
Originally Posted by
Erdos32212
Background: Let $\displaystyle K$ be a field. A discrete valuation on $\displaystyle K$ is a function $\displaystyle \nu$ on $\displaystyle K$ is a function $\displaystyle \nu$ $\displaystyle : K^* \rightarrow \mathbb{Z}$ such that
(i) $\displaystyle \nu(ab)=\nu(a)+\nu(b)$ for all $\displaystyle a, b \in K^*.$
(ii) $\displaystyle \nu$ is surjective
(iii) For all $\displaystyle x, y$ $\displaystyle \in K^*$ such that $\displaystyle y \neq -x$, $\displaystyle \nu(x+y)\geq$ min$\displaystyle \{ \nu(x), \nu(y) \}$. The set $\displaystyle R=$$\displaystyle \{ x \in K^* | \nu(x) \geq 0 \} \cup \{ 0 \}$ is called the valuation ring of $\displaystyle \nu$.
Prove:
(a) Prove that $\displaystyle R$ is a subring of $\displaystyle K$.
(b) Prove that for each nonzero element $\displaystyle x \in K$, $\displaystyle x \in R$, or $\displaystyle x^{-1} \in R$.
(c) Prove that $\displaystyle x \in R$ is a unit if and only if $\displaystyle \nu(x)=0$.
(d) Now let $\displaystyle K = \mathbb{Q}$ and let $\displaystyle p$ be a prime. Given a nonzero element $\displaystyle x \in \mathbb{Q}$, write $\displaystyle x=p^e \cdot \frac{c}{d}$ where $\displaystyle e$ is an integer and $\displaystyle p$ divides neither $\displaystyle c$ nor $\displaystyle d$; then define $\displaystyle \nu_p=e$. Prove $\displaystyle e$ is uniquely defined and that $\displaystyle \nu_p$ is a discrete valuation on $\displaystyle \mathbb{Q}.$