The previous post got somewhat detailed and long, so today’s will be somewhat lighter. I’ll use completions to illustrate a well-known categorical trick using finite presentations.

The finite presentation trick

Our goal here is:

Theorem 1  Let ${A}$ be a Noetherian ring, and ${I}$ an ideal. If we take all completions with respect to the ${I}$-adic topology,

$\displaystyle \hat{M} = \hat{A} \otimes_A M$

for any f.g. ${A}$-module ${M}$.

We know this is true for free modules since ${\hat{A} \otimes_A A = \hat{A}}$ and completion commutes with direct sums (since inverse limits do, or more generally any additive functor). With many such theorems that ask to prove the isomorphism of two quantities, we can use the following general trick:

Proposition 2

Let ${F, G: \mathcal{A}, \mathcal{A} \rightarrow \mathcal{B}}$ be right-exact (resp. left exact) functors between abelian categories ${\mathcal{A}, \mathcal{B}}$. Suppose ${\mathcal{C} \subset \mathcal{A}}$ is a full subcategory and there is a natural isomorphism ${F |_{\mathcal{C}} \simeq G |_{\mathcal{C}} }$. Finally, suppose every element ${A \in \mathcal{A}}$ admits a presentation  $C' \to C \to A \to 0$ with $C, C' \in \mathcal{C}$.  Then for all $A \in A$, we have $F(A) \simeq G(A)$.

We only handle the right-exact case. The left-exact case follows by dualization (or considering the opposite category). To prove this, pick ${A \in \mathcal{A}}$, choose a presentation as in (1), and note that by right-exactness we have a commutative diagram:

The universal property of the cokernel induces a map ${F(A) \rightarrow G(A)}$ as follows: there is a map ${F(C) \rightarrow G(A)}$ from composition, which when pulled back to ${F(C')}$ is just ${F(C') \rightarrow G(C') \rightarrow G(C) \rightarrow G(A)}$ which is zero. Hence the map ${F(C) \rightarrow G(A)}$ factors uniquely through the cokernel, ${F(A) \rightarrow G(A)}$. Similarly, since there is a commutative diagram

we get a map ${G(A) \rightarrow F(A)}$ making this diagram commutative; it follows from a similar uniqueness argument that this map is the inverse to the map ${F(A) \rightarrow G(A)}$ described previously.

Here are two well-known examples of this trick:

Example 1 Let ${A}$ be a commutative ring, ${S}$ a multiplicative set, and ${M}$ an ${A}$-module. Then ${S^{-1} M = S^{-1} A \otimes_A M}$

As is well-known, the tensor product is right exact and localization is even exact. Take ${\mathcal{A}}$ to be the category of ${A}$-modules, ${\mathcal{B}}$ the category of ${S^{-1}A}$-modules. Let ${F(X) := S^{-1}X}$, ${G(X) = S^{-1}A \otimes_A X}$. Finally, as is common, let ${\mathcal{B}}$ be the full subcategory coming from free ${A}$-modules. Then it follows that there is a functorial isomorphism

$\displaystyle F(Y) \simeq G(Y), \quad Y \in \mathcal{B},$

and since any module admits a free presentation (proof: if ${X}$ is an ${A}$-module, choose an epimorphism ${F \rightarrow X \rightarrow 0}$ for ${F}$ free, let ${K}$ be the kernel, and choose an epimorphism ${F' \rightarrow K \rightarrow 0}$; this induces an exact sequence ${F' \rightarrow F \rightarrow X \rightarrow 0}$), the trick implies the assertion.

The example implies that tensoring by ${S^{-1}A}$ is an exact functor, so ${S^{-1}A}$ is flat

Example 2 Let ${A}$ be a commutative ring, ${S}$ a multiplicative set, and ${M,N}$ ${A}$-modules with ${M}$ finitely presented. Let ${B}$ be a flat ${A}$-algebra. Then ${B \otimes_A \mathrm{Hom} _A(M,N) = \mathrm{Hom} _B( B \otimes_A M, B \otimes_A N)}$.

Using the notation ${X_B := B \otimes_A X}$, we can write this more elegantly as

$\displaystyle \mathrm{Hom} (M,N)_B = \mathrm{Hom} (M_B, N_B).$

Fix ${N}$; the result is true for ${M}$ a finitely generated free module, since it is true for ${M=A}$. The functors ${F: M \rightarrow \mathrm{Hom} (M,N)_B}$, ${G: M \rightarrow \mathrm{Hom} (M_B, N_B)}$ are left exact because ${\mathrm{Hom} }$ is and because tensoring by ${B}$ is left exact. (Note that the composition of left-exact functors is left-exact.) Let ${\mathcal{A}}$ be the category of finitely presented ${A}$-modules, ${\mathcal{B}}$ the category of ${B}$-modules, and ${\mathcal{C} \subset \mathcal{A}}$ the full subcategory of finitely generated free ${A}$-modules. Now the trick implies the result.

Now, finally, let’s do the proof of Theorem 1. Take ${\mathcal{A}}$ to be the category of finitely presented ${A}$-modules, ${\mathcal{B}}$ the category of ${\hat{A}}$-modules, and ${\mathcal{C}}$ the full subcategory of finitely generated free ${A}$-modules. Let ${F}$ be completion and ${G}$ be tensoring with ${\hat{A}}$. We know from the previous post that ${F}$ is right-exact (even exact) while so is ${G}$ by the general tensor product properties. As I remarked, there is a functorial isomorphism ${F|_{\mathcal{C}} \simeq G|_{\mathcal{C}}}$. The trick now implies the theorem.

Here are two quick corollaries:

Corollary 3 Hypotheses as in Theorem 1, ${\hat{A}}$ is flat over ${A}$.

Indeed, completion is an exact functor.

Corollary 4 Hypotheses as in Theorem 1, if ${M}$ is finitely generated as an ${A}$-module, ${\hat{M}}$ is finitely generated as an ${\hat{A}}$-module.

There is a surjection ${A^r \rightarrow M \rightarrow 0}$, which leads to a surjection ${\hat{A}^r \rightarrow \hat{M} \rightarrow 0}$.

For now we’re almost done with discussing completions; the main part we haven’t covered is Hensel’s lemma, which will be my next post. Then I plan to talk about some other areas of mathematics (besides algebra).

[Edit- I’ve had to make numerous fixes to this post, due to my messing up the spacing. AM]