Let $E/\Bbb C : y^2=x^3+ax+b$ be an elliptic curve over $\Bbb C$. Let $$E(\Bbb C)=\{(x,y)\in\Bbb C^2:y^2=x^3+ax+b\}$$ be the affine variety generated by $E$. Finally, let $$I(E(\Bbb C))=\{f(x,y)\in\Bbb C[x,y]: f(p)=0,\, \forall p\in E(\Bbb C)\}.$$ I am trying to show that $I(E(\Bbb C))=I_E$ is the ideal generated by $E$, that is, $$I_E=\left< y^2-x^3-ax-b\right>\subset\Bbb C[x,y].$$
My attempt so far has been the following.
Suppose $f\in\Bbb C[x,y]$. Then we can divide by $y^2-x^3-ax-b$ and get $$f(x,y)=q(x,y)(y^2-x^3-ax-b)+r(x,y),$$ where the total degree of $r\in\Bbb C[x,y]$ is less than $3$. Suppose then, that $f$ vanishes on $E(\Bbb C)$. That is, $f\in I_E$. From the uniformalization theorem, there is a lattice $L\subset \Bbb C$ such that $a=-15G_4(L),$ and $b=-35G_6(L)$, where $G_4,G_6$ are Eisenstein series corresponding to $L$. Because of this, the point $(\wp(z,L),\tfrac12\wp'(z,L))$ is in $E(\Bbb C)$ for all $z\in\Bbb C$. Here, $\wp(z,L)$ is the Weierstrass Elliptic function corresponding to the lattice $L$. Thus, $$0=f(\wp,\tfrac12\wp')=q(\wp,\tfrac12\wp')(\tfrac14\wp'^2-\wp^3-a\wp-b)+r(\wp,\tfrac12\wp').$$ Due to the differential equation $$\frac14\wp'(z,L)^2=\wp(z,L)^3-15G_4(L)\wp(z,L)-35G_6(L),$$ we then have that $$r(\wp,\tfrac12\wp')=0.$$ All I need to show now is that $r(x,y)$ is identically equal to $0$, but I am not quite sure how. One thing I considered is to write $$r(x,y)=\beta_1x^2+\beta_2xy+\beta_3y^2+\beta_3x+\beta_5y+\beta_6,$$ so that $$\begin{align} r(\wp,\tfrac12\wp')&=\beta_1\wp^2+\tfrac12\beta_2\wp\wp'+\tfrac14\beta_3\wp'^2+\beta_4\wp+\tfrac12\beta_5\wp'+\beta_6\\ &=\beta_1\wp^2+\tfrac12\beta_2\wp\wp'+\beta_3(\wp^3+a\wp+b)+\beta_4\wp+\tfrac12\beta_5\wp'+\beta_6\\ &=\beta_3\wp^3+\beta_1\wp^2+(a\beta_3+\beta_4)\wp+\tfrac12(\beta_2\wp+\beta_5)\wp'+\beta_6+b\beta_3\\ &= 0. \end{align}$$ Then using the expansion $$\wp(z,L)=\frac1{z^2}+\sum_{n\ge1}(2n+1)G_{2n+2}(L)z^{2n},$$ the expansion of $\beta_3\wp^3+\beta_1\wp^2+(a\beta_3+\beta_4)\wp$ would only contain even powers of $z$, while the expansion of $\tfrac12(\beta_2\wp+\beta_5)\wp'$ would only contain odd powers of $z$, making the coefficient of $z^n$ of $r(\wp,\tfrac12\wp')$ be $0$ for all $n\ne0$, and the constant term of $r(\wp,\tfrac12\wp')$ equal to $-\beta_6-b\beta_3$.
I was hoping that this would then imply that $r(x,y)\equiv 0$, but I am not exactly sure if that works.
Is there a way to show that $r\equiv 0$ from here? Or perhaps an easier way to show that $f\in\left<y^2-x^3-ax-b\right>$? Thanks.
Edit for context:
I encountered this problem by looking through my old notes from my undergrad algebraic geometry class. I was wondering how I could apply what I learned from the class to elliptic curves.
Suppose $p$ is a polynomial vanishing on $E$. Then after dividing $p$ by $y^2-(x^3+ax+b)$, we have $p=(y^2-(x^3+ax+b))q+r$, where $r$ has degree at most one in $y$. So we can write $r(x,y)=c(x)y+d(x)$, and $p$ vanishes on $E$ iff $r$ does (clearly $y^2-(x^3+ax+b)$ vanishes on $E$).
Now let $\overline{r}=-c(x)y+d(x)$. Then $r\overline{r}$ vanishes on $E$, and therefore $$r\overline{r} = c(x)^2-d(x)^2y^2 = c(x)^2 - d(x)^2(x^3+ax+b)$$ is a polynomial in $x$ which vanishes for all $x\in\Bbb C$. But this implies it is the zero polynomial, and for reasons of degree, this gives that $c=d=0$. So $r=0$ and in fact $p$ is divisible by $y^2-(x^3+ax+b)$.
One can get an awful lot of mileage out of this conjugation technique for varieties with equations of the form $z^2+pz+q=0$, just like one can get a lot of mileage out of conjugation for the complex numbers.