A rational curve of degree 4 in \(\PP^3\) is contained in a unique quadric surface \(Q\), and \(Q\) is necessarily nonsingular.
Solution
Let \(X\subseteq\PP^3\) be the given rational quartic, and put \(H=\OO_X(1)\).
1. The curve \(X\) is not contained in a plane.
If \(X\) lay in a plane, then, being a nonsingular plane curve of degree \(4\), it would have genus \[ \frac{(4-1)(4-2)}2=3. \] But \(X\) is rational, so its genus is \(0\). Hence \(X\) is nondegenerate in \(\PP^3\).
2. At least one quadric surface contains \(X\).
Since \(X\cong\PP^1\) and \(\deg H=4\), \[ \deg(2H)=8, \qquad h^0(X,\OO_X(2))=8+1=9. \] Twisting the ideal sequence of \(X\) by \(\OO_{\PP^3}(2)\) gives \[ 0\longrightarrow\mathcal I_X(2) \longrightarrow\OO_{\PP^3}(2) \longrightarrow\OO_X(2) \longrightarrow0. \] On global sections, \[ h^0(\PP^3,\OO_{\PP^3}(2))=\binom{5}{3}=10>9, \] so the restriction map cannot be injective. Thus \[ H^0(\PP^3,\mathcal I_X(2))\neq0, \] and a nonzero element is the equation of a quadric containing \(X\).
3. That containing quadric is unique.
1. Hence\[ Y=Q_1\cap Q_2 \]is a complete-intersection curve of type \((2,2)\).
By Bezout, \[ \deg Y=2\cdot2=4=\deg X. \] The surjection \(\OO_Y\twoheadrightarrow\OO_X\) therefore has kernel supported in dimension \(0\): its Hilbert polynomial has zero leading coefficient because \(X\) and \(Y\) have the same degree. But the complete-intersection curve \(Y\) is Cohen–Macaulay of pure dimension \(1\), so \(\OO_Y\) has no nonzero zero-dimensional subsheaf. Thus the kernel is zero and \(Y=X\) scheme-theoretically. But the complete-intersection genus formula gives \[ p_a(Y) =1+\frac{(2)(2)(2+2-4)}2 =1, \] whereas the nonsingular rational curve \(X\) has \(p_a(X)=0\). This contradiction proves uniqueness.
4. The unique quadric \(Q\) is nonsingular.
Over the algebraically closed ground field, a singular quadric surface in \(\PP^3\) is either a union (possibly doubled) of planes or an irreducible quadric cone. The first possibility would put the irreducible curve \(X\) in a plane, contradicting step 1.
Suppose therefore that \(Q\) is a quadric cone. The quadric-cone calculation on FE-CRVQUAD says that an integral curve of even degree \(2a\) on \(Q\) has \[ p_a=(a-1)^2. \] Here \(\deg X=4=2\cdot2\), so this gives \[ p_a(X)=(2-1)^2=1, \] again contradicting the fact that the nonsingular rational curve \(X\) has genus \(0\). Thus \(Q\) cannot be singular.
5. Q.E.D.
Step 2 gives existence, step 3 gives uniqueness, and step 4 proves that the unique quadric is nonsingular.