Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 25F / a

WriteThe equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the cuspidal cubicIt is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.

If $y=0$, the other equations becomeTheir solutions are $(0,0,0)$, which already lies on $C$, and the two pointsThus the irreducible components areEquivalently, the radical ideal of the cusp with two isolated-point components is

Let $(a,b,c)$ be a tangent vector in the $x,y,z$ directions. At a point $(0,y,z)\in C$, the two isolated components are absent in a Zariski neighbourhood, so the Zariski tangent space is that of $C$. Differentiating $x$ and $y^2-z^3$ givesFor $(y,z)\ne (0,0)$ this is a line. At the cusp,which is two-dimensional.

Each of $P$ and $Q$ is a reduced isolated component with local ring $k$, soThis last conclusion must use all functions in $I(V)$, as required by the definition of the Zariski tangent space. The Jacobian of only the three displayed defining polynomials has an extra spurious tangent direction at $Q$, because those polynomials do not generate the radical ideal there; for example, $x(x^2-1)\in I(V)$ removes that direction.

The curve $C$ has dimension one. Its tangent has dimension one at every point except the origin and dimension two at the origin. By the smooth locus of a variety,and the origin is its unique singular point of an algebraic variety. Each reduced point component has dimension zero and zero-dimensional tangent space, so both $P$ and $Q$ are smooth on their respective components.

The irreducible affine hypersurface $y^2-z^3=0$ in the $(y,z)$-plane has dimension one, while a closed point has dimension zero. Therefore