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I'm writing a minor thesis about different criteria of supersingularity and I wanted to show the following from Husemöller's Elliptic Curves [Prop. 13.3.8]:

An elliptic curve $E$ in characteristic $p$ is supersingular if and only if the invariant differential $\omega$ is exact.

The invariant differential is $\omega= \frac{dx}{2y+a_1x+a_3}$.

The proof starts by showing this proposition for $p=2$. My problems start in the second part, where Husemöller wants to show it for $p>2$. I know that for $p\neq 2$ we can put any elliptic curve into Legendre-Form $y^2=x(x-1)(x-\lambda)\Leftrightarrow y=[x(x-1)(x-\lambda)]^{\frac{1}{2}} $. So the invariant differential becomes

$\omega=\frac{dx}{2y}=\frac{dx}{2[x(x-1)(x-\lambda)]^{\frac{1}{2}}}$.

Furthermore it is my understanding that this differential is exact iff there exists a function F satisfying $\omega=\frac{dF}{dx}\cdot dx$.

First of all it is clear to me, that $\frac{dx}{2y}$ ist exact iff $\frac{dx}{y}$ is exact.

But why is

1) $\frac{dx}{y}$ exact equivalent to $y^{p-1}dx$ exact ?

I see that one can write $\frac{1}{y}dx$ as $\frac{y^{p-1}}{y^p}dx$ and I have (with y=... put into the equation as above) integrated both functions $\frac{1}{y}$ and $y^{p-1}$ with respect to $x$ using wolfram alpha to find the searched $F$ for each differential. But I do not see this equivalence. From my point of view the antiderivatives do not have the same singularities.

2) $y^{p-1}dx$ exact equivalent to the coefficient of $x^{p-1}$ in $y^{p-1}$ being zero (I know that the latter means supersingularity)? I do not have a real idea here.

Thanks in advance for any form of help!

Kind regards SC1912

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  • 1
    $\begingroup$ I would have thought that exact means that $$\omega=dF:=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy.$$ Do observe that the equation of the elliptic curve creates a linear dependence between $dx$ and $dy$: $$y^2=x^3+a_2x^2+a_1x+a_0\implies 2y\,dy=(3x^2+2a_2x+a_1)\,dx.$$ Anyway, if $$\frac{dx}y=dF(x,y)$$ for some $F(x,y)$, what is $dG(x,y)$, where $G(x,y)=y^pF(x,y)?$ $\endgroup$ – Jyrki Lahtonen Sep 22 '13 at 17:27
  • $\begingroup$ First of all: thanks for your help! So you mean for $G(x,y)=y^pF(x,y)$ with $\frac{\partial F}{\partial x}$ it is $dG=y^p\cdot \frac{1}{y}dx=y^{p-1}$. This means $G$ exists iff $F$ exists? I would have never thought of that ;) Do you have any idea regarding 2)? Is there a possibility like in your first answer? $\endgroup$ – SC1912 Sep 22 '13 at 17:49
  • $\begingroup$ Haven't thought about 2), but I still think that you need to compute both partial derivatives, so I think you need $$\frac{\partial G}{\partial x}\,dx+\frac{\partial G}{\partial y}\,dy=y^p\left(\frac{\partial F}{\partial x}\,dx+\frac{\partial F}{\partial y}\,dy\right).$$ $\endgroup$ – Jyrki Lahtonen Sep 22 '13 at 18:43
  • $\begingroup$ (2): if the coefficient of $x^{p-1}$ in $y^{p-1}$ is zero, then $y^{p-1}=(x^3+a_2x^2+...)^{(p-1)/2}$ is a sum of $c_kx^k$ with $k\le 3(p-1)/2 < 2p-1$ and $k\ne p-1$. In this range of $k$, we have $k+1$ prime to $p$. So $$y^{p-1}dx=d(\sum_k c_k \frac{x^{k+1}}{k+1}).$$ Conversely, if $(x^3+a_2x^2+...)^{(p-1)/2}dx=dF$, write $F=f(x)+g(x)y$. We have $$dy=\frac{3x^2+a_2x+a_1}{2y}dx=y\frac{3x^2+a_2x+a_1}{x^3+a_2x^2+...}dx.$$ So $$dF=df(x)+y(dg(x)+g(x)h(x)dx), \quad h(x)\in K(x).$$ Therefore $$(x^3+a_2x^2+...)^{(p-1)/2}dx=df(x).$$ $\endgroup$ – Cantlog Oct 21 '13 at 23:36
  • $\begingroup$ Continued: But in characteristic $p$, in an exact differential form, there is no term in $x^{p-1}$ because it would come from $x^p$, but $dx^p=0$. So there is $x^{p-1}$ term in the left-hand side, and the latter is equal to $y^{p-1}dx$. $\endgroup$ – Cantlog Oct 21 '13 at 23:37

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