Past exam of the algebraic curves course of the University of Oxford / 2023 (B3.3, A12157W1)

Solutions to all three questions of the University of Oxford Part B algebraic curve paper, sat on 6 June 2023. Source: Oxford's released examination PDF, compilation pages 218–220; paper pages 1–3. The solutions use the paper's labels. Curves in Question 1 are understood to be reduced plane curves; the qualification concerning lines in 2(a), and the unnecessary restriction on the half-period values in 3(d), are discussed where they arise.

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.

A nonsingular point of an algebraic curve $p$ is an inflection point of an algebraic plane curve, or flex, if its projective tangent line has contact of order at least three:Here the projective tangent line must not be a component of $C$. Ordinary tangency has contact order two.

Let $C=V(F)$, where $F$ is a homogeneous polynomial of degree $d\ge 2$. Its Hessian curve is defined byEach entry has degree $d-2$, so a nonzero $H_F$ has degree $3(d-2)$. By the Hessian criterion for a flex, the flexes are exactly the points of $C\cap V(H_F)$. The stated no-common-component assumption and Bézout's theorem therefore giveThis counts distinct points; higher contact can use more than one unit of intersection multiplicity at a single point.

For completeness, the Hessian criterion for a flex follows by putting $p=[0:0:1]$ and $T_{p}C=\{y=0\}$, with $F_y(p)\ne 0$. The Euler theorem for homogeneous functions gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whenceThe restriction $F(x,0,1)$ already has vanishing constant and linear terms. Its quadratic term vanishes exactly when the intersection multiplicity with the tangent is at least three, proving the criterion.

For $d\ge 3$ the bound is the actual maximum number of flexes of a smooth plane curve: a general nonsingular curve has only contact of order three, and the corresponding intersections with its Hessian curve are simple. One way to see genericity is to vary a point and a line through it, a three-dimensional family. Contact of order at least four imposes four independent linear conditions on the coefficients of a degree-$d$ equation for $d\ge 3$. Thus the locus of equations with such contact is a proper exceptional locus. Locally, if the curve is $y=h(x)$, its Hessian curve restricted to $C$ is a nonvanishing factor times $h^{\prime \prime }(x)$; contact exactly three gives a simple zero.

A nonsingular plane conic has no flexes, agreeing with the formula at $d=2$. A line is excluded from this formulation: its tangent is itself and its Hessian matrix is zero, so the no-common-component hypothesis fails. The formula is therefore intended for $d\ge 2$, rather than giving the meaningless negative value at $d=1$.

For $F=x^3+y^3+3x{z}^2$, the first partial derivatives areTheir simultaneous vanishing would force $y=0$, $xz=0$ and $x^2+z^2=0$, hence $x=y=z=0$, which is not a projective point. This also verifies that the given plane cubic is nonsingular.

The Hessian matrix and its determinant areThus the Hessian curve is the union of $y=0$, $x=z$ and $x=-z$. On $y=0$, the equation of $C$ reduces to $x(x^2+3z^2)=0$, givingOn $x=z$ one must have $z\ne 0$, so put $z=x=1$ and obtain $y^3=-4$. On $x=-z$ similarly put $z=1$, $x=-1$, obtaining $y^3=4$. Choose $r\in \mathbb{C}$ with $r^3=4$ and let $\omega =\exp (2\pi \mathrm{i}/3)$. The remaining points areAll nine listed projective points are distinct. They exhaust the intersection with the Hessian curve, so the Hessian criterion for a flex proves that these, and only these, are the flexes in this example. There are none at infinity: $z=0$ and $H_F=0$ would force $x=y=0$ on $C$.

First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex $[0:0:1]$ to the point at infinity $[0:1:0]$ of a Weierstrass model.

Put $h=\omega -1$ and choose $s\in \mathbb{C}$ with $s^2=h^3$. Both are nonzero. DefineSince $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, $\lambda \ne 0,1$.

Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is $-\sqrt{3}/(hs)\ne 0$, so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.

Let $(r_0,r_1,r_2)$ be any ordering of the three distinct numbers $0,1,\lambda$. Set $h=r_1-r_0$ and choose $s\ne 0$ with $s^2=h^3$. The projective transformationhas the required matrix form. As $x=hX+r_{0}Z$ and $y=sY$, substitution givesThe choice $s^2=h^3$ is exactly what cancels the coefficient introduced by the change in $x$.

The six orderings give the Legendre parameter orbit:Every entry is different from $0,1$ when $\lambda \ne 0,1$.

There are no further possibilities for a matrix of the stated form. Its determinant is $acd$, so $a,c,d$ must be nonzero. On $z=1$ it induces the affine change $u\mapsto (a/d)u+b/d$ and scales $v=y/z$ by $c/d$. It preserves the line $y=0$, whose three intersections with the plane cubic have $u$-coordinates $0,1,\lambda$. Their images must have coordinates $0,1,\tilde{\lambda }$. Choosing the two roots sent to $0,1$ determines the affine change and leaves exactly the six orderings above.

For general $\lambda$ the six values are distinct. They coincide in some special cases: $\lambda \in \{-1,2,1/2\}$ gives three values, and ${\lambda }^2-\lambda +1=0$ gives two. In particular, for the value obtained in part (c), the two possible values are $(1\pm \mathrm{i}\sqrt{3})/2$. This does not contradict the paper's qualification that there are six for general $\lambda$.

A divisor on an algebraic curve is a finite formal sumIts degree of a divisor is $\mathrm{deg}D={\sum }_p{n}_p$. It is an effective divisor, written $D\ge 0$, when every $n_p\ge 0$.

A meromorphic differential on a Riemann surface is a one-form represented in every local holomorphic coordinate $t$ by $a(t)dt$, where $a$ is a meromorphic function. Under a coordinate change $t=t(u)$, its coefficient becomes $a(t(u))t^{\prime }(u)$, so these expressions agree on overlapping charts. Since $t^{\prime }(u)$ is a nonvanishing holomorphic function for a valid coordinate change, the order of a zero or pole of the differential is independent of coordinate. Equivalently, for a fixed nonconstant meromorphic function $h$, any such differential can be written $fdh$ with $f$ meromorphic.

A canonical divisor is the divisor on an algebraic curveof a nonzero meromorphic differential on a Riemann surface $\eta$. If ${\eta }^{\prime }=f\eta$, then $({\eta }^{\prime })=(\eta )+(f)$, so all canonical divisors are related by linear equivalence of divisors. Their degree is

The Riemann-Roch theorem states that, for every divisor on an algebraic curve $D$ and any canonical divisor $K$ on the nonsingular projective curve $C$ of genus $g$,The term $\ell (K-D)$ measures the failure of the naive count of conditions on meromorphic functions.

Choose a nonzero meromorphic differential on a Riemann surface $\eta$ and write $K=(\eta )$. Multiplication defines a linear mapIt takes values in the holomorphic differential forms because $(f\eta )=(f)+K\ge 0$. It is injective since $\eta \ne 0$. It is surjective because, for any holomorphic differential form $\omega$, the ratio $\omega /\eta$ is a meromorphic function and has $(\omega /\eta )+K=(\omega )\ge 0$. Thus this map identifies $\mathrm{HD}(C)$ with the canonical Riemann-Roch space.

Furthermore, $\mathcal{L}(0)=\mathbb{C}$: a meromorphic function with no poles is holomorphic on the compact Riemann surface $C$, and hence constant by the maximum modulus principle. Applying the Riemann-Roch theorem with $D=0$ givesTherefore

The generator $\eta =fdh$ is a nonzero holomorphic differential form. Its canonical divisor $(\eta )$ is an effective divisor, since there are no poles. In genus one, the degree of a divisor formula in part (a) givesAn effective divisor of degree zero has every coefficient zero. ThusThis proves nonvanishing of holomorphic differentials in genus one. It concerns the differential itself: a zero or pole of the particular coefficient $f$ may be cancelled by the local behavior of $dh$.

An explicit answer, initially written where $yz\ne 0$, is the meromorphic differential on a Riemann surfaceThe ratios $x/z$ and $y/z$ are functions on $C$, so the expression does not depend on the representative of its homogeneous coordinates. It is a scalar multiple of an invariant differential on an elliptic curve.

On the complex torus $\mathbb{C}/\Lambda$, the differential $dw$ descends from $\mathbb{C}$ because translation by any element of the period lattice leaves $dw$ unchanged. It is holomorphic and nowhere zero. The given biholomorphism satisfies, wherever the displayed quotient is initially defined,Consequently, $\eta =({\Phi }^{-1}{)}^{\ast }(dw)$ extends across every apparent denominator zero and across $[0:1:0]$ as a nonzero holomorphic differential form on all of $C$.

One can also check the apparent exceptional points directly. On $z=1$, put $P(X)=4{\prod }_i(X-e_i)$, so $Y^2=P(X)$. At $Y=0$ the roots are distinct, and differentiating giveswhich is regular and nonvanishing there. At $[0:1:0]$, use $u=x/y$, $v=z/y$. The equation is $v=4{\prod }_i(u-e_{i}v)$, so $u$ is a local parameter on a smooth algebraic curve and $v=4u^3+O(u^5)$. Hencealso regular and nonvanishing.

Only distinctness of the $e_i$ is needed. The paper additionally writes $e_i\ne 0$, which is not true for every period lattice: for $\Lambda =\mathbb{Z}+\mathrm{i}\mathbb{Z}$, the symmetry $\wp (\mathrm{i}w)=-\wp (w)$ makes the half-period value at $w=(1+\mathrm{i})/2$ zero, since $\mathrm{i}w\equiv w(\mathrm{mod}\Lambda )$. This unnecessary restriction has no effect on the answer.

First fix distinct $q_1,\dots ,q_k$ and let $Q=q_1+\cdots +q_k$. A negative-degree divisor has no sections: if $0\ne f\in \mathcal{L}(D)$, then $(f)+D$ is an effective divisor of degree $\mathrm{deg}D$, since a principal divisor has degree zero, so $\mathrm{deg}D$ cannot be negative. As $\mathrm{deg}K=0$ and $\mathrm{deg}(K-Q)=-k<0$, the Riemann-Roch theorem therefore gives

Now choose distinct $p_1,\dots ,p_k$ outside the support of $Q$, and put $P=p_1+\cdots +p_k$. A function with the required zeros and poles must be a nonzero element of $\mathcal{L}(Q-P)$. Conversely, if $0\ne f\in \mathcal{L}(Q-P)$, then $(f)+Q-P$ is an effective divisor of degree zero, hence is zero. It follows that $(f)=P-Q$, giving precisely the required simple zeros and simple poles, with no others. Thus the existence question is exactly whether $\mathcal{L}(Q-P)\ne 0$.

Take a basis $f_1,\dots ,f_k$ of the Riemann-Roch space $\mathcal{L}(Q)$. Evaluation at the $p_i$ defines the linear mapwhose kernel is $\mathcal{L}(Q-P)$. The evaluation criterion for a prescribed principal divisor becomesThe crucial point is that this determinant does not vanish identically; a count of unknowns and equations alone would not prove this.

To verify it, begin with $V_0=\mathcal{L}(Q)$, of dimension $k$. Whenever $V_{j-1}\ne 0$, choose a nonzero $f\in V_{j-1}$. Its zeros are finite. Choose $p_j$ outside those zeros, the support of $Q$ and the previously chosen points. Evaluation at $p_j$ is then a nonzero linear functional on $V_{j-1}$, soAfter $k$ steps, $V_k=0$. Thus there exists a tuple of distinct points with $\Delta \ne 0$.

The entries of this determinant vary holomorphically away from the $q_i$, and are rational functions on the nonsingular projective curve. Its vanishing is therefore a proper closed exceptional locus on this configuration space; its complement is open and dense in the Zariski topology. For a generic choice of the $p_i$, $E_P$ is invertible andThis holds for every fixed tuple of distinct $q_i$, which proves the assertion for generic choices of both sets of points. At $k=1$, it is especially transparent: $\mathcal{L}(q_1)=\mathbb{C}$, so no nonzero member can vanish at a distinct $p_1$.