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.