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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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It depends on what you mean by "finite Galois". If you mean that the morphism is finite and the function field extension is Galois, then it is not necessarily unramified.

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.

A "practical" way to check whether something is etale at a point by the way, is to count the number of points in each fibre. If this is the degree of the finite morphism you win.

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    Yes. Check chapter IV, paragraph 2. This is the paragraph about Riemann-Hurwitz.2012-11-20