Past exam of the algebraic curves course of the University of Oxford / 2023 / 1

Write $C=V(P)$, where $P\in \mathbb{C}[x,y,z]$ is a squarefree polynomial and a homogeneous polynomial of degree $d$. A projective point $p\in C$ is a singular point precisely whenIf $p=[p_0:p_1:p_2]$ is a nonsingular point of an algebraic curve, its projective tangent line isThe coefficients are not all zero, and the Euler theorem for homogeneous functions gives $p_0{P}_x(p)+p_1{P}_y(p)+p_2{P}_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the projective point only rescales the equation.

The strong form of Bézout's theorem says that two projective plane curves $C,D$ of degrees $d,e$, with no common irreducible component, have finitely many intersections andEach intersection multiplicity is a positive integer. At $p\in C\cap D$,This is the criterion for a transverse intersection of plane curves. In particular, an intersection at a singular point of either curve contributes at least two.

If $d=1$, $C$ is a line and has no singular points. Suppose $d>1$ and take $P$ to be an irreducible polynomial. First, $P_x$ is not the zero polynomial: otherwise $P\in \mathbb{C}[y,z]$, and a homogeneous polynomial in two variables of degree greater than one factors into linear factors over $\mathbb{C}$, contrary to irreducibility.

Let $D=V(P_x)$ be the polar curve in this coordinate direction. Since $\mathrm{deg}{P}_x=d-1<d$, the irreducible polynomial $P$ cannot divide $P_x$. Thus $C,D$ have no common irreducible component, and Bézout's theorem applies. One may count repeated factors of $P_x$ with their multiplicities; alternatively, passing to its reduced equation only decreases $\mathrm{deg}D$.

Every singular point $p$ of $C$ satisfies $P_x(p)=0$ and has $I_p(C,D)\ge 2$ by part (a). Consequently, if $s$ is the number of distinct singular points,This argument also proves that the singular locus is finite.

There is a nonsingular point of an algebraic curve on $C$, since part (b) makes its singular locus finite whereas a projective curve has infinitely many complex points. Apply a projective transformation taking one such point to $p=[1:0:0]$, and use $P$ for the transformed homogeneous polynomial. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common irreducible component with $C$.

The Euler theorem for homogeneous functions and its differentiated identities giveandSince $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the polar curve $D$ is also nonsingular at $p$, and these identities show that $T_{p}D=T_{p}C$. Hence $I_p(C,D)\ge 2$.

This additional intersection is separate from all the singular points of $C$. Thus Bézout's theorem givesThe improvement comes from making the polar curve tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.

Let $d_i=\mathrm{deg}{C}_i$. Under the reduced plane curve convention, the distinct irreducible components have ${\sum }_i{d}_i=d$. Away from intersections with other components, a point of $C_i$ is nonsingular on the union exactly when it is nonsingular on $C_i$. For example, a local equation of the union is the equation of $C_i$ multiplied by a nonvanishing factor. At an intersection of two components, differentiating their product makes every first partial derivative vanish. Therefore

Part (b) bounds the number of singular points of $C_i$ by $d_i(d_i-1)/2$. For each distinct pair of irreducible components, Bézout's theorem bounds the number of distinct intersections by $d_i{d}_j$. The cardinality of a union is at most the sum of the cardinalities, even if some points have been counted several times. Hence

The reduced convention matters. If one instead allows the equation $L^2=0$ as a nonreduced curve and uses its equation to test singularity, every point on its supporting line has vanishing first partial derivatives. A finite bound would then be false.

Take a union of $d$ distinct lines in general position for projective lines, meaning that no three pass through one point. Each pair contributes one distinct singular point, and every other point lies on just one nonsingular line. Thus the union has degree of a projective curve $d$ and exactlydistinct singular points. For $d=1$, this gives a line with zero singular points, as required.

For a concrete construction, choose distinct $t_1,\dots ,t_d\in \mathbb{C}$ and setFor $i\ne j$, the intersection is $[-(t_i+t_j):-t_i{t}_j:1]$. Substitution into the equation of $L_k$ gives $-(t_k-t_i)(t_k-t_j)$, which is nonzero when $k$ is distinct from $i,j$. Thus there are no triple intersections, including at infinity.