1
$\begingroup$

Is there a general method to calculate the Hasse Invariant for any elliptic curve over any finite field?

I have read about the Hasse Invariant on page 140 in 'The Arithmetic of Elliptic Curves' but I would like some more explanation. Thanks

$\endgroup$
3
$\begingroup$

Yes, there is. In a word, if your curve is $y^2=x^3+ax^2+bx+c=f(x)$, then you look at the coefficient of $x^{p-1}$ in $f^{(p-1)/2}$. That’s the Hasse invariant, and it’s actually there in Th. 4.1(a) on p. 140.
It seems to me that there’s a shorter proof of this single fact — if you’re interested, let me know in a comment, or e-mail me.

$\endgroup$
  • $\begingroup$ Thanks for your reply, for an elliptic curve to be supersingular its Hasse invariant must be 0. Over a finite field, in order to have a supersingular curve, does the cefficient of $x^{p-1}$ have to be exactly 0 or is it 0 modulo $p$? $\endgroup$ – Junsworth Mar 18 '17 at 16:20
  • $\begingroup$ The elliiptic curve is to be defined over a field (or ring) of characteristic $p$. So there is, in effect, no “modulo $p$”. $\endgroup$ – Lubin Mar 18 '17 at 16:53
  • $\begingroup$ Oh I see, so if I am working with a field of characteristic $p$ with $p^2$ elements and calculating the coefficient of $x^{p-1}$. For the curve to be supersingular should the coefficient be exactly 0 or 0 modulo $p^2$? Thanks $\endgroup$ – Junsworth Mar 18 '17 at 17:29
  • $\begingroup$ You’re doing your calculation in a field of characteristic $p$. In your field with $p^2$ elements, all computations have $p=0$. You’re not confusing $\Bbb F_{p^2}$ with $\Bbb Z/(p^2)$, are you? $\endgroup$ – Lubin Mar 18 '17 at 18:18
  • 1
    $\begingroup$ Yes, precisely. $\endgroup$ – Lubin Mar 19 '17 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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