2
$\begingroup$

In the paper "Links between Stable elliptic curves and Diophantine equations" for an elliptic curve $E$ with normal Weierstrass form $$y^2 = x^3 -g_2x -g_3$$ with $g_i \in \mathbb{Z}$ w.l.o.g. Then he states the Hasse invariant $\delta$ for $E$ is defined as $$\delta_E := -\frac{1}{2} g_2 g_3^{-1} \: (mod \: \mathbb{Q}^{*2}) .$$

In all other references to the Hasse Invariant of elliptic curves, Arithmetic of Elliptic Curves by Silverman etc., it is stated that for an elliptic curve over a finite field the invariant is either $0$ or $1$.

Furthermore later in the paper Frey begins to talk about the extension of $\mathbb{Q}$ by the square root of the Hasse invariant i.e. $ \mathbb{Q}(\sqrt{\delta_E})$ where $E$ is the 'Frey curve.' In one case this is trivial but in other not.

Are these two seperate quantities that I am getting confused, or is there additional structure behind the Hasse invariant which I have not encountered?

$\endgroup$
1
  • 1
    $\begingroup$ Two different Hasse invariants. $\endgroup$
    – Lubin
    Commented Mar 19, 2015 at 0:29

1 Answer 1

1
$\begingroup$

This is one of those subjects where I used to know a lot more than I do now; and I wasn’t an expert even then. About the Hasse invariant that Frey is talking about, if I ever knew anything, I forgot it entirely. But I do (did) know something serious about the Hasse invariant $H(E)$ for an elliptic curve $E$ in characteristic $p$.

Joe Silverman simplified the story considerably for these Hasse invariants. In fact, $H(E)$ is an element of the base well-defined up to $(p-1)$-th powers. If you set the weight of $g_2$ to be $4$ and of $g_3$ to be $6$, then the H-invariant of $y^2=x^3-g_2x-g_3$ will be a polynomial in which every monomial has weight $p-1$. For instance, in characteristic $5$, it’s $3g_2$, in characteristic $7$, it’s $4g_3$, and in characteristic $13$, it’s $2g_3^2 + 6g_2^3$. I think I recall that N. Katz has a way of looking at this according to which $H(E)=E_{p-1}$, the Eisenstein series. I’ll have to beg one of the better-educated correspondents here to fill in the details.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .