# Degree of extension of the field of coefficients of modular forms

I am beginning to study modular forms and I came across an inequality defining the bound for the degree of extension of the field of coefficients of a modular form $$f\in S_k(\Gamma_1(N))$$. This goes as follows

$$[K_f/\mathbb{Q}]\leq \text{rank}_{\mathbb{Z}}\mathbb{T}(S_k^{\text{}new}(\Gamma_1(N)))$$ where $$\mathbb{T}$$ is the algebra of Hecke operators acting on $$S_k^{\text{}new}(\Gamma_1(N))$$.

Is this true for any level and weight? If it is can anyone give or refer me to the proof of this?

• Where did you come across this inequality? And what do you mean by $\mathbb{T}$? Commented Jun 10 at 16:01
• Sorry for my unclear question earlier. I have edited it now. By \mathcal{T}, I meant the algebra of Hecke operators. You can find it here math.leidenuniv.nl/~pbruin/coefficients.pdf. It doesn't gives a proof or mentions any reference though. Commented Jun 10 at 21:06
• I'm sure there's a better way, but this implicitly follows from methods of computing modular forms using modular symbols. See for instance Algorithm 9.14 at wstein.org/books/modform/modform/newforms.html. Commented Jun 13 at 19:56

Yes, this is true regardless of level and weight, provided that $$f$$ is a normalized newform. The point is that $$T_n \in \mathbb{T}\otimes \mathbb{Q} \longmapsto a_n(f) \in K_f$$ is a surjective ring homomorphism.
It is (much) more difficult to show that $$\mathbb{T}\otimes \mathbb{Q}$$ is a finite dimensional $$\mathbb{Q}$$-algebra (in fact, that $$\mathbb{T}$$ is a finite free $$\mathbb{Z}$$-algebra), and that newforms correspond exactly to ring homomorphisms $$\mathbb{T} \rightarrow \mathbb{C}$$ (whence, for instance, the fact that the Galois-conjugate of a newform is a newform).
One way to go about this is to use modular symbols as hinted in davidlowryduda’s comments: you construct a finitely generated abelian group of modular symbols $$S := H^1_{par}(\Gamma_1(N),V_{k-2}(\mathbb{Z}))$$ and an isomorphism $$S \otimes_{\mathbb{Z}} \mathbb{C} \simeq \mathcal{S}_k(\Gamma_1(N))$$. Then you can show that under this isomorphism, the action of $$\mathbb{T}$$ on the RHS comes from an action of $$\mathbb{T}$$ on $$S$$.