 Solution

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