If \(D\) is any divisor of degree \(\geq 3\) on the elliptic curve \(X\), and if we embed \(X\) in \(\PP^n\) by the complete linear system \(\abs{D}\), show that the image of \(X\) in \(\PP^n\) is projectively normal.
Note. It is true more generally that if \(D\) is a divisor of degree \(\geq 2g+1\) on a curve of genus \(g\), then the embedding of \(X\) by \(\abs{D}\) is projectively normal (Mumford, p. 55).
Solution
Put \[ L=\mco_X(D), \qquad d=\deg L\ge3. \]
1. Every line bundle \(A\) on \(X\) of degree at least \(2\) is globallygenerated, and it contains a two-dimensional base-point-free space ofsections.
For any point \(Q\in X\), Serre duality and \(K_X\sim0\) give \[ H^1(X,A(-Q)) \cong H^0(X,A^{-1}(Q))^*. \] Since \[ \deg A^{-1}(Q)=1-\deg A<0, \] the group on the right vanishes. The exact sequence \[ 0\longrightarrow A(-Q) \longrightarrow A \longrightarrow A|_Q \longrightarrow0 \] therefore shows that \[ H^0(X,A)\longrightarrow A|_Q \] is surjective for every \(Q\). Thus \(A\) is globally generated.
Choose a nonzero section \(s_0\). Its zero divisor has finitely many points. For each such point, the sections vanishing there form a proper hyperplane of \(H^0(X,A)\). Since \(k\) is infinite, choose \(s_1\) outside the finite union of these hyperplanes. Then \(s_0\) and \(s_1\) have no common zero, so \[ V=\langle s_0,s_1\rangle \] is a base-point-free pencil.
2. The quadratic multiplication map\[ \operatorname{Sym}^2H^0(X,L) \longrightarrow H^0(X,L^2) \]is surjective.
Fix a point \(P\in X\) and set \[ A=L(-P). \] Since \(\deg A=d-1\ge2\), step 1 supplies a base-point-free pencil \[ V\subseteq H^0(X,A). \] The base-point-free pencil sequence is \[ 0\longrightarrow A^{-1} \longrightarrow V\otimes\mco_X \longrightarrow A \longrightarrow0. \] Tensoring by \(L\) gives \[ 0\longrightarrow\mco_X(P) \longrightarrow V\otimes L \longrightarrow L^2(-P) \longrightarrow0, \] because \(L\otimes A^{-1}\cong\mco_X(P)\). The line bundle \(\mco_X(P)\) has positive degree, hence \[ H^1(X,\mco_X(P))=0. \] Taking global sections therefore gives a surjection \[ V\otimes H^0(X,L) \twoheadrightarrow H^0(X,L^2(-P)). \] Since \(H^0(X,A)\subseteq H^0(X,L)\), the image of the full quadratic multiplication map contains \(H^0(X,L^2(-P))\).
By step 1, \(L\) is globally generated. Choose \[ s\in H^0(X,L) \] with \(s(P)\ne0\). Then \(s^2\) does not vanish at \(P\), so \[ s^2\notin H^0(X,L^2(-P)). \] The latter space has codimension one in \(H^0(X,L^2)\), because \(H^1(X,L^2(-P))=0\). Thus the quadratic multiplication image contains a codimension-one subspace and an element outside it. It is all of \(H^0(X,L^2)\).
3. For every \(n\ge2\), multiplication gives a surjection\[ H^0(X,L)\otimes H^0(X,L^n) \twoheadrightarrow H^0(X,L^{n+1}). \]
By step 1 choose a base-point-free pencil \[ W\subseteq H^0(X,L). \] Its pencil sequence \[ 0\longrightarrow L^{-1} \longrightarrow W\otimes\mco_X \longrightarrow L \longrightarrow0 \] becomes, after tensoring by \(L^n\), \[ 0\longrightarrow L^{n-1} \longrightarrow W\otimes L^n \longrightarrow L^{n+1} \longrightarrow0. \] For \(n\ge2\) the line bundle \(L^{n-1}\) has positive degree, so \[ H^1(X,L^{n-1})=0. \] Hence \[ W\otimes H^0(X,L^n) \twoheadrightarrow H^0(X,L^{n+1}). \] The same is therefore true with \(H^0(X,L)\) in place of \(W\).
4. For every \(m\ge1\), the natural map\[ \operatorname{Sym}^mH^0(X,L) \longrightarrow H^0(X,L^m) \]is surjective.
The assertion is tautological for \(m=1\) and is step 2 for \(m=2\). Assume it for some \(m\ge2\). Compose its tensor product with \(H^0(X,L)\) with the multiplication map of step 3: \[ H^0(X,L)\otimes\operatorname{Sym}^mH^0(X,L) \twoheadrightarrow H^0(X,L)\otimes H^0(X,L^m) \twoheadrightarrow H^0(X,L^{m+1}). \] This composition is multiplication and therefore factors through \[ \operatorname{Sym}^{m+1}H^0(X,L). \] Hence the latter maps surjectively to \(H^0(X,L^{m+1})\). Induction proves the assertion for all \(m\).
5. The embedding defined by \(\abs{D}\) is projectively normal.
For the complete linear-series embedding \[ X\hookrightarrow\PP(H^0(X,L)^*), \] the restriction map on degree-\(m\) homogeneous forms is precisely \[ \operatorname{Sym}^mH^0(X,L) \longrightarrow H^0(X,L^m). \] Step 4 proves that it is surjective for every \(m\ge1\). For \(m=0\), the restriction map is \[ k\longrightarrow H^0(X,\mco_X), \] which is an isomorphism because \(X\) is an integral projective curve over the algebraically closed field \(k\). Since \(X\) is nonsingular and hence normal, the criterion of Exercise II.5.14(d) now gives \[ \boxed{X\subseteq\PP^n\text{ is projectively normal}.} \]
6. Q.E.D.
Steps <1>2–<1>4 prove normal generation of \(L=\mco_X(D)\), and step 5 identifies this with projective normality of the complete-linear-series embedding.