If \(X\) is a curve of genus \(\geq 2\) which is a complete intersection (II, Ex. 8.4) in some \(\PP^n\), show that the canonical divisor \(K\) is very ample. Conclude that a curve of genus 2 can never be a complete intersection in any \(\PP^n\). Cf. (Ex. 5.1).
Solution
Suppose \[ X=H_1\cap\cdots\cap H_{n-1}\subset\PP^n \] is a nonsingular complete-intersection curve, where \[ \deg H_i=d_i. \]
1. The canonical sheaf of \(X\) is\[ \omega_X \cong \OO_X(m), \qquad m=\sum_{i=1}^{n-1}d_i-n-1. \]
The complete-intersection adjunction formula of Hartshorne II.8.4(d) gives \[ \omega_X \cong \left( \omega_{\PP^n} \tensor \OO_{\PP^n}\!\left(\sum_i d_i\right) \right)|_X. \] Since \[ \omega_{\PP^n}\cong\OO_{\PP^n}(-n-1), \] this becomes \[ \omega_X \cong \OO_X\!\left(\sum_i d_i-n-1\right) =\OO_X(m). \]
2. If \(g(X)\ge2\), then\[ m>0. \]
The hyperplane bundle \(\OO_X(1)\) has positive degree \[ \deg X=\prod_i d_i>0. \] By step 1, \[ 2g-2 = \deg\omega_X = m\deg\OO_X(1). \] The left-hand side is positive because \(g\ge2\), and \(\deg\OO_X(1)>0\). Hence \(m>0\).
3. The canonical divisor of \(X\) is very ample.
The line bundle \(\OO_X(1)\) is very ample because it is the restriction of the hyperplane bundle for the given closed immersion \[ X\hookrightarrow\PP^n. \] By step 2, \(m\ge1\). Every positive tensor power of a very ample line bundle is very ample: the corresponding morphism is the given embedding followed by the \(m\)-uple Veronese embedding. Thus \[ \omega_X\cong\OO_X(1)^{\tensor m} \] is very ample.
4. A curve of genus \(2\) cannot be a complete intersection in any\(\PP^n\).
If \(g(X)=2\), then every canonical divisor has degree \[ \deg K_X=2g-2=2. \] If \(X\) were a complete intersection, step 3 would make \(K_X\) very ample. But Exercise IV.3.1 proves that on a genus-\(2\) curve every very ample divisor has degree at least \(5\). This contradicts \(\deg K_X=2\).
5. Q.E.D.
Steps <1>1–<1>3 prove that the canonical divisor of a complete-intersection curve of genus at least \(2\) is very ample, and step 4 proves the genus-\(2\) conclusion.