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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.

Every projective line on the smooth algebraic surface through $P$ lies in the projective tangent plane $T_{P}S$, since its tangent direction is contained in the surface's Zariski tangent space. The plane section $S\cap T_{P}S$ is a nonzero plane cubic. It cannot be the entire plane, since a smooth cubic surface is irreducible and has no plane component.

Each distinct projective line through $P$ is therefore a distinct degree-one factor of the cubic restricted to $T_{P}S$. A degree-three polynomial has at most three such factors. Hence there are at most three lines through any point. This is the lines through a point of a smooth cubic surface bound; equality is allowed when the tangent section consists of three concurrent lines.

Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.