Commutative algebra | Lovely little lemmas

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.

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.

Lovely little lemmas

Or maybe ‘amusing observations’

Category Archives: Commutative algebra

Flat and projective

Ext and localisation