Smooth cubic surface

A cubic surface with empty singular locus. Over an algebraically closed field, it is irreducible: distinct positive-degree hypersurface components would meet, and every point of their intersection would be singular in their product equation.

For two skew lines $L,M$ on a smooth cubic surface, their joining line meets the surface at a third point generically. In coordinates with $L=\{x_2=x_3=0\}$ and $M=\{x_0=x_1=0\}$, write $F(\alpha A+\beta B)=\alpha \beta (\alpha U+\beta V)$. The third point is $[VA-UB]$ whenever $UV\ne 0$. Neither $U$ nor $V$ is identically zero, because that would make one of the two lines singular on the surface. The inverse records $[x_0:x_1]$ and $[x_2:x_3]$ of a general surface point. Thus the surface is birational to ${\mathbb{P}}^1\times {\mathbb{P}}^1$, hence to ${\mathbb{P}}^2$.

Write a line as $L=\{x_2=x_3=0\}$ and the surface equation as $F=x_2{Q}_2+x_3{Q}_3$. Smoothness says $A=Q_2{|}_L$, $B=Q_3{|}_L$ have no common zero. Projection $[x_2:x_3]$ extends over $L$ by $[-B:A]$. Its fibres are the residual plane conics in the planes through $L$, since substituting $x_2=sw$, $x_3=tw$ gives $F=w(s{Q}_2+t{Q}_3)$. Different residual conics are disjoint everywhere, including their points on $L$, because they are fibres of one morphism of algebraic varieties. The family is flat: over the smooth base curve its local rings are torsion-free over the corresponding discrete valuation ring; smoothness of the total surface supplies the reduced-fibre conclusions.

The matrix of the residual plane conic in coordinates $(u,v,w)$ has coefficient degrees $\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 3\end{array}\right)$ in the plane parameter $(s,t)$. Every term of its determinant has degree five. The reduced singular fibres of a smooth conic bundle lemma rules out an identically zero determinant and makes each zero simple. Over an algebraically closed field, there are therefore exactly five singular fibres, each a pair of distinct coplanar projective lines. No residual conic contains the original line, because $sA+tB\equiv 0$ would make $A,B$ proportional and force a common zero. Pairs from different fibres are disjoint.

Every projective line through $P$ on a smooth cubic surface lies in its projective tangent plane. Each is a distinct linear factor of the cubic plane section. There can therefore be at most three. Equality occurs when the tangent section consists of three concurrent lines.