Curves of Degree 4.
- If \(X\) is a curve of degree 4 in
some \(\PP^n\), show that either
- \(g=0\), in which case \(X\) is either the rational normal quartic in \(\PP^4\) (Ex. 3.4) or the rational quartic curve in \(\PP^3\) (II, 7.8.6), or
- \(X \subseteq \PP^2\), in which case \(g=3\), or
- \(X \subseteq \PP^3\) and \(g=1\).
- In the case \(g=1\), show that \(X\) is a complete intersection of two irreducible quadric surfaces in \(\PP^3\) (I, Ex. 5.11). Hint: Use the exact sequence \(0 \to \mci_X \to \OO_{\PP^3} \to \OO_X \to 0\) to compute \(\dim H^0(\PP^3, \mci_X(2))\), and thus conclude that \(X\) is contained in at least two irreducible quadric surfaces.
Solution
Let \[ \PP^r=\langle X\rangle \] be the linear span of \(X\). Then \(X\) is nondegenerate in \(\PP^r\), and its degree in the span is still \(4\).
1. One has\[ 2\le r\le4. \]
The case \(r=1\) is impossible: an integral one-dimensional closed subscheme of \(\PP^1\) is \(\PP^1\) itself and has degree \(1\).
If \(r\ge4\), then \[ \deg X=4\le r. \] By Exercise IV.3.4(b), a nondegenerate curve of degree at most its ambient dimension must have degree equal to that dimension. Hence \(r=4\). Thus \(2\le r\le4\).
2. If \(r=4\), then\[ g(X)=0 \]and \(X\) is the rational normal quartic in \(\PP^4\) up to projectiveautomorphism.
This is exactly Exercise IV.3.4(b) with \[ d=r=4. \]
3. If \(r=2\), then\[ g(X)=3. \]
In this case \(X\) is a nonsingular plane curve of degree \(4\). The plane-curve genus formula gives \[ g(X) = \frac{(4-1)(4-2)}2 =3. \]
4. If \(r=3\), then\[ g(X)\in\{0,1\}. \]
Since \(X\subseteq\PP^3\) is nondegenerate, Exercise IV.3.5(b) gives \[ g(X) < \frac12(4-1)(4-2) =3. \] Thus \[ g(X)\in\{0,1,2\}. \]
Suppose \(g(X)=2\). Put \[ \mcl=\mco_X(1). \] Then \[ \deg\mcl=4. \] For a canonical divisor \(K\) on a genus-\(2\) curve, \[ \deg(K-\mcl)=-2, \] so Riemann–Roch gives \[ h^0(X,\mcl) = 4+1-2 =3. \] But nondegeneracy in \(\PP^3\) makes the restriction map \[ H^0(\PP^3,\mco(1)) \hookrightarrow H^0(X,\mcl) \] injective, so the right side has dimension at least \(4\). This contradiction excludes genus \(2\).
5. The alternatives in part (a) are exactly:
- \(g=0\), with \(X\) a rational normal quartic in \(\PP^4\) or a nonsingular rational quartic in \(\PP^3\);
- \(X\subseteq\PP^2\), in which case \(g=3\);
- \(X\subseteq\PP^3\) and \(g=1\).
Steps <1>1–<1>4 leave only the stated cases.
If \(r=4\), step 2 gives the rational normal quartic. If \(r=3\) and \(g=0\), then \(X\cong\PP^1\) and the given degree-\(4\) embedding is precisely a nonsingular rational quartic in \(\PP^3\) in the sense of Hartshorne II.7.8.6. The remaining \(r=3\) possibility is \(g=1\), and step 3 handles \(r=2\). This proves (a).
6. Assume now that \(g(X)=1\). Then\[ h^0(X,\mco_X(2))=8. \]
Part (a) puts \(X\) nondegenerately in \(\PP^3\). Let \[ H=\mco_X(1). \] Then \[ \deg(2H)=8. \] On a genus-\(1\) curve the canonical divisor has degree \(0\), so \[ \deg(K_X-2H)=-8 \] and hence \[ h^0(K_X-2H)=0. \] Riemann–Roch therefore gives \[ h^0(X,\mco_X(2)) = 8+1-1 =8. \]
7. The vector space of quadrics containing \(X\) has dimension at least\(2\):\[ h^0(\PP^3,\mci_X(2))\ge2. \]
Twist \[ 0 \longrightarrow \mci_X \longrightarrow \mco_{\PP^3} \longrightarrow \mco_X \longrightarrow0 \] by \(\mco_{\PP^3}(2)\) and take global sections. Exactness gives \[ 0 \longrightarrow H^0(\PP^3,\mci_X(2)) \longrightarrow H^0(\PP^3,\mco(2)) \longrightarrow H^0(X,\mco_X(2)). \] The middle space has dimension \[ \binom{3+2}{2}=10, \] whereas step 6 gives dimension \(8\) for the target. Therefore the kernel has dimension at least \[ 10-8=2. \]
8. Every nonzero quadric surface containing \(X\) is irreducible.
A reducible quadric surface in \(\PP^3\) is a union of two planes, allowing the two planes to coincide. If such a quadric contained the irreducible curve \(X\), irreducibility of \(X\) would force \(X\) to lie in one of those planes. This contradicts part (a), which places the genus-\(1\) case nondegenerately in \(\PP^3\). Hence every containing quadric is irreducible.
9. Choose two linearly independent quadrics\[ Q_1,Q_2\in H^0(\PP^3,\mci_X(2)). \]Then\[ Z=V(Q_1,Q_2) \]is a complete-intersection curve with Hilbert polynomial\[ P_Z(m)=4m. \]
Step 7 supplies the independent quadrics, and step 8 makes them irreducible. Since they are distinct irreducible polynomials, they have no common factor. Thus they form a regular sequence in \[ k[x_0,x_1,x_2,x_3], \] so \(Z\) is the complete intersection of two quadrics.
Its Koszul resolution is \[ 0 \longrightarrow \mco_{\PP^3}(-4) \longrightarrow \mco_{\PP^3}(-2)^{\oplus2} \longrightarrow \mco_{\PP^3} \longrightarrow \mco_Z \longrightarrow0. \] Hence \[ \begin{aligned} P_Z(m) &= \binom{m+3}{3} -2\binom{m+1}{3} +\binom{m-1}{3}\\ &=4m. \end{aligned} \]
10. One has\[ X=Z \]scheme-theoretically.
Because both quadrics contain \(X\), there is a closed immersion \[ X\hookrightarrow Z. \] The Hilbert polynomial of the degree-\(4\), genus-\(1\) curve \(X\) is \[ P_X(m) = 4m+1-g =4m. \] Step 9 gives the same polynomial for \(Z\).
Let \(\mathcal K\) be the kernel of the induced surjection \[ \mco_Z\twoheadrightarrow\mco_X. \] Additivity of Hilbert polynomials gives \[ P_{\mathcal K}=P_Z-P_X=0. \] A nonzero coherent sheaf on a projective scheme has a nonzero Hilbert polynomial, so \[ \mathcal K=0. \] Therefore \(X=Z\) as schemes.
11. The genus-\(1\) quartic is a complete intersection of two irreduciblequadric surfaces in \(\PP^3\).
Steps <1>8–<1>10 produce irreducible quadric surfaces \(V(Q_1)\) and \(V(Q_2)\) and prove \[ \boxed{X=V(Q_1)\cap V(Q_2)} \] scheme-theoretically. This proves (b).
12. Q.E.D.
Step 5 proves (a), and step 11 proves (b).