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258 13. Elliptic Curves over Finite Fields
(2.2) Proposition. For N m = #C(k m ), where C is an algebraic curve over k 1 ,the
zeta function satisfies the following relation
∞
N m −ms
ζ C (s) = exp q .
m
m=1
m
(2.3) Remark. Since N m ≤ 1+q +q 2m = #P 2 (k m ) for all m, the previous formula
for the zeta function shows that it is a convergent series or convergent product for
Re(s)> 2. In fact, we can do much better and show that it is a rational function with
poles at s = 0and s = 1, and its zeros are on the line Re(s) = 1/2.
Returning to the case of an elliptic curve E, we know that
m m m m −m
N m = 1 + q − Tr(π ) = 1 + q − α − α ,
where α and ¯α are the two imaginary conjugate roots of the characteristic polynomial
det(1 − π E T ) as an element of End(E). There is also another interpretation where α
and ¯α are the eigenvalues of the inverse of Frobenius endomorphism acting on any
Tate module V (E) where is any prime number different from the characteristic p.
This follows from the fact that the inverse of Frobenius has the same characteristic
polynomial on V (E). Using the above expression for N m , we calculate the log of
the zeta function
∞
m m m −ms
log ζ C (s) = (1 + q − α −¯α )q
m=1
∞ ∞ ∞ ∞
−ms −m(s−1) −s m −s m
= q + q − (αq ) − (¯αq )
m=1 m=1 m=1 m=1
=− log(1 − q −s ) − log(1 − q 1−s ) + log[(1 − αq −s )(1 −¯αq −s )].
Hence the exponential is the zeta function as a rational function of q −s
1 − (α +¯α)q −s + q 1−2s f E (q −s )
ζ C (s) = −s 1−s = −s −s ,
(1 − q )(1 − q ) (1 − q )(1 − q · q )
2
where f E (T ) = det(1 − π E T ) = 1 − (α +¯α)T + qT = (1 − αT )(1 −¯αT ).
(2.4) Remark. From the above calculation we see that the two definitions (1.5) and
(2.1) for the zeta function of an elliptic curve yield the same function.
Further, the above discussion gives the framework for describing the result in the
general case of any curve.
(2.5) Theorem. Let C be a smooth projective curve of genus g over F q = k m for
m
m = 1, and let N m = #C(k m ). Then there are algebraic integers α 1 ,α 2 ,... ,α 2g
√
with |α j |= q and