If $K$ is a number field and $\mathfrak{O}_K$ the ring of algebraic integers. Let $\mathfrak{p}$ a prime ideal, for each $\alpha\in{\mathfrak{O}_K}$ if
$(\alpha)=\mathfrak{p}_1^{n_1}\cdots{\mathfrak{p}_r^{n_r}}$ with $n_i\in{\mathbb{N}}$ (ideal factorization)
define:
$v_{\mathfrak{p}}(\alpha)=n_s $ if $\mathfrak{p}=\mathfrak{p}_s$; $0$ in otherwise.
My question is: Exists $\pi\in{\mathfrak{p}\setminus{\mathfrak{p}^2}}$ such that $v_{\mathfrak{p}}(\pi)=1$?
Thanks you all.
Yes, there is $\pi \in \mathfrak p \setminus \mathfrak p^2$, since otherwise we would have $\mathfrak p = \mathfrak p^2$, contradicting unique factorization into prime ideals (the ring of integers is a Dedekind domain). That such a $\pi$ satisfies $v_{\mathfrak p}(\pi) = 1$ follows from your definition of the valuation, since $\mathfrak p$ divides $\pi$ but $\mathfrak p^2$ does not.