Past exam of the geometry course of the University of Oxford / 2010 / B / 5

The strong Bézout theorem says that if two projective plane algebraic curves $C,D$ over $\mathbb{C}$ have degrees $m,n$ and no common irreducible component, their intersection is finite andEach intersection multiplicity is a positive integer. Scheme-theoretic multiplicities, rather than just distinct intersection points, are essential: tangencies and singularities can make individual terms larger than one.

For curves with no common local component at $p$,In local affine coordinates, their defining equations have linearly independent gradients exactly in this case. These independent linear terms form local coordinates, so the local intersection quotient has dimension one. Conversely, if either equation lacks a nonzero linear term or the two linear terms are dependent, the local intersection quotient cannot have dimension one. This is transverse intersection of plane curves of two smooth plane curves.

Write the irreducible curve as $F=0$, with $F$ homogeneous of degree $n$. For $n=1$ there are no singular points. Otherwise choose a nonzero partial derivative, say $F_x$, which has degree $n-1$. Such a derivative exists in characteristic zero. Irreducibility and its smaller degree show that $F$ and $F_x$ have no common component.

At every singular point $p$ of $C$, $F_x(p)=0$ and the curve multiplicity of $F$ is at least two. The standard local lower bound for intersection multiplicity gives $I_p(F,F_x)\ge 2$. Applying the Bézout theorem,ThereforeThis is the polar bound for singular points of a plane curve.

Take the union of $n$ distinct projective lines in general position, with no three concurrent. Its defining polynomial is the product of their linear equations, so its degree is $n$. Exactly two lines meet at each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections, producing an ordinary double point; elsewhere just one line passes through and the curve is smooth. HenceFor $n=1$, the line itself gives zero singular points, as required.

The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda ,\mu )\ne (0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.

The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the Bézout theorem would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. ThusAt every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. ThereforeThis is residual intersection through a curve pencil. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.

Let the six distinct conic points be $p_1,\dots ,p_6$, and write ${\ell }_{ij}$ for their joining line. Form two reducible cubicsNo three distinct points of an irreducible conic are collinear, by the Bézout theorem applied to a line and the conic. Thus all six side lines are distinct, the cubics have no common component, and six of their intersections are precisely the vertices. The remaining pairings giveThese three points are off the conic and distinct. For example, equality of $q_1$ and $q_2$ would put that point on both ${\ell }_{34}$ and ${\ell }_{45}$, making it $p_4$ and then placing $p_4$ on ${\ell }_{12}$, impossible. The other cases are identical. Hence the two cubics meet in nine distinct points.

Apply residual intersection through a curve pencil with $n=3$, $m=2$ and $E$ the conic: six intersections lie on $E$, so the other three lie on a curve of degree at most one. Therefore $q_1,q_2,q_3$ are collinear, proving Pascal's theorem.