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 1Let be a Noetherian ring, and an ideal. If we take all completions with respect to the -adic topology,

for any f.g. -module .

We know this is true for free modules since 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 2Let be right-exact (resp. left exact) functors between abelian categories . Suppose is a full subcategory and there is a natural isomorphism . Finally, suppose every element admits a presentation with . Then for all , we have .

We only handle the right-exact case. The left-exact case follows by dualization (or considering the opposite category). To prove this, pick , 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 as follows: there is a map from composition, which when pulled back to is just which is zero. Hence the map factors uniquely through the cokernel, . Similarly, since there is a commutative diagram

we get a map making this diagram commutative; it follows from a similar uniqueness argument that this map is the inverse to the map described previously.

Here are two well-known examples of this trick:

Example 1Let be a commutative ring, a multiplicative set, and an -module. Then .

As is well-known, the tensor product is right exact and localization is even exact. Take to be the category of -modules, the category of -modules. Let , . Finally, as is common, let be the full subcategory coming from free -modules. Then it follows that there is a functorial isomorphism

and since any module admits a free presentation (proof: if is an -module, choose an epimorphism for free, let be the kernel, and choose an epimorphism ; this induces an exact sequence ), the trick implies the assertion.

The example implies that tensoring by is an *exact functor*, so is flat.

Example 2Let be a commutative ring, a multiplicative set, and -modules with finitely presented. Let be a flat -algebra. Then .

Using the notation , we can write this more elegantly as

Fix ; the result is true for a finitely generated free module, since it is true for . The functors , are left exact because is and because tensoring by is left exact. (Note that the composition of left-exact functors is left-exact.) Let be the category of finitely presented -modules, the category of -modules, and the full subcategory of finitely generated free -modules. Now the trick implies the result.

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

Here are two quick corollaries:

Corollary 3Hypotheses as in Theorem 1, is flat over .

Indeed, completion is an exact functor.

Corollary 4Hypotheses as in Theorem 1, if is finitely generated as an -module, is finitely generated as an -module.

There is a surjection , which leads to a surjection .

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]*

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