Commutative Algebra 57

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 \mathfrak a-stable for a fixed ideal \mathfrak a \subseteq A.


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

Lemma 1.

If \mathfrak a = (a_1, \ldots, a_n), then

\hat A \cong A[[X_1, \ldots, X_n]] / (X_1 - a_1, \ldots, X_n - a_n).


Let B = A[X_1, \ldots, X_n], still noetherian, with ideal \mathfrak a' = (X_1 - a_1, \ldots, X_n - a_n). We have a ring isomorphism f : B/\mathfrak a' \stackrel \cong\to A taking X_i \mapsto a_i. Let \mathfrak b = (X_1, \ldots, X_n) be an ideal of B; we will take the \mathfrak b-adic completion on both sides of f, treated as B-modules.

  • On the LHS, \hat B = A[[X_1, \ldots, X_n]] and \hat{\mathfrak a}' is generated by X_1 - a_1, \ldots, X_n - a_n by proposition 5 here.
  • On the RHS, we get the f(\mathfrak{b})-adic completion on A; but f(\mathfrak b) = (a_1, \ldots, a_n), so we get the \mathfrak a-adic completion.

This completes our proof. ♦

Now all it remains is to prove this.

Proposition 1.

If A is a noetherian ring, so is A[[X]].


For a formal power series f\in A[[X]] where f = b_n X^n + b_{n+1} X^{n+1} + \ldots with b_n \ne 0, we say the lowest coefficient of f is b_n and its lowest term is b_n X^n. We also write \deg f = n.

Now suppose \mathfrak b\subseteq A[[X]] is a non-zero ideal.

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

Now for each n = 0, 1, …, let \mathfrak a_n \subseteq A be the set of all b\in A for which b=0 or b X^n occurs as a lowest term of some f \in \mathfrak b. We get an ascending chain of ideals \mathfrak a_0 \subseteq \mathfrak a_1 \subseteq \ldots. Since A is noetherian, for some n we have \mathfrak a_n = \mathfrak a_{n+1} = \ldots.

For each of 0\le i \le n, pick a finite generating set S_i of \mathfrak a_i comprising of non-zero elements; for each b\in S_i pick f \in \mathfrak b whose lowest term is bX^i. This gives a finite subset T_i \subseteq \mathfrak b of degree-i power series whose lowest coefficients generate \mathfrak a_i. Note that if \mathfrak a_i = 0 then T_i = \emptyset.

Let T := \cup_{i=0}^n T_i.

Step 2: prove that T generates 𝔟.

Now suppose f\in \mathfrak b has lowest term bX^m so b\in \mathfrak a_m. By our choice of T we can find g_1, \ldots, g_k \in T such that

f - (a_1 X^{d_1}) g_1 -\ldots - (a_k X^{d_k})g_k = b' X^{m+1} + (\text{higher terms}),

for some a_i \in A, d_i = m - \deg g_i \ge 0. Since T is finite, in fact we can assume T = \{g_1, \ldots, g_k\}, setting a_i = 0 for unneeded g_i. Repeating the process with the RHS polynomial, we obtain

f - (a_1 X^{d_1} + b_1 X^{d_1 + 1}) g_1 - \ldots - (a_k X^{d_k} + b_k X^{d_k + 1}) g_k = b'' X^{m+2} + (\text{higher terms}).

Repeating this inductively, we obtain formal power series h_1, \ldots, h_k \in A[[X]] such that f - h_1 g_1 - \ldots - h_k g_k = 0. ♦

Hence, by proposition 1 and lemma 1 we have:

Corollary 1.

The \mathfrak a-adic completion of A is noetherian.


Completion of Local Rings

Next suppose (A,\mathfrak m) is a noetherian local ring.


The completion of A is its \mathfrak m-adic completion.

Since \hat A/\hat {\mathfrak m} \cong A/\mathfrak m =: k is a field, \hat{\mathfrak m} is a maximal ideal of \hat A. Also we have:

Lemma 2.

(\hat A, \hat {\mathfrak m}) is a local ring.


By lemma 2 here, \hat{\mathfrak m} is contained in the Jacobson radical of \hat A, so it is contained in all maximal ideals of \hat A. But \hat{\mathfrak m} is already maximal. ♦

Hence, we see that the noetherian local ring (\hat A, \hat {\mathfrak m}) inherits many of the properties of (A, \mathfrak m). E.g. they have the same Hilbert polynomial

P(r) = \dim_{A/\mathfrak m} \mathfrak m^r/\mathfrak m^{r++1}

since \hat{\mathfrak m}^n / \hat{\mathfrak m}^{n+1} \cong \mathfrak m^n /\mathfrak m^{n+1} 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 (A, \mathfrak m) is a complete local ring. Let f(X) \in A[X] be a polynomial. If there exists \alpha\in A such that f(\alpha) \equiv 0 \pmod {\mathfrak m} and f'(\alpha) \not\equiv 0 \pmod {\mathfrak m}, then there is a unique a\in A such that

a\equiv \alpha \pmod {\mathfrak m}, \quad f(a) = 0.


Hence most of the time, if we can find a root \alpha for f(X) \in A[X] in the residue field A/\mathfrak m, then \alpha lifts to a root a\in A.

There are more refined versions of Hensel’s lemma to consider the case where f'(\alpha) \equiv 0 \pmod {\mathfrak m}. One can even generalize it to multivariate polynomials. For our purpose, we will only consider the simplest case.


Fix y\in A such that f'(\alpha)y \equiv 1 \pmod {\mathfrak m}.

Set a_1 = \alpha. It suffices to show: we can find a_2, a_3, \ldots \in A such that

i\ge 1 \implies a_{i+1} \equiv a_i \pmod {\mathfrak m^i}, \ f(a_i) \equiv 0 \pmod {\mathfrak m^i},

so that \lim_{n\to\infty} a_n \in A gives us the desired element. We construct this sequence recursively; suppose we have a_1, \ldots, a_n (n\ge 1). Write

f(X) = c + d(X - a_n) + (X - a_n)^2 g(X),\quad g(X) \in A[X],\ c = f(a_n), \ d = f'(a_n).

Since a_n \equiv \alpha \pmod {\mathfrak m} we have f'(a_n) \equiv f'(\alpha) \pmod {\mathfrak m} so f'(a_n)y \equiv 1 \pmod {\mathfrak m}. Hence setting a_{n+1} := a_n + x with x = -f(a_n)y \in \mathfrak m^n gives

\begin{aligned} f(a_{n+1}) &\equiv \overbrace{f(a_n)}^c + \overbrace{f'(a_n)}^d x \pmod {\mathfrak m^{2n}} \\ &= f(a_n) - f'(a_n) f(a_n) y \\ &\equiv 0 \pmod {\mathfrak m^{n+1}}.\end{aligned},

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 \mathbb C[[X, Y]] with maximal ideal \mathfrak m = (X, Y), consider the equation f(T) = T^2 - (1+X). Modulo \mathfrak m, we obtain f(T) \equiv T^2 - 1 which has two roots: +1 and -1. Since f'(1) = 2 \ne 0, by Hensel’s lemma there is a unique g \in \mathbb C[[X, Y]] with constant term 1 such that g^2 = 1+X.

2. Similarly, consider the ring \mathbb Z_p of p-adic integers with p > 2. Let f(X) = X^2 - a for a \in \mathbb Z_p outside p\mathbb Z_p. If a mod p has a square root b, then there is a Hensel lift of b to a square root of a in \mathbb Z_p.

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

\mathbb C[[Y]] \longrightarrow \mathbb C[[X, Y]]/(Y^2 - X^3 + X)

is an isomorphism. Consider the polynomial f(X) = X^3 - X - Y^2 as a polynomial in X with coefficients in \mathbb C[[Y]]. Modulo \mathfrak m, we have f(X) \equiv X^3 - X which has roots -1, 0, +1. Hence

X^3 - X - Y^2 = (X - \alpha_1) (X - \alpha_2)(X - \alpha_3) where \alpha_i \in \mathbb C[[Y]]

with \alpha_1, \alpha_2, \alpha_3 \equiv -1, 0, +1 \pmod Y respectively. But X-\alpha_1 and X-\alpha_3 are invertible in \mathbb C[[X, Y]] so

\mathbb C[[X, Y]]/(Y^2 - X^3 +X) \cong \mathbb C[[X, Y]]/(X - \alpha_2) \cong \mathbb C[[Y]].

4. In \mathbb Z_p, consider the equation f(X) = X^{p-1} - 1. Modulo p, this has exactly p – 1 roots; in fact any a \in \mathbb F_p - \{0\} is a root of f. Now f'(X) = (p-1)X^{p-2} so f'(a) \ne 0 in \mathbb F_p for all a \in \mathbb F_p - \{0\}.

Hence by Hensel’s lemma, for each 1 \le a \le p-1, there is a unique lift of a to an \omega_a \in \mathbb Z_p which is a (p – 1)-th root of unity. We call \omega_a the Teichmuller lift of a\in \mathbb F_p - \{0\}.


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 \mathbb C[[X]], we can compute the square root of (1+X) with the binomial expansion:

\begin{aligned}(1+X)^{\frac 1 2} &= 1+ \tfrac 1 2 X + \tfrac{\frac 1 2 (\frac 1 2 - 1)}{2!} X^2 + \tfrac{\frac 1 2 (\frac 1 2 -1)(\frac 1 2 - 2)}{3!} X^3 + \ldots\\ &= 1 + \tfrac 1 2 X - \tfrac 1 8 X^2 + \tfrac 1 {16} X^3 + \ldots \in \mathbb C[[X]].\end{aligned}

By the same token, we can compute square roots in \mathbb Z_p by taking binomial expansion of (1+\alpha)^n, as long as the convergence is “fast enough”. For example, to compute \sqrt 2 \in \mathbb Z_7, binomial expansion gives

2\sqrt 2 = \sqrt 8 = (1 + 7)^{\frac 1 2} = 1 + \frac 1 2 (7) + \frac{\frac 1 2 (\frac 1 2 - 1)}{2!} (7^2) + \frac{\frac 1 2 (\frac 1 2 -1)(\frac 1 2 - 2)}{3!} (7^3) + \ldots \in \mathbb Z_7.

Taking the first four terms we have 2\sqrt 2 \equiv 470 \pmod {7^4} so that \sqrt 2 \equiv 235 \pmod {7^4}. Indeed, we can easily check that m = 235 is a solution to m^2 \equiv 2 \pmod {7^4}.

Example 2: Fixed-Point Method

While solving equations of the form x = f(x) in analysis, it is sometimes effective to start with a good estimate x_0 then iteratively compute x_{n+1} = f(x_n). We can do this in complete local rings too.

For example, let us solve X^3 - X = Y^2 as a polynomial in X with coefficients in \mathbb C[[Y]]. We saw above there is a unique root x \equiv 0 \pmod Y. Start with x_0 = 0 then iteratively compute x_{n+1} = x_n^3 - Y^2. This gives

\begin{aligned} x_0 &= 0, \\ x_1 &= -Y^2,\\ x_2 &= -Y^6 - Y^2, \\ x_3 &= -Y^{18} - 3Y^{14} - 3Y^{10} - Y^6 - Y^2,\end{aligned}

where x_3 is accurate up to Y^{13}.

Example 3: Newton Method

To solve an equation of the form f(x) = 0, one starts with a good estimate x_0 then iterate x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Exercise A

Use the Newton root-finding method to obtain \sqrt 2 \in \mathbb Z_7 to high precision (in Python).

Completion in Geometry

As another application of completion, consider A = \mathbb C[X, Y]/(Y^2 - X^3 - X^2) with \mathfrak m = (X, Y). Taking the \mathfrak m-adic completion, we obtain

\hat A = \mathbb C[[X, Y]]/(Y^2 - X^3 - X^2)

by proposition 5 here. Note that since 1+X has a square root in \mathbb C[[X, Y]], the ring \hat A is no longer an integral domain, but a “union of two lines” since Y^2 - X^3 - X^2 = (Y - \alpha X)(Y + \alpha X) where \alpha = \sqrt{1+X} \in \mathbb C[[X, Y]] 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 (A,\mathfrak m) be a local ring. Prove that if the \mathfrak m-adic completion of A is an integral domain, then so is A.

[ Hint: use a one-line proof. ]


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