Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 25J / Solution

For $X\subseteq {\mathbb{A}}^n$ and $P\in X$, the Zariski tangent space isEquivalently, it is the kernel of the Jacobian of generators of $I(X)$ at $P$. With the tangent-space definition used here,and $P$ is smooth when $\dim T_{X,P}=\dim X$.

Now let $X=Z(f)$. Write $g$ for the square-free product of the distinct irreducible factors of $f$. Hilbert's Nullstellensatz givesThus $T_{X,P}=\ker (d{g}_P)$, whose dimension is at least $n-1$. There is some $P\in X$ with $d{g}_P\ne 0$. Otherwise every partial derivative of $g$ would vanish on $X$, hence belong to $(g)$ by the Nullstellensatz. Its degree is smaller than $\mathrm{deg}g$, so every partial derivative would be zero; in characteristic zero this would make $g$ constant, a contradiction. At such a point the tangent space has dimension $n-1$, and therefore

Next suppose $f$ is irreducible of degree at least two and $P$ is smooth. Translate $P$ to the origin and writeThen $H=T_{X,P}=Z(\ell )$ is a hyperplane. The restriction $f{|}_H$ is nonzero: otherwise $\ell$ would divide $f$, contradicting irreducibility and $\mathrm{deg}f\ge 2$. Since $f(P)=0$, the restriction is nonconstant, so the first part givesFor the natural intersection cut out by $(f,\ell )$, the two differentials at $P$ are $d{f}_P=\ell$ and $d\ell =\ell$. Hence they impose only one independent tangent equation andTherefore $X\cap T_{X,P}$ is singular at $P$.

Finally, represent $\phi :{\mathbb{A}}^2\to {\mathbb{A}}^3$ byIt is not injective exactly when $\mathrm{rank}A\le 1$, equivalently when all its $2\times 2$ minors vanish. Thusan ideal generated by three quadrics. By the stated assumption this is the full radical ideal of $Y$.

Every nonzero point of $Y$ has rank one. The action of $G{L}_3(k)\times G{L}_2(k)$ by row and column operations is transitive on rank-one matrices, so it suffices to compute atAt $A_0$, the differentials of the three minors are respectivelyHence $\dim T_{Y,A_0}=6-2=4$, and the same holds at every rank-one point. At the zero matrix all three quadrics have zero differential, so $T_{Y,0}={\mathbb{A}}^6$ has dimension six. It follows that