# Weierstrass Equation of an Elliptic Curve

I am trying to prove that any elliptic curve E over a field $$K$$ has a corresponding Weierstrass equation by proving that there is an isomorphism \begin{align*} \phi: &E \rightarrow \mathbb{P}^2\\ & P \mapsto [x(P),y(P),1] \end{align*} such that $$\phi(O)=[0,1,0]$$, where $$O$$ is the point at infinity and $$P$$ is a point on our elliptic curve. I understand the idea of the proof: we start by considering the space of function on $$E$$ that are given by $$\mathscr{L}$$ spaces, this is, $$\mathscr{L}(n(O))$$ for $$n \geq 1$$. We then contrust the basis for these spaces and we reach one ($$\mathscr{L}(6(O))$$ with the elements $$1,x,y,x^2,xy,x^3,y^2$$) where we have 7 elements in a 6 dimension space.

Therefore we can find a linear relation between these elements given by: $$\begin{equation*} A_1+A_2x+A_3y+A_4x^2+A_5xy+A_6x^3+A_7y^2=0, \end{equation*}$$ for any $$A_i \in K$$. What I don't understand is why $$A_6A_7 \neq 0$$. Can anyone give me a brief explanation?

Any help is very much appreciated. Thank you!

• Assume for example that $A_6=0$. Then, you have that $$A_1+A_2x+A_3y+A_4x^2+A_5xy+A_7y^2=0$$. But now you have a linear combination of $6$ linearly independent elements equal to $0$, so this equality can only hold only when $A_1=\dots=A_5=A_7=0$. Similarly, if $A_7=0$, if any linear combination of the other $6$ elements sums to $0$, by linear independance, you must have that all the other coefficients are also $0$. Hence, the only way to have such a combination sum to $0$ for any $A_i$, is if $A_6A_7\neq 0$ Commented Jul 28, 2023 at 11:06
• @Fotis thank you so much! I now can see it clearly :)
– babu
Commented Jul 28, 2023 at 14:22

If only one of $$A_6$$ is zero and $$A_7$$ is not, you can write $$y^2$$ in terms of $$\{1,x,y, x^2, xy\}$$. Now there is a problem here. $$y^2$$ is a poles of order $$6$$ at $$O$$ (i.e. it's in $$L(6\mathcal{O})$$) but we have written it as a sum of elements with poles to order $$5$$! When you add two functions together, the poles can't get worse (i.e. pole of order at most $$5$$ + poles of order at most $$5$$ should not result in poles of order $$6$$).
Likewise we have a similar issue if $$A_7=0$$ and $$A_6\neq 0$$.
If $$A_6=A_7=0$$ then we have a different contradiction. This time we now that $$\{1,x,y,x^2,xy\}$$ are linearly dependent! But these formed a basis for $$L(5\mathcal{O})$$.
In other words, $$A_6\neq 0$$ and $$A_7\neq 0$$