Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 24G / iii / Solution

The ideal $(f)$ is prime, since an irreducible element of the polynomial ring is prime. By the Hilbert Nullstellensatz, it is the full vanishing ideal of $Z(f)$. Its Zariski tangent space at a point $a$ is therefore the kernel of the single linear form $d{f}_a$, and has dimension $n-1$ or $n$.

At least one partial derivative $f_j$ is nonzero. In characteristic zero this is immediate for a nonconstant polynomial. In characteristic $p$, if all partial derivatives vanished, all exponents would be multiples of $p$. Since an algebraically closed field is perfect, $f$ would be a $p$th power of a nonconstant polynomial, contradicting irreducibility.

The nonzero $f_j$ cannot vanish everywhere on $Z(f)$: the Nullstellensatz would give $f_j^r\in (f)$, hence $f\mid f_j$, contradicting $\mathrm{deg}{f}_j<\mathrm{deg}f$. Thus some point has $d{f}_a\ne 0$, and the minimum tangent-space dimension is