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