Hartshorne Exercise IV.3.3.
Let $X$ be a plane curve of degree 4.
(a) Show that the effective canonical divisors on $X$ are exactly the divisors $X.L$, where $L$ is a line in $\PP{k}{2}$.
(b) If $D$ is any effective divisor of degree 2 on $X$, show that $\dim |D|=0$.
(c) Conclude that $X$ is not hyperelliptic.
(a) Let $X$ be a smooth projective plan curve of degree 4. Its genus is 3. Using the adjunction formula, $\omega_X\cong\OO_X(1)$. That is, $K_X$ is linearly equivalent to a hyperplane section $X.L$ of $X$. because $\omega_{\PP{k}{2}}\cong\OO_{\PP{k}{2}}(-3)$ is isomorphic to the ideal sheaf of $X$, we have a SES
$$0\ra \OO_{\PP{k}{2}}(-3)\ra\OO_{\PP{k}{2}}(1)\ra \iota_*\omega_X\ra 0,$$
where $\iota$ is the embedding of $X$. Considering the LES in cohomology, we see all the $H^1$’s vanish, so in particular, $H^0(\PP{k}{2},\OO_{\PP{k}{2}}(1))\ra H^0(X,\OO_X(X.L))$ is an isomorphism. Thus, if $D\in |K_X|$ then $D\in |X.L|$ and so $D=X.L’$ for some unique $L’\sim L$ (because $D\sim X.L\implies (\exists ! L’\ge 0)(X.L’=D)$).
(b) and (c) We will solve (b) and (c) at once. The effective divisors $D$ of degree 2 are exactly the sums $D=p+q$ for any two points $p,q\in X$. Recall that on a smooth projective curve $X$, $\mathcal{L} \in\op{Pic}(X)$, $h^0(\mathcal{L} (-p-q))=h^0(\mathcal{L} )-2\:\:\forall p,q\in X$ $\iff$ $\mathcal{L} $ is very ample, and if $X$ has genus $\ge 2$, then the canonical divisor $K_X$ is very ample $\iff$ $X$ is not hyperelliptic.
By RR and SD we find that
$$h^0(\OO_X(D))=h^0(\OO_X(K_X-D)).$$
Thus, it suffices to show that $K_X$ is very ample. Then $X$ is not hyperelliptic and $h^0(\OO_X(K_X-D))=3-2=1$. Let $L\subset\PP{k}{2}$ be the line containing $D=p+q$. Then we know (by Bezout) that $X.L=p+q+r+s$ for some other points $r,s$ on $X$. By (a), $X.L\sim K_X$. But $L$ is a very ample divisor on $\PP{k}{2}$ and $X.L$ is the pullback of $L$ under the closed immersion of $X$, so it must also be a very ample divisor, which completes the proof.
Remark: Because the equality involving hyperellipticity isn’t introduced until later in Chapter IV, we will give a different proof of (c). If $X$ were hyperelliptic, then we could find a divisor $D’$ of degree 2 with no base points and whose linear system is a pencil: $h^0(\OO_X(D’))=2$. But then there would be at least one effective divisor $D\in|D’|$ and then we’d have $h^0(\OO_X(D’))=h^0(\OO_X(D))=h^0(\OO_X(K_X-D))=1$, a contradiction.
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