Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 24I (Algebraic Geometry)

For an affine variety $X\subseteq {\mathbb{A}}_k^n$ with vanishing ideal $I(X)$, its Zariski tangent space at $p$ isThe dimension from minimum tangent dimension is

Now let $X=Z(f)$ with $f$ irreducible. The Strong Hilbert Nullstellensatz gives $I(X)=(f)$, soIts dimension is at least $n-1$, and it is exactly $n-1$ wherever $\nabla f(p)\ne 0$.

It remains to find such a point in positive characteristic $q$. If every partial derivative of $f$ were the zero polynomial, every exponent occurring in $f$ would be divisible by $q$. Since the algebraically closed field $k$ is perfect, all coefficients have $q$th roots and $f$ would be a $q$th power, contradicting irreducibility. Thus some partial derivative $f_{x_i}$ is nonzero. If the gradient nevertheless vanished at every point of $Z(f)$, the Nullstellensatz would imply $f_{x_i}\in (f)$. This is impossible because $0\le \mathrm{deg}{f}_{x_i}<\mathrm{deg}f$. The smooth point on an irreducible hypersurface in positive characteristic therefore exists, and

Write a polynomial $f\in W$ by its coefficients. Evaluation of $f$ and of each first partial derivative at $p$ is polynomial in these coefficients and in the coordinates of $p$. Consequently the bad incidence locusis Zariski closed, and $U=(W\times k^3)\setminus B$ is Zariski open. Here a hypersurface is regarded as smooth at an ambient point outside it automatically.

In fact $B$ is empty. For homogeneous $f$ of degree $d>0$, the Euler homogeneous function theorem saysIf $(f,p)\in B$, its left side is zero while its right side is $d$, which is nonzero because $\mathrm{char}k=0$. Thus every nonzero level is covered by the smoothness of a nonzero homogeneous level set, andwhich is certainly nonempty and Zariski open.