Continuing from the previous article, *A* denotes a noetherian ring and all *A*-modules are finitely generated. As before all completions are taken to be -stable for a fixed ideal .

# Noetherianness

We wish to prove that the -adic completion of a noetherian ring is noetherian. First we express:

Lemma 1.If , then

**Proof**

Let , still noetherian, with ideal . We have a ring isomorphism taking . Let be an ideal of *B*; we will take the -adic completion on both sides of *f*, treated as *B*-modules.

- On the LHS, and is generated by by proposition 5 here.
- On the RHS, we get the -adic completion on
*A*; but , so we get the -adic completion.

This completes our proof. ♦

Now all it remains is to prove this.

Proposition 1.If A is a noetherian ring, so is .

**Proof**

For a formal power series where with , we say the *lowest coefficient* of *f* is and its *lowest term* is . We also write .

Now suppose is a non-zero ideal.

**Step 1: find a finite set of generators of 𝔟.**

Now for each *n* = 0, 1, …, let be the set of all for which or occurs as a lowest term of some . We get an ascending chain of ideals . Since *A* is noetherian, for some *n* we have .

For each of , pick a finite generating set of comprising of non-zero elements; for each pick whose lowest term is . This gives a finite subset of degree-*i* power series whose lowest coefficients generate . Note that if then .

Let .

**Step 2: prove that T generates 𝔟.**

Now suppose has lowest term so . By our choice of *T* we can find such that

,

for some , . Since *T* is finite, in fact we can assume , setting for unneeded . Repeating the process with the RHS polynomial, we obtain

.

Repeating this inductively, we obtain formal power series such that . ♦

Hence, by proposition 1 and lemma 1 we have:

Corollary 1.The -adic completion of A is noetherian.

# Completion of Local Rings

Next suppose is a noetherian local ring.

Definition.The

completionof A is its -adic completion.

Since is a field, is a maximal ideal of . Also we have:

Lemma 2.is a local ring.

**Proof**

By lemma 2 here, is contained in the Jacobson radical of , so it is contained in all maximal ideals of . But is already maximal. ♦

Hence, we see that the noetherian local ring inherits many of the properties of . E.g. they have the same Hilbert polynomial

since as *k*-vector spaces.

# Hensel’s Lemma

Here is another key aspect of complete local rings, which distinguishes them from normal local rings.

Proposition (Hensel’s Lemma).Suppose is a complete local ring. Let be a polynomial. If there exists such that and , then there is a unique such that

.

**Note**

Hence most of the time, if we can find a root for in the residue field , then *lifts* to a root .

There are more refined versions of Hensel’s lemma to consider the case where . One can even generalize it to multivariate polynomials. For our purpose, we will only consider the simplest case.

**Proof**

Fix such that .

Set . It suffices to show: we can find such that

,

so that gives us the desired element. We construct this sequence recursively; suppose we have (). Write

.

Since we have so . Hence setting with gives

which gives us the desired sequence. ♦

# Applications of Hensel’s Lemma

Now we can justify some of our earlier claims.

**Examples**

1. In the complete local ring with maximal ideal , consider the equation . Modulo , we obtain which has two roots: +1 and -1. Since , by Hensel’s lemma there is a unique with constant term 1 such that .

2. Similarly, consider the ring of *p*-adic integers with *p* > 2. Let for outside . If *a* mod *p* has a square root *b*, then there is a Hensel lift of *b* to a square root of *a* in .

3. Next, we will prove an earlier claim that the canonical map

is an isomorphism. Consider the polynomial as a polynomial in *X* with coefficients in . Modulo , we have which has roots -1, 0, +1. Hence

where

with respectively. But and are invertible in so

4. In , consider the equation . Modulo *p*, this has exactly *p* – 1 roots; in fact any is a root of *f*. Now so in for all .

Hence by Hensel’s lemma, for each , there is a unique lift of *a* to an which is a (*p* – 1)-th root of unity. We call the **Teichmuller lift** of .

# Analysis in Completed Rings

*Warning: the purpose of this section is to give the reader a flavour of the subject matter. It is not meant to be comprehensive. In particular, there are no proofs here*.

Hensel’s lemma gives us an effective criterion to determine if a polynomial over a complete local ring have roots. Although its proof gives us a method to effectively compute these roots to arbitrary precision, there are other techniques we can borrow from real analysis.

**Example 1: Binomial Expansion**

In , we can compute the square root of (1+*X*) with the binomial expansion:

By the same token, we can compute square roots in by taking binomial expansion of , as long as the convergence is “fast enough”. For example, to compute , binomial expansion gives

Taking the first four terms we have so that . Indeed, we can easily check that is a solution to .

**Example 2: Fixed-Point Method**

While solving equations of the form in analysis, it is sometimes effective to start with a good estimate then iteratively compute . We can do this in complete local rings too.

For example, let us solve as a polynomial in *X* with coefficients in . We saw above there is a unique root . Start with then iteratively compute . This gives

where is accurate up to .

**Example 3: Newton Method**

To solve an equation of the form , one starts with a good estimate then iterate .

**Exercise A**

Use the Newton root-finding method to obtain to high precision (in Python).

### Completion in Geometry

As another application of completion, consider with . Taking the -adic completion, we obtain

by proposition 5 here. Note that since has a square root in , the ring is no longer an integral domain, but a “union of two lines” since where is a unit.

This reflects the geometrical fact that when we magnify at the origin, we obtain a union of two lines.

[ Image edited from GeoGebra plot. ]

**Exercise B**

Let be a local ring. Prove that if the -adic completion of *A* is an integral domain, then so is *A*.

[ Hint: use a one-line proof. ]