Let $(E,\mathcal{O})$ be the elliptic curve of equation $$ f=Y^{2}+a_{1}XY+a_{3}Y-X^{3}-a_{2}X^{2}-a_{4}X-a_{6}, $$
$\alpha:K(E)\rightarrow K(E)$ the derivation such that $$ \alpha(X)=\frac{\partial f}{\partial Y},\quad \alpha(Y)=-\frac{\partial f}{\partial X} $$ and $m$ a positive integer. I have to prove that, if we are working in a field $K$ of characteristic $\geq m+1$, and $\mathcal{L}(m\mathcal{O})=\langle 1,f_{1},\ldots,f_{m-1}\rangle$, then the $m$-torsion points of $E$ different from $\mathcal{O}$ coincide with the zeros of $$ \det (\alpha^{i}(f_{j}))_{1\leq i,j\leq m-1}. $$ By the way, what can be said about the number of zeros of that determinant?
$\textbf{Edit:}$ In fact, that determinant vanishes at $p\in E-\{\mathcal{O}\}$ if and only if (see the comments that follow mercio's answer) there exists $g\in\mathcal{L}(m\mathcal{O})$ such that $\alpha(g)|_{p}=\cdots=\alpha^{m-1}(g)|_{p}=0$. This reduces the problem (because of mercio's answer again) to show that this is equivalent to $g$ having a zero of order $m$ at $p$. Is this true?
Any hint would be appreciated.
Let $P$ be a point of $E$ different from $\mathcal O$.
the determinant vanishes at $P$ if and only if there is a linear combination of the $f_j$ (which means a function with poles of order at most $m$ at $\mathcal O$), such that its first $m-1$ derivatives at $P$ all vanish.
By adding a suitable constant to that function if necessary, this means you have a function with a pole of order $m$ at $\mathcal O$ and a zero of order $m$ at $P$ (and nothing else because you can't have any additional pole)
But this means exactly that $m(\mathcal O - P)$ is a principal divisor, so that once you fix $\mathcal O$ as the origin of the group law, $P$ is a point of order $m$.