Notations: Let $k$ be a field of characteristic $\neq 2$, $k(X)$ be a function field. Suppose $p(X)\in k[X]$ is an irreducible polynomial and $\alpha$ be a root of $p(X)$. Let $F=k(\alpha)=k[X]/p(X).k[X]$. We can define $p(X)$-adic valuation on $k(X)$ as : for $a(X)\in k(X)$, it can be written as $a(X)=p(X)^n.\frac{h(X)}{r(X)}$, for some $n\in\mathbb Z$ ,$(h(X),r(X))=1$ and $p(X)$ does not divide $r(X)$. Then $p(X)$-adic valution of $a(X)$ is $n$.
Question: If $g(X)\in k(X)$ is such that $p(X)$-adic valuation of $g(X)$ is odd. Then we need to show that $(X-\alpha)$-addic valuation (this is over $F(X)$) of $g(X)$ is also odd.
Thanks!
Write $$g(X) = p(X)^n \frac{h(X)}{r(X)}$$ where $p(X)$ does not divide $h(X)$ or $r(X)$. This implies that $X-\alpha$ does not divide $h(X)$ nor $r(X)$, since $p(X)$ was the minimal polynomial of $\alpha$.
So the $X-\alpha$ valuation of $g$ is $n$ times the $X-\alpha$ valuation of $p$. Why is the latter odd?
Essentially, this is because $\mathrm{char} \, k \ne 2$ is odd. If we were over a field of characteristic zero, $p$ would be separable (because irreducible) and the valuation would be one. We aren't necessarily, though, but in the positive characteristic case, the ($X-\alpha$)-valuation of $p$ is the inseparability degree of $k(\alpha)/k$, which is a power of $\mathrm{char} \, k$ and therefore odd.