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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

Junsworth
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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.

Lubin
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  • 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 $p$? – Junsworth Mar 18 '17 at 16:20
  • The elliiptic curve is to be defined over a field (or ring) of characteristic $p$. So there is, in effect, no “modulo $p$”. – Lubin Mar 18 '17 at 16:53
  • 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 – Junsworth Mar 18 '17 at 17:29
  • 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? – Lubin Mar 18 '17 at 18:18
  • Let me give an example for a bit more context. Let's say that I'm trying to determine whether $y^2=x^3-33x+1$ over $F_{11^2}$ is supersingular. I have calculated the coefficient of $x^{10}$ in $(x^3-33x+1)^5$ to be -660. Do I need to apply any modulo arithmetic to my calculations? Sorry for my confusion.. I am new to elliptic curves. – Junsworth Mar 18 '17 at 21:08
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    Over $\Bbb F_{121}$, your curve is $y^2=x^3+1$. I see that you really are confusing $\Bbb F_{p^2}$ with arithmetic modulo $p^2$. Calling $\Bbb F_{11}=k$, we notice that $2$ is not a square in $k$, so that you can get $\Bbb F_{11^2}$ as $k(\sqrt2,)$. You should play around in this verdant field. – Lubin Mar 19 '17 at 12:31
  • Hmm, ok. So over a field $F_{p^2}$ calculations are done modulo $p$? – Junsworth Mar 19 '17 at 14:03
  • Over any field that contains $\Bbb F_p$, calculations are done modulo $p$, because $1+1+\cdots+1$ ($p$ times) is equal to zero there. – Lubin Mar 19 '17 at 18:33
  • Ok I understand, so for my example that I stated above, $y^2=x^3-33x+1=x^3+1$ since $F_{11^2}$ contains $F_{11}$ . Therefore $y^2=x^3+1$ is supersingular since the coefficient of $x^{10}$ in $(x^3+1)^5$ is $0$? – Junsworth Mar 19 '17 at 19:01
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    Yes, precisely. – Lubin Mar 19 '17 at 19:03
  • Perfect, thank you for your help Lubin! – Junsworth Mar 19 '17 at 19:04