Show that a hyperelliptic curve can never be a complete intersection in any projective space. Cf. (Ex. 3.3).
Solution
Let \(X\) be hyperelliptic of genus \(g\ge2\). Choose a finite morphism \[ f:X\longrightarrow\PP^1 \] of degree \(2\), and let \[ D=f^*(Q) \] for some point \(Q\in\PP^1\).
1. The divisor \(D\) satisfies\[ \deg D=2, \qquad \ell(D)\ge2. \]
Since \(f\) has degree \(2\), \[ \deg f^*(Q)=2. \] Moreover \[ \OO_X(D)\cong f^*\OO_{\PP^1}(1). \] Pullback of the two-dimensional space \[ H^0(\PP^1,\OO_{\PP^1}(1)) \] injects into \(H^0(X,\OO_X(D))\), because \(f\) is surjective. Hence \[ \ell(D)\ge2. \]
2. For every canonical divisor \(K\) on \(X\),\[ \boxed{\ell(K-D)\ge g-1.} \]
Riemann–Roch applied to \(D\) gives \[ \ell(D)-\ell(K-D) = \deg D+1-g = 3-g. \] Therefore \[ \ell(K-D) = \ell(D)+g-3 \ge 2+g-3 = g-1. \]
3. The canonical divisor of a hyperelliptic curve is not very ample.
Suppose \(K\) were very ample. Then the embedding defined by the complete linear system \(|K|\) separates every subscheme of length \(2\). In particular, the restriction map \[ H^0(X,\OO_X(K)) \longrightarrow H^0(D,\OO_X(K)|_D) \] is surjective.
The divisor \(D\) has length \(2\), so the target has dimension \(2\). Since \[ \ell(K)=g, \] the exact sequence \[ 0\longrightarrow\OO_X(K-D) \longrightarrow\OO_X(K) \longrightarrow\OO_X(K)|_D \longrightarrow0 \] would therefore give \[ \ell(K-D)=g-2. \] This contradicts step 2, which gives \[ \ell(K-D)\ge g-1. \] Hence \(K\) is not very ample.
4. The curve \(X\) cannot be a complete intersection in any projectivespace.
Suppose, to the contrary, that \[ X\subseteq\PP^n \] were a complete-intersection curve. Since \(g\ge2\), Exercise IV.3.3 says that its canonical divisor is very ample. This contradicts step 3.
5. Q.E.D.
Steps <1>1–<1>3 show that a hyperelliptic curve has non-very-ample canonical divisor, while step 4 contradicts the complete-intersection very-ampleness theorem.