# Branch points lie in $\mathbb P^1\left(\overline{\mathbb Q}\right)$

Let $X$ be a complex smooth projective curve and suppose moreover that $X$ is defined over $\overline{\mathbb Q}$. Now consider a finite map $f:X\longrightarrow\mathbb P^1(\mathbb C)$ of degree $d$ and defined over $\overline{\mathbb Q}$.

First question: I don't understand why the branch points of $f$ (if they exist) must lie in $\mathbb P^1\left(\overline{\mathbb Q}\right)\subseteq\mathbb P^1(\mathbb C)$. In particular I'd like to see why one requires that both the map $f$ and the curve $X$ must be defined over $\overline{\mathbb Q}$.

This should be an evident fact since many books don't explain this point, but simply they assume it.

Second question: Suppose that $X$ is fixed and defined over $\overline{\mathbb Q}$ as above. It is always possible to find a map $f:X\longrightarrow\mathbb P^1(\mathbb C)$ defined over $\overline{\mathbb Q}$ with branch points?

Why the bounty: Despite before the start of the bounty there were some comments and an answer, after some days of work I don't understand the solution to my problems. In particular in a comment under the answer I've pointed out where is my trouble (I will appreciate detailed answers). However I apologize for my stubbornness.

• Hint: Think about the action of $\text{Gal}(\mathbb{C}/\mathbb{Q})$ on the branch points. The orbit is finite (why?) and this solves the problem (why?). – Alex Youcis Mar 28 '14 at 7:13
• Ok, branch points form a finite set, by the way in don't understand why the "image" of a branch point under tha action of $\operatorname{Gal}(\mathbb C/\mathbb Q)$ should be a branch point. – Dubious Mar 28 '14 at 7:17
• Because they act as automorphisms, and so map a stalk to zero if and only if the original stalk was zero. Use this and the fact that being unramified at $x$ is equivalent to $\Omega_X$ not being supported there. – Alex Youcis Mar 29 '14 at 11:11
• Does my comment make sense to you? – Alex Youcis Mar 30 '14 at 12:30
• I'm working about it. Many thanks. – Dubious Mar 30 '14 at 12:36

The equations cutting out the branch points will be defined over $\overline{\mathbb Q}$, since $f$ is. The branch locus is also zero-dimensional (there are only finitely many branch points).

The Nullstellensatz shows that for any zero-dimensional variety over an alg. closed field $k$, all the points are defined over $k$.

Regarding the second question: any non-constant rational function on $X$ over $\overline{\mathbb Q}$ gives a map to $\mathbb P^1$ (which will have branch points, unless $X$ is also $\mathbb P^1$, and we chose our morphism to be of degree $1$).

• For me the only non clear point is when you say: "The equations cutting out the branch points will be defined over $\overline{\mathbb Q}$, since $f$ is". I know only that locally $f$ is a polynomial in $\overline{ \mathbb Q}$. I apologize for my stupidity, maybe it is an obviuos thing that I don't see. – Dubious Apr 1 '14 at 7:23

Please do not accept this as the answer. I think this question is a great example of how Grothendieck's theory of schemes helps one think about this kind of thing. Namely, when you work with schemes you can succinctly say what it means to have a morphism $f : X \to Y$ of varieties over any field $k$. Assuming that $f$ is finite and $X$ and $Y$ are smooth, the ramification locus of $f$ is a closed subscheme $R \subset X$ also defined over $k$. Technically, $R$ is the scheme theoretic support of $\Omega_{X/Y}$, which means that locally the equations of $R$ are given by the determinant of a matrix of partial derivatives of certain regular functions (everything over $k$). [The simplest case is to consider a morphism $(f_1, \ldots, f_n) : \mathbf{A}^n_k \to \mathbf{A}^n_k$ given by $n$ polynomials in $n$ variables and to consider the determinant of the matrix of partial derivatives.] Then finally the branch locus is $B = f(Z)$ which is a closed subscheme of $X$ defined over $k$.

This doesn't yet answer your question. Assume that $k \subset K$ is an extension of algebraically closed fields. For example $k = \overline{\mathbf{Q}}$ and $K = \mathbf{C}$. We want to see that if we extend the base field from $k$ to $K$ then the ramification locus does not change (in some sense). To make this more precise we think about base change. Base change is a functor which to a variety (or scheme) $X$ over $k$ associates a variety (or scheme) $X_K$ over $K$. Then your question is: Why is it true that $Z_K$ is the ramification locus of $f_K : X_K \to Y_K$? The answer is of course, that if the morphism $f$ is locally given by some regular functions (expressed even more locally as polynomials perhaps), then the base change $f_K$ is given by those same regular functions (or polynomials but now their coefficients are seen as elements of $K$ not $k$) and therefore the relevant derivates will match as well. A more technical way of saying this is that formation of $\Omega_{X/Y}$, $R$, and $B$ commute with base change.

Moreover, in this setup a variety $W$ over $K$ can be defined over $k$ if there exists some variety $U$ over $k$ such that $U_K \cong W$ as varieties over $K$. Given varieties $X$ and $Y$ over $k$ we can ask whether a morphism $g : X_K \to Y_K$ is the base change of a morphism $f : X \to Y$, etc. In this language whether or not $W$ or $g$ is defined over $k$ is a problem of "descent". For example, there are uncountably many morphisms $\mathbf{P}^1_{\mathbf{C}} \to \mathbf{P}^1_{\mathbf{C}}$ of varieties over $\mathbf{C}$ and only countably many morphisms $\mathbf{P}^1_{\overline{\mathbf{Q}}} \to \mathbf{P}^1_{\overline{\mathbf{Q}}}$ so most of the former aren't base changes of the latter.

Hope this helps!

• In relation to your answer, I've done the following reasoning but I don't know if it is right: $f$ is locally represented by a polynomial in $\overline{\mathbb Q}[T]$. It follows that the ramification points of $f$ are those points that annihilate the derivative of some polynomial in $\overline{\mathbb Q}[T]$; since $\overline{\mathbb Q}$ is algebraically closed, then the ramification points of $f$ lie in $\overline{\mathbb Q}$ and we can conclude that $\operatorname{br}(f)\subseteq\overline{\mathbb Q}$. I'don't use the fact that $X$ is defined over $\overline{\mathbb Q}$!! – Dubious Apr 5 '14 at 9:12
• Dear Galoisfan, my answer is trying to get you to think about what it is you are asking. Your question doesn't tell us what definitions you are using (varieties a la Weil, a la Hartshorne Ch I, a la Mumford's red book, or schemes, etc), the order of quantifiers, etc, etc. If you are confused about some math question (which happens to me a lot), then it usually means there's some definition you haven't understood precisely or you aren't using it right. So I suggest taking a few days to work on formulating a really precise question and then asking again if it isn't clear yet. – answer_bot Apr 5 '14 at 13:11