This comes from Ex IV.4.2 in Hartshorne. Namely we want to show that if $D$ is a divisor of degree at least three on an elliptic curve $X$, and we embed $X$ into $\mathbb{P}^n$ via this very ample divisor, then the image is projectively normal.

I want to use the trace condition, namely that for every $d>0$, the natural map $$\Gamma(\mathbb{P}^n,\mathcal{O}(d)) \to \Gamma(X,i^*\mathcal{O}(d))$$ is surjective (this is essentially Ex II.5.14(d)). I can show this for $d=1$, but I'm not sure how to show this more generally. I figure something cohomological or the $d$-Uple embedding might help, but I'm just not sure.

Any ideas?


A solution to this is given in Eisenbud's book "The Geometry of Syzygies", section 6.D.

The upshot is that for an elliptic curve $X$ and a point $p\in X$, the natural map $H^0(X, \mathcal{O}(dp)) \otimes H^0(X, \mathcal{O}(ep)) \rightarrow H^0(X, \mathcal{O}((d+e)p))$ is surjective, for $d\ge 2$ and $e\ge 3$. The proof of this is similar to how one puts an elliptic curve in Weierstrass form. Using Riemann-Roch, choose $x\in H^0(\mathcal{O}(2p))$ with a pole at $p$ of order $2$, and $y\in H^0(\mathcal{O}(3p))$ with a pole at $p$ of order $3$. Then you can construct a basis for $H^0(\mathcal{O}(np))$ for any $n$ by considering $x^i$ and $x^jy$. Then it is not too hard to show surjectivity by counting dimensions and using that rational functions with different order of vanishing at $p$ have to be linearly independent.

Let us suppose that this is true. We will prove the trace condition by induction. Consider the following commutative diagram:

$$\begin{array}{ccc} H^0(\mathbb{P}^n, \mathcal{O}(1)) \otimes H^0(\mathbb{P}^n, \mathcal{O}(d-1)) & \longrightarrow & H^0(X, \mathcal{O}_X(1)) \otimes H^0(X, \mathcal{O}_X(d-1))\\ \downarrow && \downarrow \\ H^0(\mathbb{P}^n, \mathcal{O}(d)) & \longrightarrow & H^0(X,\mathcal{O}_X(d)) \end{array}$$

The top arrow is surjective by inductive hypothesis, and the arrow on the right is surjective by the above fact (note that for any divisor $D$ of degree $n$ on an elliptic curve $X$, we have that $D = nP$ for some $P$) which implies that the bottom arrow is surjective.

  • $\begingroup$ This is encouraging, I had discovered that the map $H^0(X,\mathcal{O}_X(p))^{\otimes k\} \to H^0(X,\mathcal{O}_X(kp))$ was surjective, but wasn't able to put it all together. So close! Thank you for your help. $\endgroup$ – DKS Apr 9 '18 at 13:10

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