I was initially planning on doing a post on Hensel’s lemma. Actually, I think I’ll leave that for later, after I’ve covered some more number theory (which may motivate it better).

So the goal for the next several posts is to cover some algebraic number theory, eventually leading into class field theory. At least in the near future, I intend to keep everything purely local. Thus, the appropriate place to start is to discuss discrete valuation rings rather than Dedekind domains.

Absolute Values

Actually, it is perhaps more logical to introduce discrete valuations as a special case of absolute values, which in turn generalize the standard absolute value on ${\mathbb{R}}$

Definition 1 Let ${F}$ be a field. An absolute value on ${F}$ is a function ${\left\lvert{\cdot}\right\rvert: F \rightarrow \mathbb{R}}$, satisfying the following conditions:

1. ${\left\lvert{x}\right\rvert > 0}$ for ${x\in F}$ with ${x \neq 0}$, and ${\left\lvert{0}\right\rvert=0}$.
2. ${\left\lvert{xy}\right\rvert=\left\lvert{x}\right\rvert\left\lvert{y}\right\rvert}$ for all ${x,y \in F}$.
3. ${\left\lvert{x+y}\right\rvert \leq \left\lvert{x}\right\rvert + \left\lvert{y}\right\rvert.}$ (Triangle inequality.)

So for instance ${\left\lvert{1}\right\rvert^2 = \left\lvert{1^2}\right\rvert = \left\lvert{1}\right\rvert}$, i.e. ${\left\lvert{1}\right\rvert = 1}$.

The standard example is of course the normal absolute value on ${\mathbb{R}}$ or ${\mathbb{C}}$, but here is another:

Example 1 For ${p}$ prime, let ${\left\lvert{\cdot}\right\rvert_p}$ be the ${p}$-adic absolute value on ${\mathbb{Q}}$ defined as follows: if ${r = \frac{x}{y}}$ with ${p^r \mid x, p^s \mid y}$, and ${p^r,p^s}$ are the highest powers of ${p}$ dividing ${x,y}$ respectively, then

$\displaystyle \left\lvert{\frac{x}{y}}\right\rvert_p = p^{s-r}. \ \ \ \ \ (1)$

(Also ${\left\lvert{0}\right\rvert_p= 0}$.)

It can be checked directly from the definition that the ${p}$-adic absolute value is indeed an absolute value, though there are some strange properties: a number has a small ${p}$-adic absolute value precisely when it is divisible by a high power of ${p}$.

Moreover, by elementary number theory, it satisfies the nonarchimedean property:

Definition 2 An absolute value ${\left\lvert{\cdot}\right\rvert}$ on a field is nonarchimedean if ${\left\lvert{x+y}\right\rvert \leq \max(\left\lvert{x}\right\rvert,\left\lvert{y}\right\rvert)}$

This is a key property of the ${p}$-adic absolute value, and what distinguishes it fundamentally from the regular absolute value restricted to ${\mathbb{Q}}$. In general, there is an easy way to check for this:

Proposition 3 The absolute value ${\left\lvert{\cdot}\right\rvert}$ on ${F}$ is non-archimedean if and only if there is a ${C}$ with ${\left\lvert{n}\right\rvert \leq C}$ for all ${n \in \mathbb{N}}$ (by abuse of notation, we regard ${n}$ as an element of ${F}$ as well, even when ${F}$ is of nonzero characteristic and the map ${\mathbb{N} \rightarrow F}$ is not injective). In this case, we can even take ${C=1}$

One way is straightforward: if ${\left\lvert{\cdot}\right\rvert}$ is non-archimedean, then ${\left\lvert{1}\right\rvert =1}$, ${\left\lvert{2}\right\rvert \leq \max(\left\lvert{1}\right\rvert, \left\lvert{1}\right\rvert) = 1}$, ${\left\lvert{3}\right\rvert \leq \max(\left\lvert{2}\right\rvert, \left\lvert{1}\right\rvert) = 1}$, and so on inductively.

The other way is slightly more subtle. Suppose ${\left\lvert{v}\right\rvert \leq C}$ for ${v \in \mathbb{N}}$. Then fix ${x,y \in F}$. We have:

$\displaystyle \left\lvert{x+y}\right\rvert^n = \left\lvert{ \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}}\right\rvert \leq \sum_{k=0}^n \left\lvert{ \binom{n}{k} }\right\rvert \max( \left\lvert{x}\right\rvert,\left\lvert{y}\right\rvert)^n.$

Now by the hypothesis,

$\displaystyle \left\lvert{x+y}\right\rvert^n \leq C n \max( \left\lvert{x}\right\rvert,\left\lvert{y}\right\rvert)^n.$

Taking ${n}$-th roots and letting ${n \rightarrow \infty}$ gives the result.

This also shows that the ${p}$-adic absolute value is nonarchimedean, since it it automatically ${\leq 1}$ on the integers.

Corollary 4 If ${F}$ has nonzero characteristic, then any absolute value on ${F}$ is non-archimedean.

Indeed, if ${F}$ is of characteristic ${p}$, take ${C := \max( \left\lvert{1}\right\rvert, \left\lvert{2}\right\rvert, \dots, \left\lvert{p-1}\right\rvert)}$

Discrete Valuation Rings

The absolute values we are primarily interested in are

Definition 5 A discrete valuation is an absolute value ${\left\lvert{\cdot}\right\rvert}$ on a field ${F}$ such that ${\left\lvert{F^*}\right\rvert}$ is a cyclic group.

In other words, there is ${c \in \mathbb{R}}$ such that, for each ${x \neq 0}$, we can write ${\left\lvert{x}\right\rvert = c^v}$, where ${v=v(x)}$ is an integer depending on ${x}$. We assume without loss of generality that ${c<1}$, in which case ${v: F^* \rightarrow \mathbb{Z}}$ is the order function (sometimes itself called a valuation). Furthermore we assume ${v}$ surjective by choosing ${c}$ as a generator of the cyclic group ${\left\lvert{F^*}\right\rvert}$.

Now, if ${\left\lvert{\cdot}\right\rvert}$ is any nonarchimedean absolute value on a field ${F}$, define the ring of integers ${R}$ as

$\displaystyle R := \{ x \in F: \left\lvert{x}\right\rvert \leq 1 \}.$

(This is a ring.) Note that ${x \in R}$ is a non-unit if and only if ${\left\lvert{x}\right\rvert<1}$, so the sum of two non-units is a non-unit and ${R}$ is a local ring with maximal ideal

$\displaystyle \mathfrak{m} := \{ x \in F: \left\lvert{x}\right\rvert < 1 \}.$

When ${\left\lvert{\cdot}\right\rvert}$ is a discrete valuation, we call the ring of integers so obtained a discrete valuation ring (DVR). The first thing to notice is:

Proposition 6 A discrete valuation ring is a principal ideal domain. Conversely, a local principal ideal domain is a discrete valuation ring.

Indeed, if ${R}$ is a DVR and if ${I \subset R}$ is an ideal, let ${x \in I}$ be an element of maximal order ${v(x)}$; then ${I = Rx}$, since ${R}$ consists of the elements of ${F}$ of nonnegative order.

Conversely, if ${R}$ is a local PID, then let ${\pi}$ generate the maximal ideal ${\mathfrak{m} \subset R}$; then since the Krull intersection theorem (Cor. 6 here) implies

$\displaystyle \bigcap \mathfrak{m}^n = \bigcap \mathfrak (\pi^n) = 0,$

we can write each nonzero ${x \in R}$, say ${x \in \mathfrak{m}^n - \mathfrak{m}^{n+1}}$ as ${x = \pi^n u}$ for ${u \notin \mathfrak{m}}$, i.e. ${u}$ a unit. This is unique and we can define a discrete valuation ${\left\lvert{\cdot}\right\rvert}$ by ${\left\lvert{x}\right\rvert = 2^{-n}}$ for ${n}$ as above. This extends to the quotient field and makes ${R}$ a DVR.

A much more interesting (and nontrivial) result is the following:

Theorem 7 If ${R}$ is Noetherian, integrally closed (in its quotient field), and has a unique nonzero prime ideal ${\mathfrak{m}}$, then ${R}$ is a DVR.

This is equivalent to the fact that Dedekind domains have unique factorization, but I’m only going to be able to get to it in the next post.