# 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 the 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.