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Let $K\subset L$ be a finite field extension.

Let $X$ and $Y$ be (smooth projective geometrically connected) curves over $L$.

Let $f:X\to Y$ be a finite morphism of curves over $L$.

Assume that $Y$ can not be defined over $K$.

Is it possible that $X$ can still be defined over $K$?

Equivalently, suppose that $X$ can be defined over $K$. Then is it true that $Y$ can be defined over $K$?

So basically, I'm asking about the following. If you have a curve $X$ over some field $K$ and a finite morphism $X_L\to Y$ over some extension $L/K$, can we define the curve $Y$ over $K$?

I suspect the answer to be no. But I can't seem to find an easy counterexample.

I was thinking about taking $Y$ to be an elliptic curve which can't be defined over $\mathbf{Q}$ (irrational $j$-invariant) and constructing a suitable branched cover...

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2 Answers

up vote 3 down vote accepted

Yes. Take $Y$ to be the elliptic curve $y^2=x^3-x$. Take $L$ to be large enough that $Y[3]$ is defined over $L$. Then $Y$ has four quotients, obtained as $Y/\langle z \rangle$ for various $3$-torsion points $z$. We will take one of these curves to be our $X$.

More specifically, working over $\mathbb{C}$, the curve $Y$ is $\mathbb{C}/\langle 1, i \rangle$. The $3$-torsion points are the images of $\pm i/3$, $\pm (i+1)/3$, $\pm (i-1)/3$ and $\pm 1/3$. The curve $Y$ has complex multiplication by $i$, which takes the torsion subgroup $\langle i/3$ to $\langle 1/3$ and $\langle (i+1)/3$ to $\langle (i-1)/3 \rangle$. So there are two different quotients: One with $j$ invariant $(i/3)$ and one with $j$ invariant $j((i+1)/3)$.

Mathematica gives $$\begin{array} {r l} j(i/3) =& \phantom{-}153553679.396728884585209285932\ldots \\ j((i+1)/3) =& \phantom{000}-11663.396728884585209285932 \ldots \end{array}$$

So $j(i/3) + j((i+1)/3) = 153542016$ and $j(i/3) \cdot j((i+1)/3) = -1790957481984$. Although I computed these quantities using floating point arithmetic, the last two answers are exact, because the theory of elliptic curves with complex multiplication tells me that they must be integers. So $j(i/3)$ and $j((i+1)/3)$ are the roots of $$x^2 - 153542016 x -1790957481984=0$$ which we can compute to be $76771008 \pm 44330496 \sqrt{3}$.

In short, we have shown that, over a large enough base field, there is an isogeny from $y^2=x^3-x$ to the elliptic curves with $j$-invariants $76771008 \pm 44330496 \sqrt{3}$.

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This answers my question completely. Is it clear how one can construct an example in higher genus? I'm guessing the same strategy should work. We take a certain curve $Y$ over $\mathbf{Q}$ with a certain action. I had the following in mind. Consider the $1$-dimensional moduli space of covers of $\mathbf{P}^1$ of degree $n$ ramified over precisely $0,\infty,1,\lambda$, where $\lambda$ varies over $\bar{\mathbf{Q}}$, the ramification over $0$ and $\infty$ is "total" and the ramification over $1$ and $\lambda$ is "simple". to be continued... –  seporhau Mar 7 '12 at 16:56
...This "Hurwitz space", denoted by $H(n)$, can be defined over $\mathbf{Q}$ (I believe...). It maps to $X_1(n)$ by sending the cover $f:X\to \mathbf{P}^1$ to the elliptic curve $(X,f^{-1}(0))$ with the $n$-torsion point $f^{-1}(\infty)$. The curve $X_1(n)$ cannot be defined over $\mathbf{Q}$. –  seporhau Mar 7 '12 at 16:56
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I have wondered this question myself. A reference I found that might be helpful is Lang's Elliptic Functions section 5.2 It states two elliptic curves $E$ and $E'$ curves have a cyclic isogeny over $\mathbb{C}$ of degree $l$ if and only if $\varphi_l(j(E),j(E'))=0$. where $\varphi_l(x,y)\in \mathbb{Z}[x,y]$. One particular $\varphi_l$ is $$\varphi_2(x,y) = x^3 + y^3 − x^2y^2 + 1488(x^2y + xy^2) − 162000(x^2 + y^2) +40773375xy + 8748000000(x+ y) − 157464000000000$$ So if you want a counterexample with number fields you can just find some $a$ and $b$ such that $\varphi_2(a,b)=0$ but $\mathbb{Q}(a)\neq \mathbb{Q}(b)$.

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For example, plugging in $x=1$ gives $-157455252161999 + 8788774863 y - 160513 y^2 + y^3$, an irreducible cubic. –  David Speyer Mar 7 '12 at 14:38
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