## 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 $\mathfrak a \subseteq A$.

# Noetherianness

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

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

Proof

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 $A[[X]]$.

Proof

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.

Definition.

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.

Proof

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

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

Note

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.

Proof

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.

Examples

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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### 1 Response to Commutative Algebra 57

1. Vanya says:

What about uniqueness part in Hensels lemma.