I have a question and I am stuck, although it should not be too difficult.
We consider $K$ a field, $v$ a discrete valuation on $K$ and $O=\{x \in K:v(x)\geq 0\}$ the valuation ring of $v$. Let $\pi\in K^{\times}$. Then TFAE:
i) $$v(\pi)=\min\{v(x):x\in K^{\times} \; \text{satisfies} \; v(x)>0\}$$
ii) The ideal $\pi O$ is a prime ideal of $O$.
Can somebody please help me. I do not know how to relate these two affirmations. Thank you in advance!
(i) $\Rightarrow$ (ii): Take $a,b\in O_v$ such that $ab\in\pi O_v$. We have $v(ab)\ge v(\pi)>0$, and therefore can't have $v(a)=v(b)=0$. If $v(a)>0$, then $v(a)\ge v(\pi)$, so $v(a\pi^{-1})\ge 0$ and this shows that $a\pi^{-1}\in O_v$ hence $a\in\pi O_v$.
(ii) $\Rightarrow$ (i): Assume that there exists $a\in K^{\times}$ with $v(a)>0$ and $v(a)<v(\pi)$. Set $b=\pi a^{-1}$. Then $v(b)>0$. Since $ab=\pi\in\pi O_v$ we must have $a\in\pi O_v$ or $b\in\pi O_v$ which means that $v(a)\ge v(\pi)$ or $v(b)\ge v(\pi)$, but both of there are false.