Let $k$ be a number field and let $E$ and $E'$ be elliptic curve over $k$.
There is a genus two curve $X$ over $\overline{k}$ which dominates $E$ and $E'$.
Question. Is there a genus two curve $X$ over $k$ which dominates $E$ and $E'$?
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$\begingroup$ The genus 2 curve over $\bar{k}$ descends to $L$ (including the morphisms), a finite extension of $L$. Replace $k$ by $L$. So, I don't see why not. $\endgroup$– MohanCommented Sep 11, 2017 at 22:28
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1$\begingroup$ It seems unlikely that this would always be true. The question amount to finding an isogeny from $E \times E'$ to a principally polarised abelian surface which is not a product of elliptic curves (so will be the Jacobian of a genus 2 curve); if the image of the Galois representation on the Tate module of $E \times E'$ is very big, then such an isogeny probably does not exist. $\endgroup$– nafCommented Sep 12, 2017 at 5:21
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$\begingroup$ @Mohan Yes, there is a finite field extension $L/k$ and a genus two (geometrically connected) curve $X$ over $L$ which dominates $E_L$ and $E_L'$. But this curve might not descend to $k$, and even if it does, the maps to $E_L$ and $E'_L$ might not descend to $k$. $\endgroup$– JayCommented Sep 12, 2017 at 19:57
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$\begingroup$ @ulrich Yes, I thought it might be unlikely. Riemann-Hurwitz tells me that if the maps to $E$ and $E'$ have degree two, then there are precisely two branch points. If the map is of degree three, then there is precisely one branch point (and the ramification is total). I was thinking of just writing down two elliptic curves over $\mathbb{Q}$ with only one rational point, and playing around a bit. Would that work? $\endgroup$– JayCommented Sep 12, 2017 at 20:00
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$\begingroup$ Yes, probably that or something similar might work. Actually it is not even clear that given any elliptic curve $E$ there exists a genus two curve dominating it (with the curve and map defined over $k$). $\endgroup$– nafCommented Sep 13, 2017 at 5:26
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