Let \(X\) be a plane curve of degree 4.
Show that the effective canonical divisors on \(X\) are exactly the divisors \(X.L\), where \(L\) is a line in \(\PP^2\).
If \(D\) is any effective divisor of degree 2 on \(X\), show that \(\dim \abs{D}=0\).
Conclude that \(X\) is not hyperelliptic (Ex. 1.7).
Solution
By Exercise I.7.2, since \(X\) is a nonsingular plane quartic, its genus is \[ g(X)=\frac{(4-1)(4-2)}2=3, \] and therefore every canonical divisor has degree \[ 2g-2=4. \]
1. The canonical sheaf of \(X\) is\[ \boxed{\omega_X\cong\OO_X(1)}. \]
Adjunction for a nonsingular plane curve of degree \(d\) gives \[ \omega_X \cong \bigl(\omega_{\PP^2}\tensor\OO_{\PP^2}(d)\bigr)|_X. \] Here \[ \omega_{\PP^2}\cong\OO_{\PP^2}(-3) \] and \(d=4\), so \[ \omega_X\cong\OO_X(4-3)=\OO_X(1). \]
2. Restriction of linear forms gives an isomorphism\[ H^0(\PP^2,\OO_{\PP^2}(1)) \xrightarrow{\sim} H^0(X,\omega_X). \]
By step 1 the target is \(H^0(X,\OO_X(1))\). No nonzero linear form vanishes identically on the quartic \(X\), so the restriction map \[ H^0(\PP^2,\OO_{\PP^2}(1)) \longrightarrow H^0(X,\OO_X(1)) \] is injective.
The source has dimension \(3\). Since \(\OO_X(1)\cong\omega_X\) and \(g(X)=3\), the target also has dimension \[ h^0(X,\omega_X)=g(X)=3. \] Thus the restriction map is an isomorphism.
3. The effective canonical divisors on \(X\) are exactly the divisors\[ \boxed{X.L} \]where \(L\subseteq\PP^2\) is a line.
An effective canonical divisor is the zero divisor of a nonzero section of \(\omega_X\). By step 2, every such section is the restriction of a nonzero linear form on \(\PP^2\), whose zero locus is a line \(L\). Its zero divisor on \(X\) is precisely the scheme-theoretic intersection \(X.L\).
Conversely, restricting the defining linear form of any line \(L\) gives a nonzero section of \(\OO_X(1)\cong\omega_X\), so \(X.L\) is an effective canonical divisor. This proves part (a).
4. Let \(D\) be an effective divisor of degree \(2\). Then\[ \ell(D)=\ell(K-D) \]for any canonical divisor \(K\).
Riemann-Roch gives \[ \ell(D)-\ell(K-D) =\deg D+1-g =2+1-3 =0. \] Hence \(\ell(D)=\ell(K-D)\).
5. At most one effective canonical divisor contains \(D\).
Because the ground field is algebraically closed and \(\deg D=2\), either \[ D=P+Q \] with \(P\ne Q\), or \[ D=2P. \]
Suppose an effective canonical divisor contains \(D\). By step 3 it has the form \(X.L\) for a line \(L\).
If \(D=P+Q\) with \(P\ne Q\), then \(L\) must contain both \(P\) and \(Q\), so it is the unique secant line \(\overline{PQ}\).
If \(D=2P\), then the intersection multiplicity of \(L\) and \(X\) at \(P\) is at least \(2\). Since \(X\) is nonsingular at \(P\), this says exactly that \(L\) is the tangent line \(T_PX\), which is unique.
Thus in either case there is at most one effective canonical divisor containing \(D\).
6. Every effective divisor \(D\) of degree \(2\) satisfies\[ \boxed{\dim|D|=0}. \]
Since \(D\) is effective, \[ \ell(D)\geq1. \] By step 4, the same is true of \(\ell(K-D)\), so there exists at least one effective canonical divisor containing \(D\).
5.Hence\[ \ell(K-D)=1. \]Step 4 gives \(\ell(D)=1\), and therefore\[ \dim|D|=\ell(D)-1=0. \]This proves part (b).
7. The plane quartic \(X\) is not hyperelliptic.
6. Thus \(X\) is not hyperelliptic, proving part (c).
8. Q.E.D.
Step 3 proves part (a), step 6 proves part (b), and step 7 proves part (c).