## Completions of rings and modules August 25, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
Tags: , , , ,

So, we saw in the previous post that completion can be defined generally for abelian groups. Now, to specialize to rings and modules.

Rings

The case in which we are primarily interested comes from a ring ${A}$ with a descending filtration (satisfying ${A_0 =A}$), which implies the ${A_i}$ are ideals; as we saw, the completion will also be a ring. Most often, there will be an ideal ${I}$ such that ${A_i = I^i}$, i.e. the filtration is ${I}$-adic. We have a completion functor from filtered rings to rings, sending ${A \rightarrow \hat{A}}$. Given a filtered ${A}$-module ${M}$, there is a completion ${\hat{M}}$, which is also a ${\hat{A}}$-module; this gives a functor from filtered ${A}$-modules to ${\hat{A}}$-modules.

We’ll also primarily restrict attention to the case when ${A}$ is Noetherian and ${M}$ finitely generated. The following is thus useful:

Theorem 1 If ${A}$ is Noetherian and ${I}$ an ideal, then the ${I}$-adic completion ${\hat{A}}$ is too.

As I mentioned above, the trick is to use the above result to get a filtration ${\hat{I}^i \subset \hat{A}}$, with (by the end of Sunday’s post)

$\displaystyle \mathrm{gr} \ \hat{A} = A/I \oplus I/I^2 \oplus \dots = \mathrm{gr} \ A.$

But ${\mathrm{gr} \ A}$ is an ${A}$-algebra, and if ${a_1, \dots, a_n \in I}$ generate ${I}$ as an ideal, then ${\mathrm{gr} \ A}$ is generated as an ${A}$-algebra by ${\overline{a_1}, \dots, \overline{a_n} \in I/I^2}$, so ${\mathrm{gr} \ A = \mathrm{gr} \ \hat{A}}$ is Noetherian.

Given ${x \in \hat{A}}$, we can consider the initial form ${i(x) \in \mathrm{gr} \ A}$ defined as follows: if ${x}$ is such that ${x \in \hat{I}^n, x \notin \hat{I}^{n+1}}$, then ${i(x)}$ is the image of ${x }$ in ${\hat{I}^n/\hat{I}^{n+1} = I^n/I^{n+1}}$ in the ${n}$-th part of the direct sum defining ${\mathrm{gr} \ A}$. Note that ${i(x)}$ is well-defined except when ${x=0}$, since ${\bigcap \hat{I}^n = 0}$ since a complete ring is Hausdorff. Now fix an ideal ${J \subset \hat{A}}$. Consider the ideal ${Q \subset \mathrm{gr} \ A}$ generated by ${i(J)}$; ${\mathrm{gr} \ A}$ being Noetherian, we can find a finite number of generators ${i(x_1), i(x_2), \dots, i(x_v)}$ for ${x_1, \dots, x_v \in J}$.

Set ${M := \max_k \deg i(x_k)}$. I now claim that ${J = (x_1, \dots, x_v)}$, which will prove the theorem.

This in turn will follow from

Claim 1

If ${y \in J \cap \hat{I}^n}$, we can find ${c_1, \dots, c_v \in \hat{I}^{n-M}}$ with

$\displaystyle y - \sum c_k x_k \in I^{n+1} \cap J.\ \ \ \ \ (1)$

Wlog, ${y \notin \hat{I}^{n+1}}$ so ${i(y)}$ is homogeneous of degree ${n}$. Now ${i(y) \in Q}$, which means we can find ${a_k \in \mathrm{gr} \ \hat{A}, 1 \leq k \leq v}$, with

$\displaystyle i(y) = \sum_k a_k i(x_k) \in \mathrm{gr} \ \hat{A}.$

By taking homogeneous parts, assume that the ${a_k}$ are homogeneous of degree ${n - \deg i(x_k)}$, so lift each ${a_k}$ to ${c_k \in \hat{I}^{n-M}}$. Then ${i(c_k) = a_k}$, so

$\displaystyle i(y) = \sum_k i( c_k x_k) \in \mathrm{gr} \ \hat{A},$

which implies (1).

Now, to prove ${J=(x_1, \dots, x_v)}$, proceed as follows. If ${y \in J}$, find ${d_k^{(1)} \in \hat{A}, 1 \leq k \leq v}$, with ${y - \sum d_k^{(1)} x_k \in J \cap I}$; then find ${d_k^{(2)} \in J \cap \hat{I}^{1 - M}}$ with

$\displaystyle y - \sum d_k^{(1)} x_k - \sum d_k^{(2)} x_k \in J \cap \hat{I}^2.$

Repeat this to get ${d_k^{(m)} \in \hat{I}^{m-M}}$, and note that ${\sum_m d_k^{(m)}}$ converges for each ${k}$. This proves ${y \in (x_1, \dots, x_v)}$.

The previous proof highlighted an important technique which we will use often: successive approximation. But first, here is a corollary.

Corollary 2 If ${A}$ is Noetherian, the power series ring ${A[[X_1, \dots, X_n]]}$ is Noetherian.

Indeed, Hilbert’s basis theorem implies the polynomial ring ${B := A[X_1, \dots, X_n]}$ is Noetherian; now, as is easily checked, the power series ring is the completion of ${B}$ with respect to the ideal ${(X_1, \dots, X_n)}$. So the previous result implies the theorem.

Incidentally, Zariski-Samuels prove this corollary by imitating the proof of Hilbert’s basis theorem itself; using completions is more general though. Even so, there is a useful structure theorem stating that in many cases, complete rings are actually of this form.

In view of the somewhat technical proof above, the question arises: Why does this all matter? Well, one of the most important examples is the ${p}$-adic integers ${\mathbb{Z}_p}$, defined as the completion of ${\mathbb{Z}}$ with respect to the ${p}$-adic (i.e. ${(p)}$-adic) topology, for ${p}$ prime. The ${p}$-adic integers and their quotient field, the ${p}$-adic numbers are important in number theory, as I hope to eventually discuss (when I get to class field theory). One of their important properties, shared generally by completions, is Hensel’s lemma: using a similar successive approximation technique, we can lift “approximate” solutions of equations to exact ones in a complete ring. Thus, it becomes much easier to check whether a polynomial has a root in a complete ring. And since “local-global” results such as the Hasse-Minkowski theorem imply that properties over ${\mathbb{Q}}$ can be studied by the completions, these become very useful indeed.

Modules

Let ${A}$ be a commutative ring and ${I}$ an ideal. If ${M}$ is an ${A}$-module, we can consider the ${I}$-adic completion ${\hat{M} := \varprojlim M/I^n M.}$ If ${A}$ is Noetherian, then ${I}$-adic completion is actually an exact functor on the category of finitely generated ${A}$-modules.

Theorem 3 If ${A}$ is a Noetherian ring and ${0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0}$ an exact sequence of f.g. ${A}$-modules, then ${0 \rightarrow \hat{M'} \rightarrow \hat{M} \rightarrow \hat{M''} \rightarrow 0}$ is exact.

First of all, I claim that ${\hat{M'} = \varprojlim M'/ M' \cap I^n M}$. This follows from the Artin-Rees lemma: the projective limit is the completion of ${M'}$ with respect to the topology from the filtration ${M' \cap I^n M}$. But Artin-Rees implies this is the same topology as that induced by the ${I}$-adic filtration ${I^n M'}$. Since completion only depends on the topology, the claim follows.

This is one reason I like to think in terms of topologies and not merely algebraically via inverse limits, though some books don’t seem to indicate this (in my opinion, cleaner) proof.

Now by the claim, for each ${n}$ we have an exact sequence

$\displaystyle 0 \rightarrow M'/M' \cap I^n M \rightarrow M/I^n M \rightarrow M''/I^n M'' \rightarrow 0.$

When we take the inverse limit of these sequences, we get the sequence of completions. The result follows from the two general lemmas:

Lemma 4

If ${0 \rightarrow A'_n \rightarrow A_n \rightarrow A''_n \rightarrow 0, \ n \in \mathbb{N}}$ is an inverse system of exact sequences, then

$\displaystyle 0 \rightarrow \varprojlim A'_n \rightarrow \varprojlim A_n \rightarrow \varprojlim A''_n$

is exact.

The proof is perhaps best thought of as a general phenomenon involving adjoint functors, but there is a quick (if less enlightening) direct proof too:

Injectivity on the left can be checked directly from the definition of ${\varprojlim}$ as a subset of some direct product. Similarly, if ${(a_n) \in \varprojlim A_n}$ has each ${a_n \rightarrow 0}$ in ${A''_n}$, then in fact each ${a_n \in A'_n}$ and satisfies appropriate compatibility conditions, so it follows that ${(a_n)}$ comes from some element in ${\varprojlim A'_n}$.

Right-exactness is not true generally, though in some special cases it is:

Lemma 5

If ${0 \rightarrow A'_n \rightarrow A_n \rightarrow A''_n \rightarrow 0, \ n \in \mathbb{M}}$ is an inverse system of exact sequences with the maps ${A'_n \rightarrow A'_m}$ for ${n \geq m}$ surjective, then

$\displaystyle 0 \rightarrow \varprojlim A'_n \rightarrow \varprojlim A_n \rightarrow \varprojlim A''_n \rightarrow 0$

is exact.

We need only to show that if we are given ${(a''_n) \in \varprojlim A''_n \subset \prod A''_n}$, we can lift this to some ${(a_n) \in \varprojlim A_n}$. For a preliminary approximation, choose for each ${n}$ some ${b_n \in A_n}$ with ${b_n \rightarrow a''_n}$; however ${b_n}$ needn’t satisfy the compatibility conditions for the inverse limit.

Define the ${a_n}$ from the ${b_n}$ inductively as follows. Set ${a_0 = b_0}$. Assume ${a_1, \dots,a_{n-1}}$ are defined, satisfy the compatibility conditions, and map to ${a''_1, \dots, a''_{n-1}}$. Then if ${f^n_{n-1}: A_n \rightarrow A_{n-1}}$ is the map defining the inverse system, we have ${f^n_{n-1}(b_n) - a_{n-1}}$ maps to zero in ${A''_{n-1}}$ (draw a commutative diagram). So ${f^n_{n-1}(b_n) - a_{n-1} \in A'_{n-1}}$. Now by surjectivity, choose ${a' \in A'_{n}}$ with

$\displaystyle f^n_{n-1}(b_n) + f^n_{n-1}(a') = a_{n-1} \ \ \ \ \ (2)$

and set ${a_n := b_n + a'}$. Since ${a' \in A'_{n}}$, ${a_n}$ still maps to ${a''_n}$ in ${A''_n}$, and (2) implies the compatibility condition holds. Now continue the induction.