Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 24I (Algebraic Geometry)

A projective variety is smooth when every local ring is regular, equivalently when its Jacobian has the expected rank at every point. The affine curve $y=x^3$ is smooth because the derivative of $y-x^3$ with respect to $y$ is $1$, while its projective closure $Y{Z}^2=X^3$ is singular at $[0:1:0]$.

The genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Intersect the Fermat cubic threefold with the plane $X_3=X_4=0$. This gives the smooth plane cubic $X_0^3+X_1^3+X_2^3=0$, whose genus of a smooth plane curve is $(3-1)(3-2)/2=1$.

An irreducible plane conic is smooth: a singular quadratic form in three variables has rank at most two and factors over the algebraically closed field, contradicting irreducibility. Projection from any point of the conic, or its degree-two complete linear system, then identifies it with the projective line.

Finally, a nonzero ternary quadratic form is determined up to nonzero scalar by its six coefficients. Thus generalized conics are parametrized bijectively by ${\mathbb{P}}^5$.

Write a conic as $x^{T}Ax=0$ for a symmetric $3\times 3$ matrix $A$, with off-diagonal entries equal to half the corresponding coefficients. A union of two distinct lines has matrix rank two. Thus $R^{\circ }$ is the rank-two part of the determinant hypersurface $\det A=0$. Its closure also contains rank-one matrices, representing double lines, so it is not Zariski closed. Since the determinant is one nonzero homogeneous equation in ${\mathbb{P}}^5$, its dense rank-two locus has dimension $4$.

The final sentence of the converted question writes $R^c$, but the natural intended set is $R^{\circ }$. The homogeneous ideal of its closure is the principal ideal generated by the cubic polynomial $\det A$: the determinant hypersurface is irreducible, and $R^{\circ }$ is dense in it.