Let $f:X \rightarrow Y$ be a finite Galois morphism of curves (curve: integral scheme of dimension 1, proper over an algebraically closed field $k$ will all local rings regular). Question: Is it true that $f$ is unramified at any $P \in X$? In other words, let $Q \in Y$ such that $Q=f(P), P \in X$ and let $v_P$ be a valuation corresponding to the discrete valuation ring $O_{X,P}$. Let $t$ be a local parameter (uniformizer) of $O_{Y,Q}$. We can view $t$ as an element of $O_{X,P}$ since there is a morphism $O_{Y,Q} \rightarrow O_{X,P}$. Is it true that if $f$ is finite Galois, then $v_P(t)=1$, i.e. a local parameter is taken to a local parameter by $O_{Y,Q} \rightarrow O_{X,P}$?
More generally, suppose $f$ is unramified. Can this be characterized in some convenient way, e.g. algebraically, maybe in terms of the field extension $K(Y) \subseteq K(X)$?