Proper + Affine = Finite

Posted on 27 November 2014 by Remy

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

Posted on 9 November 2014 by Remy

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

Posted on 9 November 2014 by Remy

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.