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An affine variety in ${\mathbb{A}}_k^n$ is the common zero set of a collection of polynomials in $k[x_1,\dots ,x_n]$, with its Zariski topology. We use the convention of this question that such a variety need not be irreducible. Its vanishing ideal and coordinate ring areThe quotient identifies polynomials that give the same function on $X$, so its elements are exactly polynomial functions on $X$.

For affine embeddings $X\subseteq {\mathbb{A}}^n$ and $Y\subseteq {\mathbb{A}}^m$, a morphism of affine varieties is a map $F=(f_1,\dots ,f_m)$ with $f_i\in A(X)$ and image in $Y$. Pullback of polynomial functions defines the contravariant homomorphismIt is well defined because any polynomial vanishing on $Y$ pulls back to zero on $X$, and it preserves addition, multiplication, one, and the constants in $k$.

Polynomial functions are regular, giving an injective map $A(X)\hookrightarrow {\mathcal{O}}_X(X)$; injectivity follows from quotienting by the full vanishing ideal. To prove surjectivity, let $f$ be globally regular and consider the denominator idealIt is an ideal because sums and polynomial multiples preserve this property.

Fix $p\in X$. Locally $f=a/b$ on a neighbourhood $V$ of $p$, with $b$ nonzero there. Principal affine opens form a basis, so choose $h\in A(X)$ with $h(p)\ne 0$ and $D(h)\subseteq V$. Thenas functions on all of $X$: on $D(h)$ it follows from the local fraction, and off $D(h)$ both sides vanish because $h=0$. Hence $hb\in J$ and $(hb)(p)\ne 0$.

Thus $J$ has no common zero on $X$. By the weak Nullstellensatz for $A(X)$, every proper ideal lies in a maximal ideal belonging to a point of $X$; therefore $J$ cannot be proper. We get $1\in J$, so $f$ itself is polynomial. This provesThe argument works for reducible affine varieties as well: it uses functions and denominator ideals rather than assuming a single fraction field.

This is the denominator ideal proof for global regular functions.

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