2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

I would try something like the following.

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.

$\endgroup$
  • $\begingroup$ Perfect! thank you very much :) $\endgroup$ – Willem Beek Jan 4 '16 at 10:38
  • $\begingroup$ 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. $\endgroup$ – Cornelius Mar 8 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.