# Elliptic curves as the zeros of a Weierstrass equation.

Let $\mathcal{L}$ be a very ample invertible sheaf on a $\mathbb{K}$-scheme $X$. In particular $\mathcal{L}$ induces a closed embedding $X\rightarrow \mathbb{P}^n$. It follows that $X\simeq \mathrm{Proj}(\mathbb{K}[X_0,\ldots,X_n]/I)$ for some homogeneous ideal $I$ of $\mathbb{K}[X_0,\ldots,X_n]$ and $\mathcal{L}\simeq \mathcal{O}_X(1)$. Hence, $$\mathbb{K}[X_0,\ldots,X_n]/I\simeq \bigoplus_{n\geq 0}H^0(X,\mathcal{O}_X(n))\simeq \bigoplus_{n\geq0}H^0(X,\mathcal{L}^{\otimes n}).$$ I want to use this to study elliptic curves.

If $X$ is an elliptic curve and $P_0\in X$, then $\mathcal{L}=\mathcal{O}_X(3\cdot P_0)$ is a very ample invertible sheaf with $h^0(X,\mathcal{L})=3$. Therefore it induces a closed embedding $X\rightarrow \mathbb{P}^2$ such that the homogeneous coordinate ring of the associated closed subscheme of $\mathbb{P}^2$ is $$\bigoplus_{n\geq 0} H^{0}(X,3n\cdot P_0).$$ Recall that $h^{0}(X,n\cdot P_0)=n$ for every $n>0$. According to Hartshorne's book, $$H^0(X,\mathcal{O}_X(2\cdot P_0))=\langle 1,x\rangle,$$ $$H^0(X,\mathcal{O}_X(3\cdot P_0))=\langle 1,x,y\rangle,$$ $$H^0(X,\mathcal{O}_X(6\cdot P_0))\supseteq \langle1,x,y,x^2,xy,x^3,y^2\rangle.$$ Since $h^{0}(X,6\cdot P_0)=6$, it follows that we have a linear relation $$F:=a+bx+cy+dx^2+exy+fx^3+gy^2=0.$$ My question is, how do we know that $1,x,y$ are the only independent elements in $\bigoplus_{n\geq 0} H^{0}(X,3n\cdot P_0)$? How do we know that the previous equation is the only relation between them?

Any help, or correction in case I was misunderstanding something, would be appreciated.

$\textbf{Edit:}$ I think that we can use the so called Castelnuevo's lemma to obtain that the map $$H^0(X,\mathcal{O}_X(3(n-1)\cdot P_0))\otimes H^0(X,\mathcal{O}_X(3\cdot P_0))\rightarrow H^0(X,\mathcal{O}_X(3n\cdot P_0))$$ is surjective for every $n\geq 2$. Therefore, the elements of $H^0(X,\mathcal{O}_X(3n\cdot P_0))$ are products of elements of $H^0(X,\mathcal{O}_X(3\cdot P_0))$ for every positive integer $n$. So, the only thing I had left to prove is that there are no more relations betwen $1,x,y$. To be more precise, how can we prove that there is no polynomial $G$ in the indeterminates $x,y$ such that it is zero as a rational function and is not in the ideal generated by $F$? In other words, how can we prove that $$\bigoplus_{n\geq 0} H^0(X,\mathcal{O}_X(3n\cdot P_0))\simeq \mathbb{K}[x,y]/\langle F\rangle ?$$

• Riemann-Roch (or more elementary observations along the way) tell you that $\dim H^0(X;\mathcal{O}_X(1\cdot P_0))=1$, so the constant function $1$ spans it. Consequently $x$ has pole order $2$. So A) together they span $H^0(2\cdot P_0)$, B) $x^2$ has pole order four, so the third basis element of $H^0(3\cdot P_0)$ must be something new. – Jyrki Lahtonen Jul 21 '17 at 11:50

I will write "$z$" instead of "$1$" for the element of $H^{0}(X,\mathcal{O}_{X}(n \cdot P_{0}))$. I guess so far you've shown that there is a graded, surjective $\mathbb{K}$-algebra morphism $$\mathbb{K}[x,y,z]/\langle F\rangle \to \bigoplus_{n\geq 0} H^0(X,\mathcal{O}_X(3n\cdot P_0))$$ so to show that it is an isomorphism it suffices to show that it is an isomorphism on each degree. Now it's just a matter of computing dimensions of $\mathbb{K}$-vector spaces. As you say we have $h^{0}(X,3n \cdot P_{0}) = 3n$ for $n > 0$, and we have $\dim_{\mathbb{K}} (\mathbb{K}[x,y,z])_{n} = \binom{n+2}{2}$ for $n \ge 0$ and $\dim_{\mathbb{K}} (\langle F \rangle)_{n} = \binom{n-1}{2}$ for $n > 0$, and so $\dim_{\mathbb{K}} (\mathbb{K}[x,y,z]/\langle F \rangle)_{n} = \binom{n+2}{2} - \binom{n-1}{2} = \frac{1}{2} ((n+2)(n+1) - (n-1)(n-2)) = 3n$ for $n > 0$. (Here if $M$ is a graded abelian group, then $M_{n}$ is the $n$th graded part.)