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Suppose that a (connected) Riemann surface $X$ is birational to a Riemann surface $Y$ which can be defined (algebraically) over the field of algebraic numbers.

Does this imply that $X$ itself can be defined over the field of algebraic numbers?

Basically, I'm asking if the property "can be defined over the field of algebraic numbers" is "birationally invariant".

Is this something that holds for general schemes?

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1 Answer 1

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It doesn't quite make sense to speak of "birational equivalence" of arbitrary Riemann surfaces. For instance, is the upper half plane "birationally equivalent" to $\mathbb{P}^1$?

Even if you restrict to nonsingular, connected algebraic curves over $\mathbb{C}$, the answer is no. Consider for instance the affine curve $C$ obtained as $\mathbb{P}^1$ with the points $0,1,\infty,\pi$ removed. The moduli space of curves of genus zero with four points removed is isomorphic to the affine line -- one can view this in terms of the cross ratio, or in terms of building an elliptic curve ramified over precisely those four points on $\mathbb{P}^1$. As soon as you have a positive dimensional moduli space, you have "generic points" which cannot be defined over $\overline{\mathbb{Q}}$. The curve $C$ has transcendental cross-ratio / $j$-invariant, so is such an example.

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How surprising (to me). Firstly, I was always considering the "algebraic" category. Secondly, if I take a smooth projective curve $X$ which can be defined over a number field, does there always exist a finite subset $B\subset X$ such that $X-B$ cannot be defined over a number field? It's true for genus zero curves by your answer. –  seporhau Mar 9 '12 at 8:27
    
@seporhau: About the "algebraic" category: okay, but then you shouldn't say "Riemann surface", you should say "complex algebraic curve". As to your question: yes, there are moduli spaces $M_{g,n}$ of genus $g$ curves with $n$ marked points, and as long as $n \geq 4$ then $\operatorname{\dim} M_{g,n} > \operatorname{\dim} M_{g,n-1}$, and this ensures that curves over $\overline{\mathbb{Q}}$ have "transcendental markings". –  Pete L. Clark Mar 9 '12 at 13:57
    
Dear Pete, unless I'm missing something, the moduli space of genus zero curves with 4 points removed should be $\Bbb{P}^1 \setminus \{0,1,\infty\}$, and not $\Bbb{A}^1$. Does that make sense? –  Nils Matthes Aug 20 '13 at 21:06
    
@Nils: I think it depends whether we mark the four points or not. If we mark them, then we get the modular curve $Y(2)$, which is $\mathbb{P}^1$ minus three cusps. If we don't mark them, we should mod out the above moduli space by an $S_4$-action. Now I'm afraid what I was actually thinking about was when you mark one of the four points: the resulting $S_3$-quotient is the $j$-line. My guess is that the $S_4$-quotient is again $\mathbb{A}^1$, but off the top of my head I'm not sure. (Note that any of these spaces answers the question in the same way.) –  Pete L. Clark Aug 20 '13 at 22:44
    
Ah, it seems I mistakenly took "four points removed" for "four points labelled". Thanks for the additional explanation. –  Nils Matthes Aug 21 '13 at 10:02

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