5
$\begingroup$

I have an ignorant question about elliptic curves which I'll be slightly imprecise about. If I have an elliptic curve $X$ defined over $\mathbb Z$, I can base change to $\mathbb C$, and then $X(\mathbb C)$ is isomorphic to the quotient of $\mathbb C$ by the lattice generated by $1, \tau$ for some $\tau \in \mathbb C$.

I can also base change to a finite field $\mathbb F_q$, and then the Frobenius operator on $H^1 (X( \bar {\mathbb{F} }_q), \bar {\mathbb{Q} }_l)$ has eigenvalues $\sigma, \bar \sigma$.

Q: Is there any relationship between $\sigma, \bar \sigma$, and $\tau$?

(I may have glossed over some things, like fixing an identification embedding of $\bar {\mathbb Q}$ into $\mathbb C$, or requiring $X(\mathbb F_q)$ to be smooth, or something else.)

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Probably you mean "fixing an identification of $\overline{\mathbb Q}$ with a subfield of $\mathbb C$," rather than with $\mathbb C$ itself. $\endgroup$
    – KConrad
    Dec 3, 2015 at 1:58
  • $\begingroup$ Hmm, I did indeed. $\endgroup$
    – Peter Samuelson
    Dec 3, 2015 at 3:39
  • $\begingroup$ You might enjoy reading about the Igusa-Manin theorem. $\endgroup$
    – Felipe Voloch
    Dec 3, 2015 at 3:55
  • 3
    $\begingroup$ What is intrinsic is not $\tau \in \mathbf{C}$ but $\tau \in \mathbf{C}/{\rm{SL}}_2(\mathbf{Z}) \simeq \mathbf{C}$ (isomorphism by $j$), and the pair $\sigma, \overline{\sigma}$ are roots of the characteristic polynomial of Frobenius $T^2 - a_q T + q$, where $a_q = q+1 - \#X(\mathbf{F}_q)$. At a good prime the reduction of $j(X)$ is the $j$-invariant of the reduction, so your question could be seen as entirely about elliptic curves $E$ over finite fields $k$: what is the relationship between $j(E) \in k$ and $\#E(k)$? Yet a quadratic twist can change $\#E(k)$ but not $j(E)$. $\endgroup$
    – nfdc23
    Dec 3, 2015 at 6:15
  • 1
    $\begingroup$ If $j(E)=j(E')$, then $E$ and $E'$ are isomorphic over an algebraic closure of their field of definition, but not necessarily over their field of definition. This is the whole subject of "twists". See any intro text on elliptic curves. For finite fields of characteristic $p\ge5$, a curve $E$ has a unique quadratic twist, and the traces of Frobenius for the two curves are negatives of each other. $\endgroup$
    – Joe Silverman
    Dec 3, 2015 at 18:45

1 Answer 1

Reset to default
13
$\begingroup$

If $E$ does not have CM, then $\tau$ will be transcendental over $\mathbb Q$, so it's hard to imagine a relation with the eigenvalues of Frobenius, which are integers in an imaginary quadratic field. However, if $E$ does have CM, then $\text{End}(E)\otimes\mathbb Q\cong\mathbb Q(\tau)$, and Frobenius lifts to an endomorphism of $E$ that is in the ring of integers of this field, so if you choose the right value for $\tau$, then they can be very closely related. If you want more details, read any standard treatment on complex multiplication and elliptic curves.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.