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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.