# Finite Galois Morphism of Curves and Ramification - Characterizing Unramified Morphisms

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 $K(Y) \subseteq K(X)$?

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For example, consider the following example $$\mathbf{P}^1\to \mathbf{P}^1, \quad x\mapsto x^n.$$ This is finite Galois of degree $n$ and ramifies over $0$ and $\infty$. Also, any finite morphism $X\to Y$ of degree $2$ is Galois. Clearly, these can be ramified. Take for example $X$ a hyperelliptic curve and $Y$ the projective line.
Consider the map $(x,y)\mapsto x$ from the curve $y^2=x^3-1$ to the affine line. This has degree 2. In fact, on the level of function fields it corresponds to the embedding $k(x) \subset k(x)[y]/(y^2 = x^3-1)$. Now, let us fix a point $a$ on the affine line. To check whether $a$ is a branch point I look at the fibre of $a$. This fibre consists of the elements $(a,y)$ such that $y^2=a^3-1$. This has precisely two solutions unless $a^3-1=0$. That is, when $a=1,\zeta_3$ or $\zeta_3^2$. These three points are the branch points of our map. – Theodore Nov 20 '12 at 22:25
This boils down to the usual formula from local fields: $\sum_{b\mapsto a} e_b = \deg f$. Here $f:Y\to X$ is a finite morphism of curves (say char zero) and $a\in X$. Also, $e_b$ is the ramification index. – Theodore Nov 20 '12 at 22:29