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?



Alternate expresion of L-series coefficients
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.


