Why does the coefficient field of a normalized Hecke form a number field?

Question

Let $f \in S_k(N,\chi)^{\mathrm{new}}$ be a normalized Hecke cusp form (i.e $a_1(f)=1$ and $f$ is an eigenform under Hecke actions). My question is, why is the coefficient field $K_f=\Bbb Q(\{a_n(f)\}_{n \geq 1})$ a number field (without using the deep fact that $f$ gives a Galois representation)? It's easy to see $K_f$ is algebraic. Many notes always state this result but don't give a proof.

Answer 1

Consider the integral Hecke algebra $T = \Bbb Z[T_n : n \geq 1]$, which is a commutative $\Bbb Z$-subalgebra of $A := \mathrm{End}_{\Bbb C\text{-lin.}}\left[ S_k(\Gamma) \right]$, where $\Gamma := \Gamma_1(N)$.

Lemma. The algebra $T$ is finite over $\Bbb Z$ (i.e. its additive group is a finitely generated $\Bbb Z$-module).

Before providing the proof of this lemma, here is how it solves your problem. Since $f$ is a normalized eigenform, we can consider the map $$\lambda : T_n \mapsto a_n(f) = \lambda_n(f)$$ such that $T_n(f) = \lambda_n(f) f$.

It provides a morphism of $\Bbb Z$-algebras $T \to \Bbb C$. The image of $\lambda$ is therefore a finitely generated subgroup of $(\Bbb C, +)$. This implies that the field generated by $\mathrm{Im}(\lambda)$ is not only an algebraic extension of $\Bbb Q$ (see prop. 5.27 for a direct proof when $N=1$), but is actually a number field (it is generated by finitely many algebraic elements).

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Sketch of the proof of the lemma.

It relies on the fact that $S_k(\Gamma_1(N))$ is finite dimensional over $\Bbb C$, so $A \cong M_n(\Bbb C)$ (for some $n \geq 1$) is a finitely generated (non-commutative) $\Bbb C$-algebra — it is even finitely generated as $\Bbb C$-vector space. But here we want to show that $T$ is a finitely generated module over $\Bbb Z$.

The $q$-expansion at $i\infty$ provides an injective map of $\Bbb C$-vector spaces $j : S_k(\Gamma) \to \Bbb C[\![q]\!]$. Define $S_k(\Gamma, \Bbb Z) := j^{-1}(\Bbb Z[\![q]\!])$ as the subgroup of cusp forms with integer Fourier coefficients.

The lemma now follows from the following three facts (which I simply quote — when $k=2$ or $N=1$, one can provide easier arguments, I guess):

1. The $\Bbb Z$-module $S_k(\Gamma, \Bbb Z)$ is preserved under the action of any Hecke operator $T_n$. [See Diamond–Im, Proposition 12.4.1].

2. The $\Bbb Z$-module $S_k(\Gamma, \Bbb Z)$ contains a basis of the $\Bbb C$-vector space $S_k(\Gamma)$. [See Diamond–Im, Proposition 12.3.8].

3. We have a natural isomorphism of $\Bbb C$-vector spaces $S_k(\Gamma, \Bbb Z) \otimes_{\Bbb Z} \Bbb C \cong S_k(\Gamma)$. [See Diamond–Im, Equation 12.3.5].

Indeed, the first fact implies the existence of a morphism of rings (i.e. $\Bbb Z$-algebras, possibly non-commutative) $$g : T \to A' := \mathrm{End}_{\Bbb Z\text{-lin.}}\left[ S_k(\Gamma, \Bbb Z) \right],$$ and the second fact ensures injectivity of $g$.

By the second and third facts, the $\Bbb Z$-module $S_k(\Gamma, \Bbb Z)$ is a free of finite rank, say $\Bbb Z^m$, it follows that $A' \cong \Bbb Z^{m^2}$ as abelian groups, and therefore $(T,+)$ is a finitely generated abelian group (see exercise 6.5.1 in Diamond–Shurman, p. 239).