Valuation Rings

I want to talk a bit about valuation rings. As a general topic, valuation rings are rather new to me, so hopefully my explanations go smoothly. The only prior knowledge I will assume is that of local rings and some elementary facts about integral elements.

To begin, let K be a field. Then a valuation on K is a map \nu:K^\times\longrightarrow\mathbb R such that the following hold for all a,b\in K^\times:

  1. \nu(ab)=\nu(a)+\nu(b),
  2. \nu(a+b)\geq\min\{\nu(a),\nu(b)\}.

Notice that the first condition means that \nu is a group homomorphism from the multiplicative group K^\times to the additive group \mathbb R, hence the image \Gamma=\nu(K^\times) is a subgroup of the real numbers, which we call the valuation group of \nu. We will set \nu(0)=\infty, which is assumed to be larger than any real number. Now any such pair (K,\nu) of a field K and valuation \nu on K is called a valuation field.

Now let (K,\nu) be any valuation field. We then easily see the following properties:

  1. \nu(\pm1)=0,
  2. \nu(-a)=\nu(a) for all a\in K,
  3. \nu(a^{-1})=-\nu(a) for all a\in K^\times.

If we define R:=\{x\in K:\nu(x)\geq0\}, we see that R is a subring of K called the valuation ring of \nu. Since R is a subring of a field, it is an integral domain. If we set R^{-1}:=\{x^{-1}\in K:0\neq x\in R\}, we see that K=R\cup R^{-1} and K is the field of fractions of R.

Notice that \mathfrak m:=\{x\in R:\nu(x)>0\} is an ideal of R, called theĀ valuation ideal of \nu. Now suppose that x is a unit of R. Then x,x^{-1}\in R, implies \nu(x),\nu(x^{-1})\geq0. But \nu(x^{-1})=-\nu(x), implying that \nu(x)=0. Therefore R^\times=\{x\in R:\nu(x)=0\}. Since the set of non-units \mathfrak m is an ideal of R, we see that R is a local ring. By property of local rings, \mathfrak m is maximal and k:=R/\mathfrak m is a field, called the residue field of \nu.

We want to show one more interesting property of valuation fields before moving forward. Suppose \nu(b)<\nu(a) for some a,b\in K. We then see that \nu(a)-\nu(b)=\nu(\frac ab)>0, implying \frac ab\in\mathfrak m. Then 1+\frac ab must be a unit of R since otherwise 1+\frac ab\in\mathfrak m would imply that 1+\frac ab-\frac ab=1\in\mathfrak m, an obvious contradiction. From this we have

0=\nu(1+\frac ab)=\nu(\frac{a+b}b)=\nu(a+b)-\nu(b).

So we have proven that \nu(b)<\nu(a) implies \nu(a+b)=\nu(b).

Next we want to discuss the idea of equivalent valuations on K. For two valuations \nu,\nu' on K, we say \nu and \nu' are equivalent and write \nu\sim\nu' if there exists a positive real number \lambda such that \nu'(a)=\lambda\nu(a) for all a\in K. This then defines an equivalence relation on valuations on K. It is easy to see that equivalent valuations given the same valuation ring, valuation ideal, and group of units. A valuation is called discrete if its value group is an infinite cyclic group, i.e., \nu(K^\times)\cong\mathbb Z. If the value group is equal to – not simply isomorphic to – the integers, then the valuations is called a normalized discrete valuation. Notice that every discrete valuation is equivalent to a normalized discrete valuation.

We want to focus for a bit on discrete valuations rings, or simply DVRs. So let \nu be a normalized discrete valuation on on a field K with valuation ring R and valuation ideal \mathfrak m. Then the following properties hold:

  1. Let \pi\in R be an element such that \nu(\pi)=1. Then any \alpha\in K can be expressed as \alpha=u\pi^i where i=\nu(\alpha) and u is a unit of R.
  2. Any nonzero ideal of R is of the form \mathfrak m^i=(\pi^i) for some i\geq0. In particular, R is a PID.
  3. R is integrally closed.

Such an element \pi in the above is called a uniformizing parameter of R. From here we can see that two discrete valuations on a fixed field are equivalent if and only if they have the same valuation ideal. We finish this section with a result that gives a characterization of DVRs. It can be shown that the following two conditions on an arbitrary ring R are equivalent:

  1. R is a DVR.
  2. R is a local PID.

Next time we discuss \mathfrak p-adic valuations and complete valuation rings.