I have been reading section 7 of Serre's "Quelques applications du théorème de densité de Chebotarev" (http://www.numdam.org/item/PMIHES_1981__54__123_0/), and in particular have been trying to understand the proof of Theorem 17(i) (pg.60-62) which gives the order of magnitude of the count $M_f(x)$ of non-zero Fourier coefficients upto $x$, of a non-CM modular form of weight $k \geq 2$ and level $\Gamma_0(N)$ with Nebentypus $\omega$.
For context, the idea of the proof is to consider the sets $R_0$ (respectively, $R_{1/2}$) of $f \in M_k(\Gamma_0(N), \omega)$ for which the lower asymptotic density of $M_f(x)/x$ (respectively, $M_f(x)\sqrt{log x}/x$) is zero, and show that each of these two subsets is stable under the action of the $\mathbb C$-subalgebra $\mathfrak H$ of $End_{\mathbb C} M_k(\Gamma_0(N), \omega)$ generated by the Hecke operators $T_p$ $(p \nmid N)$ and $U_p$ $(p|N)$, whereafter he shows by means of a contradiction argument that $R_0$ is precisely the set of CM-cusp forms (in the sense of Ribet) of weight $k$ and level $\Gamma_0(N)$, Nebentypus $\omega$ (henceforth denoted, $S_k(\Gamma_0(N), \omega)^{\text{cm}}$), while $R_{1/2}=0$.
In order to achieve the aforementioned contradiction, he assumes that there is some form $f \not\in S_k(\Gamma_0(N), \omega)^{\text{cm}}$ which lies in $R_0$, so that $f$ can be uniquely written as a sum $g+h$ of some nonzero $g \in S_k(\Gamma_0(N), \omega)^{\text{non-cm}} \oplus E_k(\Gamma_0(N), \omega)$ and some $h \in S_k(\Gamma_0(N), \omega)^{\text{cm}}$, with $S_k(\Gamma_0(N), \omega)^{\text{non-cm}}$ denoting the orthogonal complement of $S_k(\Gamma_0(N), \omega)^{\text{cm}}$ in $S_k(\Gamma_0(N), \omega)$ (i.e., the space of non-CM cusp forms) and with $E_k(\Gamma_0(N), \omega)$ denoting the space of Eisenstein series with usual parameters. He considers the $\mathfrak H$-module $\mathfrak Hg$ generated by $g$ and claims that it contains a simple submodule $\Sigma$, which must therefore be one-dimensional (as a complex vector space) since $\mathfrak H$ is a commutative algebra.
The last sentence is what I cannot seem to be able to see.
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First, as far as I know, nonzero modules over Artinian rings have simple submodules. If the ring is not Artinian, then even cyclic modules need not have simple submodules (as happens in the ring $\mathbb Z$). Is it known that $\mathfrak H$ is Artinian? If so, I would really appreciate a easy-to-understand reference with minimal machinery used. If not, how do we get a simple submodule of $\mathfrak H g$ in this case?
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I know that simple modules over finitely generated $\mathbb C$-algebras are one-dimensional complex vector spaces (and can see it as a consequence of the weak Hilbert Nullstellensatz). But our algebra $\mathfrak H$ here is clearly not finitely generated, right? So even we do end up with a simple submodule $\Sigma$ of $\mathfrak H g$, why is it one-dimensional?
I would really appreciate some help with these queries, they are the only places in the proof which I cannot fill in. I fear I might be missing something simple, in which case I apologize in advance. Thank you for your time.
Best Answer
The set $\mathfrak H g$ is a finite dimensional complex vector space on which the Hecke operators ($T_p$ for $p\nmid N$ and $U_p$ for $p\mid N$) act. The Hecke operators are normal and commute with each other, hence $\mathfrak H g$ has a basis consisting of simultaneous eigenfunctions of the Hecke operators. If $b$ is an element of such an eigenbasis, then $\Sigma=\mathfrak H b$ is a one-dimensional submodule that Serre talks about. This is straightforward linear algebra, no need to look at ring-theoretic properties of $\mathfrak H$.