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Consider a smooth projective curve $X$ over $\mathbb C$ (so $X$ is a projective $\mathbb C$-scheme, integral, of finite type....), and let $t: X\longrightarrow\mathbb P^1_{\mathbb C}=\operatorname{Proj}(\mathbb C[T_0,T_1])$ be a finite morphism of degree $d$. By the known equivalence between smooth projective curves and compact Riemann surfaces, there is a holomorphic map $t(\mathbb C): X(\mathbb C)\longrightarrow\mathbb P^1(\mathbb C)$ of compact Riemann surfaces, associated to $t$.

Now my question(s):

Is $t(\mathbb C)$ a $d$-branched covering of the projective line? In general do finite morphisms of degree $d$ in scheme theory setting correspond to $d$-branched coverings in the classical framework of compact Riemann surfaces?

Reasoning by analogy I think that the answer is yes to both questions but I'm not formally sure of this.

Thanks in advance.

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    $\begingroup$ This may not be very helpful: the answer is yes. $\endgroup$ – Cantlog Mar 29 '14 at 20:47
  • $\begingroup$ But now I have one more confirmation that the above statement is true. Thanks. $\endgroup$ – Dubious Mar 29 '14 at 21:00
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    $\begingroup$ This just follows from the basic fact that if $f:X\to Y$ is a map of Riemann surfaces, then $[M(X):M(Y)]$ is equal to the degree of the covering map associated to $X\to Y$ (i.e. the covering off of the branch points). This can be found in most texts. Tell me if you have difficulty finding it. $\endgroup$ – Alex Youcis Mar 30 '14 at 1:37
  • $\begingroup$ @Galoisfan Just to make sure, since you didn't upvote my answer, were you happy with it? If there is something I missed in your question, let me know. $\endgroup$ – Alex Youcis Mar 30 '14 at 12:30
  • $\begingroup$ Yes it's clear, I've accepted it. I've just forgot to upvote it. Your answers are always very useful. Many thanks. $\endgroup$ – Dubious Mar 30 '14 at 12:35
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Just to have the question answered.

Under the correspondence between irreducible projective smooth curves over $\mathbb{C}$, and Riemann surfaces, function fields are preserved. More precisely, $K(X)\cong M(X^\text{an})$, and that this isomorphism is functorial.

Now, if $f:X\to Y$ is finite of degree $d$, then $[K(X):K(Y)]=d$. Thus, for the associated map $f^\text{an}:X^\text{an}\to Y^\text{an}$ we have that $[M(X^\text{an}):M(Y^\text{an})]=d$.

It is then a common fact of complex analysis that if $f:X\to Y$ is map of compact Riemann surfaces, then the degree as a branched covering is $[M(X):M(Y)]$. While there is probably a more canonical reference, the one that comes to mind is Szamuely's Galois Groups and Fundamental Groups, Proposition 3.3.5.

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