Weierstrass form of elliptic curve with point with order larger than 3

L.S.,

Studying for my exam on elliptic curves, I tried to make exercise 8.13(a) of Silvermans "The Arithmatic of Elliptic Curves", which reads:

Let $E$ be an elliptic curve defined over a field $k$, and let $P \in E(k)$ be a point of order at least 4. Prove that there is a change of coordinates such that $E$ is described by Weierstrass equation $y^2 + uxy + vy = x^3 + vx^2$, with $u,v \in k$. and $P = (0,0)$.

I tried the following:

I was hoping that Riemann-Roch can help here. We know that, in notation of the R-R theorem, $g = 1$, so for some principal divisor $D$ we have that $l(D) \leq 2$. Now I would like to exploit the fact that we know that there is a point of order larger than 3. Let $f$ be the linear function that defines the line tangent at $P$. We know that $f$ cannot intersect the point at infinity $\mathcal{O}$, because than we would have $2P = \mathcal{O}$. Also, this line it must go through $P$ with multiplicity $2$, because it cannot intersect with multiplicity 3 since then $3P = \mathcal{O}$. Now denote $g$ the function that describes the line through $P$ and $O$. With the same reasoning, we have that $g$ has multiplicity precisely 1 at $P$, and multiplicity at most 2 at $\mathcal{O}$. Now we have that $F = \frac{f}{g}$ has order 1 at $P$ and order -2 or -1 at $\mathcal{O}$. So for divisor $D = P - 2\mathcal{O}$ we know that $F \in L(D)$, which is at most 2 dimensional. Now I am stuck, since I don't know how to exploit this any further, I was hoping that maybe there is some other function $G$ of which we also know it is in $L(D)$, and then we should have automatically some relation between $F$ and $G$.

I really hope someone could help me! Many thanks!

Willem

The usual application of Riemann-Roch allows us to find $x$ and $y$ such that they have respective pole divisors $$(x)_\infty=2\mathcal{O},\quad (y)_\infty=3\mathcal{O},$$ and that they are scaled in such a way that $x^3-y^2$ has a pole of order at most five. Because $P\neq\mathcal{O}$ we can replace $x$ with $x-x(P)$ and $y$ with $y-y(P)$ to move the origin to $P$. At this point the equation of the curve takes the form $$y^2+Axy+By=x^3+Cx^2+Dx\qquad(*)$$ as we killed the constant term by translating the origin to $P$.
The next task is to make $D=0$. The first thing that comes to mind is the substitution $$y=\tilde{y}+tx,$$ where $t\in k$ is a constant that we try to choose smartly, and $\tilde{y}$ would be the new $y$-coordinate. Observe that such a substitution won't disturb the above assumptions about the pole orders of $y^2$ and $x^3-y^2$. We arrive at the equation $$\tilde{y}^2+(A+2t)x\tilde{y}+B\tilde{y}=x^3+(C-t^2-At)+(D-Bt)x.$$ So we are successful with the choice $t=D/B$ unless we have $B=0$. But should we have $B=0$ in $(*)$ it follows that $P=(0,0)$ would have order two. Geometrically this amounts to picking the $x$-axis as the tangent line to $P$. Renaming coefficients we get an equation of the form $$y^2+Axy+By=x^3+Cx^2.\qquad(**)$$ How to get $B=C$? We can still make the substitution $y\leftarrow r^3y$, $x\leftarrow r^2x$, $r\neq0$. This results in the changes $$A\leftarrow A/r,\quad B\leftarrow B/r^3,\quad C\leftarrow C/r^2.$$ So we get the desired $B/r^3=C/r^2$, if the choice $r=B/C$ makes sense. We already saw that we must have $B\neq0$, so the remaining problem is whether we can have $C=0$. But if $C=0$ in $(**)$, then $y$ has a zero of order three at $P$ violating the assumption that $P$ cannot have order three.
• Why with $C=0$ we derive that $y$ has a zero of order $3$ at $P$? Sorry for the silly question but I am new at Elliptic Curves. – Cornelius Mar 8 at 15:18