If a surface \(X\) of degree \(d\) in \(\PP^3\) contains a straight line \(C=\PP^1\), show that \(C^2=2-d\).
Assume \(\operatorname{char} k=0\), and show for every \(d \geqslant 1\), there exists a nonsingular surface \(X\) of degree \(d\) in \(\PP^3\) containing the line \(x=y=0\).
Solution
Let \(H\) denote the hyperplane class on \(X\), and let \[ L=V(x,y)\subseteq\PP^3. \]
1. The canonical class of \(X\) is\[ K_X=(d-4)H, \]and \(H\cdot C=1\).
Since \(X\subseteq\PP^3\) is a nonsingular hypersurface of degree \(d\), hypersurface adjunction gives \[ K_X=(K_{\PP^3}+X)|_X=(-4H+dH)|_X=(d-4)H. \] The curve \(C\) is a line in \(\PP^3\), so its degree with respect to the hyperplane class is one. Hence \[ H\cdot C=1. \]
2. The self-intersection of \(C\) is\[ \boxed{C^2=2-d}. \]
The curve \(C\cong\PP^1\) has genus zero. The surface adjunction formula T-SRFADJ therefore gives \[ -2 = 2g(C)-2 = C\cdot(C+K_X). \] Using step 1, \[ -2 = C^2+(d-4)H\cdot C = C^2+d-4. \] Thus \[ C^2=2-d. \] This proves part (a).
3. For every \(d\geq1\), define\[ \boxed{ X_d= \begin{cases} V(x), & d=1,\\[2mm] V\!\left(xz^{d-1}+yw^{d-1}+x^d+y^d\right), & d\geq2. \end{cases}} \]Then \(X_d\) is a degree-\(d\) surface containing \(L\).
For \(d=1\), \(X_1=V(x)\) is a plane, hence has degree one, and it contains \(L=V(x,y)\).
Suppose \(d\geq2\) and set \[ F_d=xz^{d-1}+yw^{d-1}+x^d+y^d. \] Every monomial of \(F_d\) has degree \(d\), so \(X_d=V(F_d)\) is a degree-\(d\) hypersurface. On \(L\) one has \(x=y=0\), hence \(F_d=0\). Thus \[ L\subseteq X_d. \]
4. For every \(d\geq1\), the surface \(X_d\) in step 3 is nonsingular.
The case \(d=1\) is immediate because \(X_1\) is a plane.
Let \(d\geq2\). The first partial derivatives of \(F_d\) are \[ \begin{aligned} \frac{\partial F_d}{\partial x} &=z^{d-1}+d x^{d-1}, & \frac{\partial F_d}{\partial z} &=(d-1)xz^{d-2},\\ \frac{\partial F_d}{\partial y} &=w^{d-1}+d y^{d-1}, & \frac{\partial F_d}{\partial w} &=(d-1)yw^{d-2}. \end{aligned} \] Because \(\operatorname{char}k=0\), both \(d\) and \(d-1\) are nonzero in \(k\). Suppose all four partial derivatives vanish at a projective point.
If \(d=2\), the equation \[ \frac{\partial F_d}{\partial z}=x=0 \] followed by \(\partial F_d/\partial x=0\) gives \(z=0\). If \(d>2\), the equation \(xz^{d-2}=0\) gives either \(x=0\) or \(z=0\); in the first case \(\partial F_d/\partial x=0\) gives \(z=0\), while in the second case it gives \(d x^{d-1}=0\), hence \(x=0\). Thus in every case \[ x=z=0. \] The identical argument applied to the \(y,w\) derivatives gives \[ y=w=0. \] This is impossible for a point of \(\PP^3\). Hence the gradient of \(F_d\) never vanishes at a projective point, so the projective Jacobian criterion P-AGH58JACOBIANRANK shows that \(X_d\) is nonsingular.
5. Q.E.D.
Step 2 proves part (a). Steps <1>3–<1>4 construct, for every \(d\geq1\), a nonsingular degree-\(d\) surface containing the line \(x=y=0\), proving part (b).