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
\begin{equation}\label{eq}
 -A_1y+A_2xy+A_3y^2+A_4x^2+A_5x^3=0
 \end{equation}
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?
 A: For your first question, I'll repeat the argument from Lemma 2.1 from https://arxiv.org/pdf/1503.08127.pdf over here:
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.
More on this can also be found in section 2.3 of https://ia.cr/2020/1108
