Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thanks for the answer. Could you please explain a little bit why this practical way works? What does the degree of the field extension has to do with the number of points in each fiber? – Manos Nov 20 '12 at 22:14
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
What I'm using here is that the number of points over a point is the degree of the morphism if and only if this point is not a branch point. – Theodore Nov 20 '12 at 22:26
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
Very interesting example. Thank you. Is there any simple way to see why the field extension is of the form you mention? – Manos Nov 20 '12 at 22:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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