A rational curve of degree 5 in \(\PP^3\) is always contained in a cubic surface, but there are such curves which are not contained in any quadric surface.
Solution
1. Every rational quintic \(X\subseteq\PP^3\) is contained in a cubicsurface.
Since \(X\cong\PP^1\) and \(\deg\OO_X(1)=5\), \[ \deg\OO_X(3)=15, \qquad h^0(X,\OO_X(3))=16. \] Twisting the ideal sequence by \(\OO_{\PP^3}(3)\) gives \[ 0\longrightarrow\mathcal I_X(3) \longrightarrow\OO_{\PP^3}(3) \longrightarrow\OO_X(3) \longrightarrow0. \] Now \[ h^0(\PP^3,\OO_{\PP^3}(3))=\binom63=20. \] Therefore the restriction map from this \(20\)-dimensional space to the \(16\)-dimensional space \(H^0(X,\OO_X(3))\) has nonzero kernel; indeed \[ h^0(\PP^3,\mathcal I_X(3))\ge4. \] Any nonzero element of this kernel is a cubic equation vanishing on \(X\).
2. Fix \(\alpha\in k^*\) and define\[ \phi:\PP^1\longrightarrow\PP^3, \qquad [s:t]\longmapsto [f_0:f_1:f_2:f_3], \]where\[ f_0=s^5,\qquad f_1=s^4t,\qquad f_2=st^4+\alpha s^2t^3,\qquad f_3=t^5. \]Then \(\phi\) is a closed embedding and its image has degree \(5\).
The four sections have no common zero, so they define a morphism and \[ \phi^*\OO_{\PP^3}(1)\cong\OO_{\PP^1}(5). \]
They separate points. On \(s\ne0\) the ratio \[ \frac{f_1}{f_0}=\frac ts \] recovers the affine coordinate, while the only point with \(s=0\) maps to \([0:0:0:1]\) and cannot coincide with the image of a point having \(s\ne0\).
They also separate tangent vectors. On \(s\ne0\), the same ratio \(f_1/f_0\) is a local parameter. At the remaining point \([0:1]\), write \(u=s/t\). On the chart \(f_3\ne0\) one has \[ \frac{f_2}{f_3}=u+\alpha u^2, \] whose differential at \(u=0\) is \(du\ne0\). Hence the linear subsystem separates points and tangent vectors, so it gives a closed embedding.
For an embedding of a projective curve, the degree is the degree of the pulled-back hyperplane bundle. Therefore \[ \deg\phi(\PP^1)=\deg\OO_{\PP^1}(5)=5. \]
3. The curve \(C=\phi(\PP^1)\) is contained in no quadric surface.
A quadric containing \(C\) would give a nonzero relation among the ten products \(f_if_j\) with \(0\le i\le j\le3\). Suppose \[ \begin{aligned} 0={}&A f_0^2+B f_0f_1+C f_0f_2+D f_0f_3+E f_1^2\\ &+F f_1f_2+G f_1f_3+H f_2^2+I f_2f_3+J f_3^2. \end{aligned} \] Compare coefficients as homogeneous polynomials of degree \(10\) in \(s,t\). The coefficients of \[ s^{10},\qquad s^9t,\qquad s^8t^2 \] give respectively \[ A=B=E=0. \] The monomial \(s^7t^3\) occurs, among the remaining terms, only in \(C f_0f_2\), with coefficient \(C\alpha\); hence \(C=0\). Then the coefficient of \(s^6t^4\) is \(F\alpha\), so \(F=0\), and the coefficient of \(s^5t^5\) gives \(D=0\).
At the other end, the coefficient of \(st^9\) gives \(I=0\). The coefficient of \(s^2t^8\) then gives \(H=0\), the coefficient of \(s^4t^6\) gives \(G=0\), and finally the coefficient of \(t^{10}\) gives \(J=0\). Thus every coefficient is zero, so the ten products are linearly independent and no nonzero quadratic equation vanishes on \(C\).
4. Q.E.D.
Step 1 proves that every rational quintic lies on a cubic surface. Steps <1>2–<1>3 exhibit a nonsingular rational quintic lying on no quadric.