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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!

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    $\begingroup$ 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 $A_6A_7\neq 0$ $\endgroup$
    – Fotis
    Jul 28, 2023 at 11:06
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    $\begingroup$ @Fotis thank you so much! I now can see it clearly :) $\endgroup$
    – babu
    Jul 28, 2023 at 14:22

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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 $A_7\neq 0$

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    $\begingroup$ Thank you so much for your answer! $\endgroup$
    – babu
    Jul 28, 2023 at 14:23

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