# How do the Dim($K_f / \mathbf{Q}$) vary for all f in a given $S_k(\Gamma(N))$?

For the theory of classical modular forms, the space of new forms $S_k^{new}(\Gamma(N))$ has a basis of Hecke eigenforms $\{ f_i = \sum a_n q^n : a_1=1, a_n \in \bar{\mathbb{Q}}\}$

Given $k$ and $N$ (I'll take answers restricted to $\Gamma_0(N)$ or $\Gamma_1(N)$)

1) can we say anything about $\{dim_{\mathbf{Q}}[K_{f_i} : \mathbf{Q}] : f_i \}$, where $K_{f_i} = \mathbf{Q}[ a_n : n \geq 1 ]$? (Are they all equal? different? Are the fields themselves equal?)

2) Can we know an explicit $L = \cup K_{f_i}$ apriori? (My guess is that it's related to the defining equation of the projective curve $X(\Gamma)$).

We know that, for each $i$, the dimension of $K_{f_i}$ is finite, but what can we say about all of the $f_i$ collectively?

Simplified Question: I know that, via the modularity of rational elliptic curves, If $E/\mathbf{Q}$ is an elliptic curve of conductor N, then it's associated modular form $f \in S_2(\Gamma_0(N))$ has $dim_{\mathbf{Q}}[K_{f}:\mathbf{Q}] = 1$. Is it true that all of the $f_i$, new eigenforms, in $S_2(\Gamma_0(N))$ have $K_f = \mathbf{Q}$?

• If you want my advice, learn how to use the free software "sage" and go ahead and compute just one or two examples of these fields (the manual explains how to do it), and you will instantly see that about half of what you suggest is not right, and you will probably learn a lot. Your question makes it quite clear that you've never computed any examples of these things ever, and it's really easy to do so nowadays,
– eric
Nov 13, 2015 at 21:35

The answer to your last question is no. In general, if $K_f/\mathbb Q$ has degree $d$, then there is an associated factor of $J_0(N)$ defined over $\mathbb Q$ of dimension $d$. This is all in Shimura's Arithmetic Theory of Automorphic Forms, I think.
For your question (2), the curves $X_0(N)$ and $X_1(N)$ are defined over $\mathbb Q$, so their fields of definition have nothing to do with the field that you're calling $L$.
Finally, it is known that the number of isogeny classes of elliptic curves of conductor $N$ is less than $O(N^{1/2+\epsilon})$ (and results of Pierce, Helfgott, and Venkatesh reduce the exponent at bit). Since each such isogeny class appears as an elliptic factor of $J_0(N)$ (thanks to Wiles et.al.), and each elliptic factor gives a cusp form with coefficients in $\mathbb Q$, it follows that there are at most $O(N^{1/2+\epsilon})$ normalized cusp forms with coefficients in $\mathbb Q$. But the total number of normalized cusp forms with $\overline{\mathbb Q}$ coefficients is the genus of $X_0(N)$, which is "roughly" $N/12$. Conclusion: Most normalized cusp forms for $\Gamma_0(N)$ do not have coefficients in $\mathbb Q$.