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I was hoping that someone could help clarify a source of confusion for me, I must be doing and saying something wrong but I just don't know what: Let $E$ be an elliptic curve over $\mathbb{Q}$ and let $$L(s,E)=\sum_{n=1}^{\infty}a_n(E)n^{-s}$$ be the Hasse-Weil $L$-function of $E$. Finally, let $\tilde{E}$ be the reduction of $E$ mod $p$ and assume that $p$ is a prime for which $E$ has good reduction.

Then $$a_p(E)=p+1-|\tilde{E}(\mathbb{F}(p))|$$ and setting $a_1(E)=1$ the $p$ power coefficients are given by $$a_{p^e}(E)=a_p(E)a_{p^{e-1}}(E)-pa_{p^{e-2}}(E).$$

Now looking at Diamond and Shurman, for instance, I find that also we can write $$a_{p^e}(E)=p^e+1-|\tilde{E}(\mathbb{F}(p^e))|$$ but when I use this expression as a "definition" of $a_{p^e}(E)$ and do some explicit calculations I don't get the right recursion, for instance I seem to get in practice $$a_{p^2}(E)=a_p(E)^2 - 2p$$ instead of $$a_{p^2}(E)=a_p(E)^2-p.$$

I must be misunderstanding something, but I can't figure out what. Any help?

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I am sorry, I do not understand the "explicit calculations" you did for getting the last two expressions. Could you provide some detail or some reference? –  Anweshi Jan 15 '10 at 22:26
2  
Sure sorry I should have done that before. I have been using magma, my error could be there as well. Take the curve $$E: y^2+xy+y=x^3$$ and count its points over $F_7$ and $F_{7^2}$. Then calling magma I seem to get $$7+1-|E(\mathbb{F}(7))|=-1$$ $$7^2+1-|E(\mathbb{F}(7^2))|=-13$$ and $-13=(-1)^2-2*7$ but the $L$ series coefficient is $-6$. Here are my magma commands: > E1; Elliptic Curve defined by y^2 + xy + y = x^3 over GF(7) > E2; Elliptic Curve defined by y^2 + xy + y = x^3 over GF(7^2) > 7+1-Order(AbelianGroup(E1)); -1 > 7^2+1-Order(AbelianGroup(E2)); -13 Does that clarify? –  jude Jan 15 '10 at 22:44
    
I don't believe that $a_{p^e}(E) = p^{e+1} + 1 - |E(\mathbb{F}(p^e))|$. Computing with the characteristic polynomial of Frobenius gives that $|E(\mathbb{F}(p^2))| = p^2 + 1 - a_p^2 + 2p$. –  Martin Orr Jan 15 '10 at 23:41
    
I guess what confuses me then is p.360 of Diamond and Shurman. –  jude Jan 16 '10 at 0:14

2 Answers 2

up vote 3 down vote accepted

There are two different recursions involved here, one for the points of $E$ over ${\mathbb F}\_{p^n}$, and the other for the coefficients of the $L$-function.

If we write $a_p = \alpha + \beta,$ where $\alpha\beta = p$ (so $\alpha$ and $\beta$ are the two roots of the char. poly. of Frobenius), then

$$1 + p^n - E({\mathbb F}\_{p^n}) = \alpha^n + \beta^n.$$

On the other hand, the Euler factor at $p$ for the $L$-function of $E$ is $$(1 - \alpha p^{-s})^{-1}(1-\beta p^{-s})^{-1}$$ $$= (1 + \alpha p^{-s} + \alpha^2 p^{-2s} + \cdots )(1 + \beta p^{-s} + \beta^2 p^{-2s} + \cdots )$$ $$= 1 + (\alpha + \beta) p^{-s} + (\alpha^2 + \alpha\beta + \beta^2) p^{-2s} + \cdots ,$$ and so we conclude that $a_{p^n}$ (the coefficient of $p^{-ns}$ in the $L$-function) equals $$\alpha^n + \alpha^{n-1} \beta + \cdots + \alpha\beta^{n-1} + \beta^n.$$

These formulas are simply different, as soon as $n > 1.$ The recursion given in the question describes the second, and not the first.

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Yes, this is right. Note that in my answer I was working only with an elliptic curve over a finite field. –  Pete L. Clark Jan 16 '10 at 1:10
    
On the other hand, I think I am also right! On p. 361, Diamond and Shurman define $a_{p^e}(E) = p^e + 1 - |E(F_{p^e})|$, or equivalently as the trace of $p^e$-Frobenius. Then they write down the recursion that Jude mentions in his question. So they are indeed mistaken, I believe. –  Pete L. Clark Jan 16 '10 at 1:15

The mistake is acknowledged and corrected in the errata for the third printing:

http://people.reed.edu/~jerry/MF/mferrata3.pdf

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Thanks guys! I had looked for errata but I must have only looked at the errata for the first printing, oops :-\ –  jude Jan 16 '10 at 1:58
    
I am inclined to think that there should be just one file containing all the errata. –  Pete L. Clark Jan 16 '10 at 3:59

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