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 .
We wish to prove that the -adic completion of a noetherian ring is noetherian. First we express:
If , then
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.
If A is a noetherian ring, so is .
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 .
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:
The -adic completion of A is noetherian.
Completion of Local Rings
Next suppose is a noetherian local ring.
The completion of A is its -adic completion.
Since is a field, is a maximal ideal of . Also we have:
is a local ring.
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.
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
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.
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.
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
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 .
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. ]
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. ]