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If f is a weight 2 newform on $\Gamma_1(N)$ then there exists an abelian variety Af whose endomorphism algebra is isomorphic to the field generated by the coefficients of f.

I've seen this proven in Shimura's book, but was wondering if anyone knows of a different reference (perhaps one that is a bit more readable...).

Thanks.

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  • $\begingroup$ MAy I ask, how did you find the book of shimura? Especially for a beginner in automorphic functions.. I have the book, I have read a chapter of it, but then stopped as this is not really my research area. But somehow, I get the impression that the approach of the book is not a standard one.. at least not for a beginners (though for me, it was still quite readable) $\endgroup$
    – Jose Capco
    Commented Oct 31, 2009 at 21:08
  • $\begingroup$ It was the reference that Ribet gave in his paper 'Endomorphism algebras of abelian varieties attached to newforms of weight 2'. $\endgroup$
    – user1073
    Commented Oct 31, 2009 at 21:29
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    $\begingroup$ Shimura is the standard reference. Also he is the prover of all this stuff and creator of many streams of thought in the subject. Unfortunately his book is unreadable. See my answer below and the comments along with it. $\endgroup$
    – Anweshi
    Commented Jan 4, 2010 at 14:49

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Have a look at Section 6.6 of Diamond and Shurman, A First Course in Modular Forms:

As an aside, the theorem states a bit more than you have said: for instance, when the field of Fourier coefficients is $\mathbb{Q}$, you are just asserting the existence of an elliptic curve $E_{/\mathbb{Q}}$ with $\operatorname{End}_{\mathbb{Q}}(E) \otimes_{\mathbb{Z}} \mathbb{Q} = \mathbb{Q}$: every elliptic curve over $\mathbb{Q}$ has this property. You want to require an equality of L-series between the abelian variety and the modular form.

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Just to add one more thing to what Pete said: the variety A_f that one normally attaches to f might have endomorphism ring bigger than an order in the coefficient field of f: for example if E is an elliptic curve with complex multiplication, the associated modular form still only has coefficients in Z. The correct statement is that the endomorphism ring contains an order in the coefficient field

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  • $\begingroup$ Right, but I said End_Q(E) -- by which I meant (but didn't explain: sorry!) the subring of endomorphisms defined over Q. This is indeed always equal to an order in the number field generated by the Fourier coefficients. The full endomorphism ring may well be larger, as happens in the elliptic curve case if(f) there is complex multiplication. $\endgroup$
    – Pete L. Clark
    Commented Nov 4, 2009 at 15:10
  • $\begingroup$ Yes Pete---apologies, you're dead right. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 20:24
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The Anterp conference volumes, "Modular functions in one Variable - I, II, III, .... " might contain what you want. I am not sure though, as I am unable to verify it by looking into all the volumes.

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  • $\begingroup$ I really like the proceedings of the Antwerp conference and have found many of the lectures very useful, especially Ribet's (on the Galois Representation associated to a newform) and Eichler's (on the Basis problem). I'll have to check and see if this is covered in one of the volumes. $\endgroup$
    – user1073
    Commented Jan 4, 2010 at 13:23
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    $\begingroup$ I saw you complaining about Shimura and that is why I pointed this one out. Diamond and Shurman is "too soft" in a sense. But Shimura uses the hopelessly old language for algebraic geometry(Weil's foundations), and is unreadable. Hida has rewritten all that Grothendieck-style in his "Geoemtric modular forms" book; but if anything that is even more of a mess to figure out and is completely unreadable. The Antwerp volumes are by far the best I know of, though a bit old. $\endgroup$
    – Anweshi
    Commented Jan 4, 2010 at 14:43
  • $\begingroup$ I make apologies in advance for the polemics. But every person has a personal taste, and this is mine. There are people who don't like Shakespeare, though he is very reputed. If you have objections to me, please ignore me as a philistine like one of those guys. $\endgroup$
    – Anweshi
    Commented Jan 4, 2010 at 14:43
  • $\begingroup$ +1 for explaining why you don't like Shimura's text, it certainly sounds like that would make it significantly more difficult of a read. $\endgroup$
    – Sean Tilson
    Commented Jul 31, 2010 at 17:14

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