 0 over A. Then f is injective.

Proof. Since B is locally free of rank r, it is faithfully flat over A. Thus it suffices to prove that f \tens 1 \colon B \ra B \tens_A B is injective. But this map admits a contraction B \tens_A B \ra B given by b_1 \tens b_2 \rm b_1b_2. \qedsymbol

Furstenberg’s proof of the infinitude of the primes

Another very classical proof that’s just too beautiful to ignore. This time, the theorem is millennia old, but it’s really the proof that I’m interested in.

Lemma. There are infinitely many primes.

Proof. Define the sets

    \[ U_{a,n} = a + n\Z = \{x \in \Z\ |\ x \equiv a \pmod{n}\}, \]

for a, n \in \Z with n \neq 0. Note that if U_{a,n} \cap U_{b,m} \neq \varnothing; say it contains some c \in \Z, then

    \[ U_{a,n} \cap U_{b,m} = U_{c, \lcm (n,m)}. \]

Hence, the intersection of two sets of this form is again of this form (or empty). Thus, they form the basis for a topology \scr{T} on \Z. Notice that the sets

    \[ U_{a,n}\comp = \bigcup_{b \not \equiv a} U_{b,n} \]

are also open.

Now suppose that there were finitely many primes. Then the intersection

    \[ V = \bigcap_{p \text{ prime}} U_{0,p}\comp \]

is a finite intersection of opens, hence open. On the other hand, it equals the set of integers divisible by no prime number, which is \{1, -1\}. But all basic open sets are infinite, so this set can never be open. \qedsymbol

Remark. This proof is in essence the usual proof, where we consider p_1 \cdots p_n + 1 and conclude that some prime has to divide it and this prime can be none of the p_i. Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is \lcm(n,m) (which for V gives period p_1 \cdots p_n). Since 1 is in V, so should p_1 \cdots p_n + 1 be.

That said, it doesn’t seem possible to adopt this proof to reprove other theorems like ‘there are infinitely many primes congruent to 3 modulo 4’. The problem is that the above proof seems to rely on a global analysis of the set V: it has to be an infinite set. So the method is too crude to prove theorems with congruence restrictions.

Cauchy’s theorem

This is a really classical result, but this proof is just so magical that I have to include it.

Lemma. Let G be a finite group, and let p be a prime dividing \# G. Then G has an element of order p.

Proof. Consider the action of \Z/p\Z = \langle \sigma \rangle on

    \[ X = \{(g_1, \ldots, g_p) \in G^p\ |\ g_1 \cdots g_p = 1\} \]

given by

    \[ \sigma \cdot (g_1, \ldots, g_p) = (g_2, \ldots, g_p, g_1). \]

Then \# (X^{\Z/p\Z}) \equiv \# X \pmod{p}, since the orbits of size > 1 all have size p (and the fixed points X^{\Z/p\Z} are exactly the union of the orbits of size 1). The right hand side is 0 \pmod{p} as p \mid \# G. Thus,

    \[ \# (X^{\Z/p\Z}) \equiv 0 \pmod{p}. \]

But there is a bijection

    \begin{align*} \{g \in G\ |\ g^p = 1\} &\rA X^{\Z/p\Z}\\ g &\rM (g, \ldots, g). \end{align*}

The former trivially contains the element 1, hence (since its size is divisible by p) it has to contain some other element g. This element will then have (exact) order p. \qedsymbol

Remark. Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).

Proper + Affine = Finite

Another argument I really like. We first present two auxiliary lemmata.

Lemma 1. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \Spec B \ra \Spec A is closed. Then f^{-1}(B\x) = A\x.

Proof. Since f is injective, the map \phi is dominant. Hence, it is a closed surjective map. For such a map, a closed set Z \sbq \Spec A is empty if and only if \phi^{-1}(Z) is empty. Applying this to Z = V(a) for a \in A, we get V(a) = \varnothing if and only if V(f(a)) = \varnothing, i.e. a \in A\x if and only if f(a) \in B\x. \qedsymbol

Lemma 2. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \A^1_B \ra \A^1_A (induced by A[x] \ra B[x]) is closed. Then f is integral.

Proof. Let b \in B. Consider the ideal I = (xb - 1) \sbq B[x], and let J = I \cap A[x]. Note that B[x]/I = B[\tfrac{1}{b}], and A[x]/J is the image C of the composite map A[x] \ra B[x] \ra B[\tfrac{1}{b}] (first isomorphism theorem). Write g for the inclusion C \ra B[\tfrac{1}{b}]. Note that the map

    \[ \psi \colon \Spec B\left[\tfrac{1}{b}\right] \ra \Spec C \]

induced by g is just the restriction of \phi to the closed subsets V(I) \ra V(J), hence it is a closed map. Since g is injective, Lemma 1 asserts that

    \[ g^{-1}\left(B\left[\tfrac{1}{b}\right]\x\right) = C\x. \]

But \tfrac{1}{b} is invertible in B[\tfrac{1}{b}] (its inverse is b), and it lies in C since it is the image of x under A[x] \ra B[\tfrac{1}{b}]. Hence, it is invertible in C, i.e. b \in C. That is, we can write

    \[ b = a_0 + a_1 \left(\frac{1}{b}\right) + \ldots + a_n \left(\frac{1}{b}\right)^n \]

for certain a_0, \ldots, a_n \in A. Hence, b^{n+1} - a_0 b^n - \ldots - a_n = 0 in B\left[\tfrac{1}{b}\right]. Hence, some multiple b^m \left(b^{n+1} - a_0 b^n - \ldots - a_n\right) is 0 in B, proving that b is integral over A. \qedsymbol

Corollary. Let f \colon A \ra B be a ring homomorphism such that the associated map \Spec B \ra \Spec A is proper. Then f is finite.

Proof. Let I be the kernel of f; then \Spec B \ra \Spec A lands in \Spec A/I. Then \Spec B \ra \Spec A/I is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map

    \[ \bar{f} \colon A/I \ra B \]

is injective, hence integral by Lemma 2. Since \Spec B \ra \Spec A/I is proper, it is of finite type. Hence, \bar{f} is integral and of finite type, hence finite. Hence so is f. \qedsymbol

A more geometric version is the following:

Theorem. Let \phi \colon X \ra Y be a morphism of schemes that is both affine and proper. Then \phi is finite.

Proof. Let U = \Spec A be an affine open in Y. Then V = \phi^{-1}(U) is affine (since \phi is an affine morphism); say V = \Spec B. Then the restriction \phi|_V \colon V \ra U is proper (properness is local on the target), hence B is finite over A by the theorem above. This proves that \phi is finite. \qedsymbol

Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into \P^n_Y for some n, i.e. it embeds into a trivial projective bundle).

I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let K be a field of infinite degree of imperfection, i.e. \char K = p and K is an infinite extension of K^p. Then Theorem 2 in [B-McL] says that for each n \in \N there exists a finite field extension L_n of K which cannot be generated by fewer than n elements¹. Correspondingly, \Spec L_n \ra \Spec K cannot be embedded into \P^m_K for m < n. Then consider Y an infinite disjoint union of \Spec K (labelled by \N), and X the disjoint union of \Spec L_n for all n. Then Y is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any \P^n_X.

References.
[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.

¹It is probably not very hard to actually come up with an example of such a field K with extensions L_n. I think that K = \F_p(x_1, x_2, \ldots) and L_n = K(\sqrt[\uproot{3}p]{x_1}, \ldots, \sqrt[\uproot{3}p]{x_n}) should do the trick.

Flat and projective

See the previous post for the notion of k-finitely presented modules.

Lemma. Let M be a 2-finitely presented flat module. Then M is projective.

Proof. For every prime \fr{p} \sbq R, the module M_{\fr{p}} is finitely presented and flat, hence free (use Nakayama). In particular, it is projective over R_{\fr{p}}, hence

    \[ \Ext{R_{\fr{p}}}{i}(M_{\fr{p}},-)=0 \]

for all i > 0. By our previous lemma, we conclude that

    \[ \Ext{R}{1}(M,N)_{\fr{p}} = \Ext{R_{\fr{p}}}{1}(M_{\fr{p}},N_{\fr{p}}) = 0 \]

for any R-module N, as M is 2-finitely presented. Since \fr{p} is arbitrary, this forces

    \[ \Ext{R}{1}(M,N) = 0 \]

for any R-module N. Hence M is projective. \qedsymbol

Remark. Using the equational criterion for flatness, one can in fact prove that any finitely presented flat module is projective. However, I thought the above proof was nice enough to make up for this slight loss of generality.

Remark. The Stacks project gives an example of a finitely generated (but not finitely presented) flat module that is not projective.

Ext and localisation

This post and the next are related, but I found this result interesting enough for a post of its own.

Lemma. Let M be a finitely presented R-module, and let S \sbq R be a multiplicative subset. Then

    \[ \Hom_R(M,N)[S^{-1}] = \Hom_{R[S^{-1}]}(M[S^{-1}],N[S^{-1}]). \]

Proof. The result is true when M is finite free, since

    \[ \Hom_R(R^n, N)[S^{-1}] = (N^n)[S^{-1}] = N[S^{-1}]^n, \]

whereas

    \[ \Hom_{R[S^{-1}]}(R[S^{-1}]^n, N[S^{-1}]) = N[S^{-1}]^n. \]

Now consider a finite presentation F_1 \ra F_0 \ra M \ra 0 of M. Since \Hom is left exact and localisation is exact, we get a commutative diagram

Rendered by QuickLaTeX.com

with exact rows (where the \Hom in the bottom row is over R[S^{-1}]). The right two vertical maps are isomorphisms, hence so is the one on the left. \qedsymbol

Definition. Let M be an R-module. Then M is k-finitely presented if there exists finite free modules F_0, \ldots, F_k and an exact sequence

    \[ F_k \ra F_{k-1} \ra \ldots \ra F_0 \ra M \ra 0. \]

For example, M is finitely generated if and only if it is 0-finitely presented, and finitely presented if and only if it is 1-finitely presented. Over a Noetherian ring, any finitely generated module is k-finitely presented for any k \in \Z_{\geq 0}.

[I do not know if this is standard terminology, but it should be.]

Corollary. Let k \geq 1, let M be a k-finitely presented module, and let S \sbq R be a multiplicative subset. Then

    \[ \Ext{R}{k-1}(M,N)[S^{-1}] = \Ext{R[S^{-1}]}{k-1}(M[S^{-1}],N[S^{-1}]). \]

Proof. Given an exact sequence F_k \ra \ldots \ra F_0 \ra M \ra 0 with F_i finite free, let M' be the kernel of F_0 \ra M. Then M' is (k-1)-finitely presented, and we have a short exact sequence

    \[ 0 \ra M' \ra F_0 \ra M \ra 0. \]

Now the result follows by induction, using the long exact sequence for \Ext{}{}. \qedsymbol

Remark. As Sebastian pointed out to me, we never used any specific properties of localisation, and the same result (with the same proof) works for any flat R-algebra.

Separation properties for topological groups

Although this is quite a classical result, I really like it.

Lemma. Let G be a topological group. Then G is T_1 if and only if G is Hausdorff.

Proof. One implication is clear. Conversely, suppose G is T_1. Then the identity element is closed. The map

    \begin{align*} G\times G &\rA G\\ (g,h) &\rM gh^{-1} \end{align*}

is continuous. Hence, the inverse image of the identity is closed. But this is the diagonal, hence G is Hausdorff. \qedsymbol

Exercise. Prove that Hausdorff is in fact equivalent to T_0.