Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
add comment

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 $76771008 \pm 44330496 \sqrt{3}$.

share|improve this answer
    
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
add comment

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 $\mathbb{Q}(a)\neq \mathbb{Q}(b)$.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

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.