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


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.

  • $\begingroup$ 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
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    $\begingroup$ Yes, precisely. $\endgroup$ – Lubin Mar 19 '17 at 19:03

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