I'm reading ‘Rational Points on Elliptic Curves’ by Silverman and Tate, and the exercise 4.6 is about the following special case of the Eichler–Shimura theorem. Let $C$ be the elliptic curve given by the equation $C: y^2 = x^3-4x^2+16$ and let $F(q)$ be the formal power series given by the infinite product $F(q) = q \prod_{n=1}^\infty (1-q^n)^2(1-q^{11n})^2$. Let $N_n$ be the coefficient of $q^n$ in $F(q)$, so $F(q)=\sum_{n=1}^\infty N_n q^n$. The exercise is to prove that $M_p+N_p=p$, where $M_p$ is the number of points on $C$ over the field $\mathbb{F}_p$. So, what is the idea of proof? Because I have no ideas how $M_p$ and $N_p$ can be connected.
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$\begingroup$ This question is double-starred in the exercise. This is of the same difficulty as Eichler-Shimura. $\endgroup$– François BrunaultCommented Oct 15, 2023 at 17:16
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1$\begingroup$ Read Iwaniec's "topics" book (bookstore.ams.org/gsm-17), especially Chapter 8. Note also that L-functions of modular forms are more fundamental than L-functions of elliptic curves (although many colleagues would disagree). $\endgroup$– GH from MOCommented Oct 15, 2023 at 17:47
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5$\begingroup$ I believe the modularity of this elliptic curve is well explain in these notes. The curve in question here is isomorphic to $y^2+y=x^3-x$, which is shown to be the equation of $X_1(11)$. You will see this is really a bit more difficult than the level of Silverman-Tate, yet the notes give a good introduction into the subject. $\endgroup$– Chris WuthrichCommented Oct 15, 2023 at 22:28
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2$\begingroup$ @Dendrit Eichler-Shimura tells you that the $L$-function of the modular form $F(q)$ is equal to the $L$-function of the elliptic curve $X_0(11)$. This elliptic curve is isogenous to $X_1(11)$, which is itself isomorphic to $C$ (on a historical note, an equation of $X_0(11)$ has been determined by Fricke around 1920). This gives you the result. $\endgroup$– François BrunaultCommented Oct 16, 2023 at 14:11
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2$\begingroup$ @Dendrit This is explained in the notes linked to by Chris Wuthrich. $\endgroup$– François BrunaultCommented Oct 16, 2023 at 19:38
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