Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 25F (Algebraic Geometry)

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

Choose a $K$-basis $e_1,\dots ,e_n$ of $L$ and writethereby identifying $L$ with ${\mathbb{A}}_K^n$. Let $M(x)$ be the $n\times n$ matrix whose $j$th column is the coordinate vector of ${\alpha }^{j-1}$ in this basis, for $1\le j\le n$.

Write the multiplication table aswith fixed structure constants $c_{ij}^l\in K$. Repeated multiplication shows that every entry of $M(x)$ is a polynomial in $x_1,\dots ,x_n$. HenceBy the criterion supplied in the question,Consequentlywhich is a distinguished open set, and therefore a Zariski-open set. This is the determinant construction showing that primitive elements form a principal Zariski-open set. No assertion that $D$ is nonzero is needed.