Hensel’s lemma and a classification theorem September 2, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, commutative algebra, number theory.
Tags: , ,

So, I’ll discuss the proof of a classification theorem that DVRs are often power series rings, using Hensel’s lemma.

Systems of representatives

Let ${R}$ be a complete DVR with maximal ideal ${\mathfrak{m}}$ and quotient field ${F}$. We let ${k:=R/\mathfrak{m}}$; this is the residue field and is, e.g., the integers mod ${p}$ for the ${p}$-adic integers (I will discuss this more later).

The main result that we have today is:

Theorem 1 Suppose ${k}$ is of characteristic zero. Then ${R \simeq k[[X]]}$, the power series ring in one variable, with respect to the usual discrete valuation on ${k[[X]]}$.

The “usual discrete valuation” on the power series ring is the order at zero. Incidentally, this applies to the (non-complete) subring of ${\mathbb{C}[[X]]}$ consisting of power series that converge in some neighborhood of zero, which is the ring of germs of holomorphic functions at zero; the valuation again measures the zero at ${z=0}$.

For a generalization of this theorem, see Serre’s Local Fields.

To prove it, we need to introduce another concept. A system of representatives is a set ${S \subset R}$ such that the reduction map ${S \rightarrow k}$ is bijective. A uniformizer is a generator of the maximal ideal ${\mathfrak{m}}$. Then:

Proposition 2 If ${S}$ is a system of representatives and ${\pi}$ a uniformizer, we can write each ${x \in R}$ uniquely as

$\displaystyle x= \sum_{i=0}^\infty s_i \pi^i, \quad \mathrm{where} \ s_i \in S.$

Given ${x}$, we can find by the definitions ${s_0 \in S}$ with ${x-s_0 \in \pi R}$. Repeating, we can write ${{x-s_0}\ {\pi} \in R}$ as ${{x-s_0}\ {\pi} - s_1 \in \pi R}$, or ${x - s_0 - s_1 \pi \in \pi^2 R}$. Repeat the process inductively and note that the differences ${x - \sum_{i=0}^{n} s_i \pi^i \in \pi^{n+1}R}$ tend to zero.

In the ${p}$-adic numbers, we can take ${\{0, \dots, p-1\}}$ as a system of representatives, so we find each ${p}$-adic integer has a unique ${p}$-adic expansion ${x = \sum_{i=0}^\infty x_i p^i}$ for ${x_i \in \{0, \dots, p-1\}}$

Hensel’s Lemma

Hensel’s lemma, as already mentioned, allow us to lift approximate solutions of equations to exact solutions. This will enable us to construct a system of representatives which is actually a field.

Theorem 3 Let ${R}$ be a complete DVR with quotient field ${K}$. Suppose ${f \in R[X]}$ and ${x_0 \in R}$ satisfies ${\overline{f(x_0)} = 0 \in \mathfrak{m}}$ (i.e. ${\pi \mid f(x)}$) while ${\overline{ f'(x_0) } \neq 0}$. Then there is a unique ${x \in R}$ with ${\overline{x}=\overline{x_0}}$ and ${f(x)=0}$

(Here the bar denotes reduction.)

The idea is to use Newton’s method of successive approximation. Recall that given an approximate root ${r}$, Newton’s method “refines” it to

$\displaystyle r' := r - \frac{ f(r) }{f'(r) }.$

So define ${x_n \in R}$ inductively (${x_0}$ is already defined) as ${x_n = (x_{n-1})'}$, the ${'}$ notation as above. I claim that the ${x_n}$ approach a limit ${x \in R}$ which is as claimed.

For ${r \in R}$ by Taylor’s formula we can write ${f(X) = f(r) + f'(r)(X-r) + C(X)(X-r)^2}$, where ${C(X) \in R[X]}$ depends on ${r}$. Then for any ${r}$

$\displaystyle f(r') = f\left( r - \frac{ f(r) }{f'(r) }\right) = f(r) - f(r) + C(r')\left( \frac{f(r)}{f'(r)}\right)^2.$

Thus, if ${\left \lvert {f(r)} \right \rvert < c}$ and ${\left \lvert {f'(r)} \right \rvert = 1}$, we have ${\left \lvert {f(r')} \right \rvert \leq c^2}$ and ${\left \lvert {f'(r')} \right \rvert =1}$, since ${r' \equiv r \ \mathrm{mod} \ \mathfrak{m}}$. We even have ${\left \lvert {r-r'} \right \rvert \leq c}$. This enables us to claim inductively:

1. ${x_n \in R}$
2. ${\left \lvert {f(x_n)} \right \rvert \leq \left \lvert {f(x_0)} \right \rvert^{2^n}}$.
3. ${\left \lvert {x_n-x_{n-1}} \right \rvert \leq \left \lvert {f(x_0)} \right \rvert^{2^{n-1}}}$.

Now it follows that we may set ${x := \lim x_n}$ and we will have ${f(x)= 0}$. The last assertion follows because ${\overline{x_0}}$ is a simple root of ${\overline{f} \in k[X]}$.

There is a more general (Sorry, Bourbaki!) version of Hensel’s lemma that says if you have ${\left \lvert{f(x_0)} \right \rvert \leq \left \lvert f'(x_0) \right \rvert^2}$, the conclusion holds.  It is proved using a very similar argument.  Also, there’s no need for discreteness of the absolute value—just completeness is necessary.

Corollary 4 For ${n}$ fixed, any element of ${R}$ sufficiently close to 1 is a ${n}$-th power.

Use the polynomial ${X^n-1}$

Proof of the Classification Theorem

We now prove the first theorem.

Note that ${\mathbb{Z}-0 \subset R}$ gets sent to nonzero elements in the residue field ${k}$, which is of characteristic zero. This means that ${\mathbb{Z}-0 \subset R}$ consists of units, so ${\mathbb{Q} \subset R}$.

Let ${L \subset R}$ be a subfield. Then ${L \simeq \overline{L} \subset k}$; if ${t \in k - \overline{ L}}$, I claim that there is ${L' \supset R}$ containing ${L}$ with ${t \in \overline{L'}}$.

If ${t}$ is transcendental, lift it to ${T \in R}$; then ${T}$ is transcendental over ${L}$ and is invertible in ${R}$, so we can take ${L' := L(T)}$.

If the minimal polynomial of ${t}$ over ${\overline{L}}$ is ${\overline{f}(X) \in k[X]}$, we have ${\overline{f}(t) = 0}$. Moreover, ${\overline{f}'(t) \neq 0}$ because these fields are of characteristic zero and all extensions are separable. So lift ${\overline{f}(X)}$ to ${f(X) \in R[X]}$; by Hensel lift ${t}$ to ${u \in R}$ with ${f(u) = 0}$. Then ${f}$ is irreducible in ${L[X]}$ (otherwise we could reduce a factoring to get one of ${\overline{f} \in \overline{L}[X]}$), so ${L[u] = L[X]/(f(X))}$, which is a field ${L'}$.

So if ${K \subset R}$ is the maximal subfield (use Zorn), this is our system of representatives by the above argument.