## Basic properties of discrete valuations

Let $K$ be a field. A discrete valuation on $K$ is a function $v : K^\times \rightarrow \mathbb{Z}$ satisfying the following.

1. $v(ab) = v(a) + v(b)$
2. $v$ is surjective
3. Provided $x+y \neq 0$, $v(x+y) \geq \min(v(x) + v(y)$.

The set $R = \{ x \in K^\times \ |\ v(x) \geq 0 \} \cup \{0\}$ is called the valuation ring of $v$.

1. Prove that $R$ is a subring of $K$ which contains the identity.
2. Prove that for each nonzero element $x \in K$, either $x \in R$ or $x^{-1} \in R$.
3. Prove that $x \in R$ is a unit if and only if $v(x) = 0$.

First, we note a few facts about $v$.

If $a \in R$, then $v(a) = v(1 \cdot a)$ $= v(1) + v(a)$. Since $v$ is surjective, $v(a)$ is an arbitrary integer. By the uniqueness of additive identities, $v(1) = 0$. Now $0 = v(1) = v((-1) \cdot (-1))$ $= v(-1) + v(-1)$, so that $v(-1) = -v(-1)$. Thus $v(-1) = 0$. Finally, if $a \in R$, $v(-a) = v((-1)a))$ $= v(-1) + v(a)$ $= v(a)$, so that $v(-a) = v(a)$.

Now we show that $R$ is a subring. $0 \in R$ by definition, so that $R$ is not empty. Let $x,y \in R$ and consider $x-y$. If $x \neq 0$ and $y = 0$, then $v(x-y) = v(x) \geq 0$. If $x = 0$ and $y \neq 0$, then $v(x-y) = v(-y)$ $= v(y) \geq 0$. If $x = y = 0$, then $x - y = 0 \in R$. If $x,y \neq 0$, then either $x-y = 0 \in R$ or $x-y \neq 0$, and then $v(x-y) \geq \min(v(x),v(-y))$ $= \min(v(x),v(y)) \geq 0$. Now consider $xy$. If $xy = 0$, then $xy \in R$. If $xy \neq 0$, then $v(xy) = v(x) + v(y) \geq 0$, so that $xy \in R$. Thus $R$ is a subring of $K$. As we saw earlier, $v(1) = 0$, so that $1 \in R$.

Suppose $x \in K$ is nonzero. Now $0 = v(1) = v(xx^{-1})$ $= v(x) + v(x^{-1})$. Thus $v(x) = -v(x^{-1})$, and either $v(x)$ or $v(x^{-1})$ is nonnegative.

Suppose $u \in R$ is a unit; then $u^{-1} \in R$. Above, we saw that $v(u) = -v(u^{-1})$, and both $v(u)$ and $v(u^{-1})$ are nonnegative. Thus $v(u) = v(u^{-1}) = 0$. Now suppose $v(u) = 0$. Then $v(u^{-1}) = -v(u) = 0$, so that $u^{-1} \in R$ and $u \in R$ is a unit.