2
$\begingroup$

Let $\overline{\mathbf{Q}}\subset \mathbf{C}$ be the field of algebraic numbers.

Does there exist a number field $K$ with the following property?

There are embeddings $\sigma,\tau:K\to \overline{\mathbf{Q}}$, and a smooth projective geometrically connected curve $X$ over $K$, such that $X_{\overline{\mathbf{Q}},\sigma}$ is not isomorphic to $X_{\overline{\mathbf{Q}},\tau}$ in the category of curves over $\overline{\mathbf{Q}}$.

If yes, the genus of $X$ has to be positive.

Is there a difference between the above question and the following question?

Are there embeddings $\sigma,\tau:K\to \mathbf{C}$, and a smooth projective geometrically connected curve $X$ over $K$, such that $X_{\mathbf{C},\sigma}^{an}$ is not isomorphic to $X_{\mathbf{C},\tau}^{an}$ in the category of Riemann surfaces?

  • 1
    Consider an elliptic curve over $K$ and its $j$-invariant. Note that over $\bar{\mathbb Q}$, any elliptic curve is uniquely determined by its $j$-invariant.2011-11-28
  • 0
    Suppose that $X$ is an elliptic curve over $K$ with $j$-invariant $j$. The $j$-invariant of $X_\sigma$ equals the $j$-invariant of $X_\tau$, so they are isomorphic. Or am I missing something...? Ow the $j$-invariant of $X_\tau$ is $\tau(j)$ and the $j$-invariant of $X_\sigma$ is $\sigma(\tau)$, right? So this means that the property holds for all number fields, except for $K=\mathbf{Q}$ (of course).2011-11-28
  • 1
    Isomorphic as what? If you mean as complex manifolds, yes, such examples exist and are derived from elliptic curves with complex multiplication by non-PID's; see my blog post http://sbseminar.wordpress.com/2009/07/28/topology-that-algebra-cant-see/ . But there are enough other ways to interpret this question that I don't want to answer until I am sure what you are asking.2011-11-29
  • 0
    @Soka, yes the $j$-invariant of $X_{\bar{\mathbb Q}, \sigma}$ is $\sigma(j)$.2011-11-29

0 Answers 0