The usual modular parametrization says that if one takes a modular form $f$ which is a newform for $\Gamma\_0(N)$ (i.e. a form in the new subspace which is a normalized eigenfunction for the Hecke operators), then one can associate to $f$ an abelian variety $A\_f$ of dimension $d$, where $d$ is the degree of $K$ over $\mathbb Q$, where $K = \mathbb Q[a\_n]$, being $a\_n$ the Fourier coefficients of $f$ at the infinity cusp. The construction goes as follows, to $f$ one attachs the module (over $\mathbb Z$) 


$$I\_f = (T \in Hecke\_{\mathbb Z} \quad : \quad Tf =0)$$
(sorry but the usual latex bracket didn't work in the previous formula)


Then take the quotient of $J\_0(N)$ by the image of $I\_f$ (i.e. $J\_0(N)/I\_fJ\_0(N)$). My question is the following: if one considers $\mathbb Z\_K$ the ring of integers of $K$ and defines $I\_f$ in the same way but looks at the module generated over $\mathbb Z\_F$ (which distinguishes between $f$ and its Galois conjugates), and then take the quotient of $J\_0(N)$ by this ideal, do one gets an elliptic curve over $K$?


If not, I have a second question, if one considers the image of the Abel-Jacobi map for $f$, i.e. integration over all the homology of $X\_0(N)$ of the form $f(z)dz$, is the image of such map a lattice in $\mathbb C$? (I would expect to get the lattice which coincides with the first question if true). I know that if you consider all the conjugates of $f$ then you get an abelian variety over $\mathbb Q$, and I wonder if it is the restriction of scalars of an elliptic curve over $K$.



Question on the abel map and modular parametrization
No, the abelian variety $A\_f$ is typically absolutely simple, and hence does not factor
(even up to isogeny, and even over $\overline{\mathbb Q}$ or $\mathbb C$) as a product of elliptic curves. What will happen is that if you integrate $f(z)dz$ over the homology of $X\_0(N)$, you will get a finitely generated, but *non-discrete*, $\mathbb Z$-submodule of $\mathbb C$. *If* $f$ has coefficients in $\mathbb Q$,
then miraculously this submodule will actually be a lattice. If $f$ has cofficients over $K$ of degree $d$, and you integrate all the conjugates of $f$ to get a $\mathbb Z$-submodule
of $\mathbb C^d$, then again you miraculously get a lattice!


