TITLE: Is there an algebraic curve over Q which is not modular?
QUESTION [28 upvotes]: Every elliptic curve $E/\mathbf Q$ is modular, in the sense that there exists a nonconstant morphism $X_0(N) \to E$ for some $N$. 
It is tempting to extend this definition in a naïve way to an arbitrary projective curve over $\mathbf Q$; if $Y$ is such a curve, we might say that $Y$ is modular if there exists a nonconstant morphism $X_0(N) \to Y$ for some $N$. A necessary condition for $Y$ to be modular is that it should have at least one rational point, since $X_0(N)$ always has a rational point. For an elliptic curve, this condition is satisfied by definition. 
There are certainly a great deal of curves which are modular in this sense. But is there an example of a curve (with a rational point) which is not modular?

REPLY [21 votes]: It is conjectured that there are only finitely many curves over $\mathbf{Q}$ of given genus $g \geq 2$ which are covered by a modular curve, see Conjecture 1.1 in the following paper :
Baker, M. H. ;  González-Jiménez, E. ;  González, J. ;  Poonen, B. . Finiteness results for modular curves of genus at least 2. Amer. J. Math.  127  (2005),  no. 6, 1325--1387.
In fact, the authors prove a strong result towards this conjecture, namely that there are only finitely such curves which are new (in an appropriate sense).