# Elliptic curves (Tate normal form?)

I basically have two question, the other question can be found below.

Let $$E/k$$ be an elliptic curve with $$P\in E(k)$$ a point of order $$\geq 4$$. Show that $$E$$ can be described by $$\begin{equation*} y^2+uxy+vy=x^3+vx^2 \end{equation*}$$ with $$u,v\in k$$ and $$P=(0,0)$$.

That's my proof so far: I consider the points $$2O-P, 3O-2P, 6O-2P$$. Using Riemann-Roch, I get $$\ell(2O-P)=1$$, $$\ell(3O-2P)=1$$ and $$\ell(6O-2P)=4$$ with basis functions $$x$$ and $$y$$ of the vector spaces $$\mathcal{L}(2O-P)$$ and $$\mathcal{L}(3O-2P)$$ where $$\operatorname{ord}_O(x)=-2$$, $$\operatorname{ord}_P(x)=1$$ and $$\operatorname{ord}_O(y)=-3$$, $$\operatorname{ord}_P(y)=2$$. Now we want to compute the divisors of $$x$$ and $$y$$. Therfore, I consider the bijection between $$E$$ and the Picard-Group: \begin{align*} \phi:E&\longrightarrow \operatorname{Pic}^0(E)\\ P&\longmapsto [P-O] \end{align*} Since the order of $$P$$ is greater or equal to 4, the points $$P$$, $$2P$$ and $$3P$$ are not equal to the point at infinity. We can use the group-isomorpism $$\phi$$ to see that $$[P-O]$$, $$[2P-2O]$$ and $$[3P-3O]$$ are not the zero-element of the Picard-Group which are the principal divisors. From this, together with the fact that $$x\in\mathcal{L}(2O-P)$$ and the degree of the divisor of $$x$$ must be zero, it follows that $$\operatorname{div} x=P+Q-2O$$ for some $$Q$$ on $$E$$. Since the divisor of $$x$$ is a principal divisor, it is \begin{align*} \phi^{-1}(0)=\phi^{-1}([P+Q-2O])=\phi^{-1}([P-O])+\phi^{-1}([Q-O])=P+Q. \end{align*} So $$P+Q=O$$, thus $$Q=-P$$ and $$\operatorname{div} x=P+(-P)-2O$$. In the same way I get that $$\operatorname{div}(y)=2P+(-2P)-3O$$. It follows that \begin{align*} \operatorname{div}(y^2)&=4P+2(-2P)-6O\\ \operatorname{div}(xy)&=3P+(-P)+(-2P)-5O\\ \operatorname{div}(x^3)&=3P+3(-P)-6O\\ \operatorname{div}(x^2)&=2P+2(-P)-4O. \end{align*} That's why $$y,y^2,xy,x^3,x^2\in\mathcal{L}(6O-2P)$$.
The vector space $$\mathcal{L}(6O-2P)$$ is of dimension $$4$$ so these functions are linearly dependent. That's why I get the equation $$$$\label{eq} -A_1y+A_2xy+A_3y^2+A_4x^2+A_5x^3=0$$$$ where not all coefficients $$A_1,\dots,A_5\in k$$ are zero. I claim that $$A_1A_3A_4A_5\neq 0$$. Suppose $$A_1=0$$. Then I get the equation $$\begin{equation*} A_2xy+A_3y^2+A_5x^3=-A_4x^2 \end{equation*}$$ where $$\operatorname{ord}_P(-A_4x^2)=2$$ if $$A_a\neq 0$$ and $$\operatorname{ord}_P(-A_4x^2)=0$$ if $$A_4=0$$. On the left hand side I get $$\operatorname{ord}_P(A_2xy+A_3y^2+A_5x^3)\geq 3$$ thus $$A_4$$ must be $$0$$. But I can't do that in the same way to show that $$A_3$$ and $$A_5$$ are not zero because that didn't work out with the orders... What do I do know? That's the only problem with this exercise, after I can show that the coefficients are not zero as claimed, I already found the variable transformation in order to get the equation as claimed. But why is $$P=(0,0)$$? Thanks for your help!

I have another question: I have to show that there is a one-parameter family of elliptic curves with a $$k$$-rational point of order $$5$$.

If the point $$P$$ has order $$5$$, it follows that $$3P=-2P$$. Do I compute these points and compare the coefficients or is there an easier way?

• Closely related, possibly even a duplicate. I'm hesitant to vote that way because my answer there is not necessarily the most helpful. I want to keep these two questions linker anyway. Sep 23, 2021 at 11:53
• I "guessed" that this is the intended equation. In other words, there were two typos. Too much of a coincidence for it not to be. Sep 24, 2021 at 13:59

The dimensions of $$\mathcal{L}(1\cdot O)$$ up to $$\mathcal{L}(6\cdot O)$$ should give you that you can write $$E/k$$ as $$y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$$ that is, in long Weierstrassform, and when we translate $$P$$ of order $$N \geq 4$$ to be at $$(0,0)$$, the point $$(0,0)$$ must be on $$E$$. This implies $$a_6 = 0$$. The order of $$P$$ is larger than two, so $$a_3$$ cannot be zero. Hence we can scale $$y$$ with $$a_4/a_3 x$$ to get $$a_4 = 0$$. The order of $$P$$ is also larger than three, hence $$a_2$$ is non-zero, so scale $$x$$ and $$y$$ to get $$a_2 = a_3$$. For more details on all of these steps, consider III.3.1.b from The arithmetic of elliptic curves by Joe Silverman. Then pick $$u$$ and $$v$$ in the way you require.
For your second question, the family of elliptic curves with a $$k$$-rational point of order 5 corresponds to the modular curve $$X_1(5)$$ which has genus zero and is hence parametrizable by one variable (the spaces $$X_1(N)$$ have genus zero for $$N = 1, \ldots, 10$$ and $$N = 12$$).
But you can do better, we can show that it is parametrizable by either $$u$$ or $$v$$ that we have just found. Namely, the division polynomial for $$E$$ for $$N = 5$$ gives us a relation that $$u$$ and $$v$$ must satisfy in order for $$(0,0)$$ to be of order 5. If you compute this, you can express $$u$$ in terms of $$v$$, which explicitly gives us the parametrization of the elliptic curves with a point of order 5.