Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 14 / 1

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

Blow up the distinct singular points of the integral projective curve, once at each original point, and let $C^{\prime }$ be its strict transform of an algebraic subvariety. Write $H$ for the pullback of a projective line and $E_i$ for the algebraic exceptional divisors. The multiplicity of a plane curve at a point and the intersection formula for blowing up a surface giveThe canonical divisor formula for a surface blowup gives $K=-3H+{\sum }_i{E}_i$. Applying the arithmetic adjunction formula on a smooth surface, rather than a formula requiring a smooth strict transform, yieldsand henceThe strict transform of an algebraic subvariety is still an integral projective curve. Its arithmetic genus is $h^1(C^{\prime },{\mathcal{O}}_{C^{\prime }})\ge 0$, even if some singularities remain. Therefore the plane curve singularity multiplicity bound isThe argument uses only one blowup of a smooth algebraic surface at each original singular point; it does not assume these blowups resolve every singularity.

The genus-degree formula gives arithmetic genus three for a plane quartic. Each of its three double points has multiplicity of a plane curve at a point two, so the arithmetic genus drop under a point blowup is one at each point. The resulting integral projective curve $C^{\prime }$ hasFor its finite normalization of an algebraic curve $\nu :\tilde{C}\to C^{\prime }$, the arithmetic genus identity isBoth terms are nonnegative, so $g(\tilde{C})=0$. Over the algebraically closed field, the rationality of a smooth projective genus-zero curve identifies $\tilde{C}$ with the projective line: a point $P$ exists, and Riemann-Roch theorem gives a degree-one map from the two-dimensional space $L(P)$. Consequently $C$ is birational to ${\mathbb{P}}^1$. This proves the rationality of a plane quartic with three double points even when a double point is not an ordinary node.