Affine algebraic set

An affine algebraic set is the common zero set $V(I)$ of an ideal in a polynomial ring over a field.

A classical variety is affine if finitely many global regular functions generate the unit ideal and their nonvanishing opens are affine. Localization of global sections on a principal open identifies their coordinate rings with $A_{f_i}$, where $A=\Gamma (X,{\mathcal{O}}_X)$. Choose a finite-type subalgebra of $A$ containing the functions, their unit-ideal coefficients and numerators of generators of every $A_{f_i}$. Its affine variety has the same principal affine charts, whose isomorphisms glue globally. This avoids assuming finite generation of $A$ before affineness is known.

A map between affine varieties is a morphism when its coordinate functions are restrictions of ambient polynomials. Pullback identifies these morphisms contravariantly with homomorphisms of their coordinate rings.

The morphism ${\mathbb{A}}^1\to V(x^4-y^3)$ given by $t\mapsto (t^3,t^4)$ is bijective on points over an algebraically closed field, but is not an isomorphism. At a nonzero curve point, $t=y/x$ is the unique inverse parameter, while the origin has unique preimage zero. The induced map on coordinate rings has image $k[t^3,t^4]\subset k[t]$, which lacks the monomial $t$: all its monomial exponents have the form $3a+4b$, with $a,b\ge 0$. Nor can $t$ be regular at the cusp, since an equality $t=P(t^3,t^4)/Q(t^3,t^4)$ with $Q(0)\ne 0$ would produce a nonzero degree-one term in $tQ$ and none in $P$. Thus pointwise bijectivity does not guarantee an algebraic inverse, even though the inverse is regular away from one singular point.

For affine varieties $X,Y$ over an algebraically closed field $k$, their product in a category is affine, with coordinate ring $A{\otimes }_{k}B$, where $A=k[X]$ and $B=k[Y]$. Presenting $A=k[x]/J$ and $B=k[y]/L$ identifies this ring with $k[x,y]/(J,L)$. If both varieties are irreducible, this is an integral domain: express a nonzero tensor $F$ using linearly independent second factors. The points $x$ where its specialization $F_x$ is nonzero form a nonempty open subset of $X$. For two nonzero tensors $F,G$, choose $x$ in the intersection of these open subsets; then $F_x{G}_x\ne 0$ in the domain $B$. The Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of $X\times Y$. The universal property follows from the universal property of the tensor product of modules as a tensor product of commutative algebras.

The vanishing ideal of $X\subseteq k^n$ is

For an affine algebraic set $X\subseteq {\mathbb{A}}_k^n$, its coordinate ring isThe set $X$ is irreducible exactly when $k[X]$ is an integral domain.

For an affine algebraic set $X$ over an algebraically closed field,If $fg$ vanishes on $X$, then $X$ is the union of its intersections with $V(f)$ and $V(g)$. Conversely, a nontrivial decomposition into closed subsets supplies functions vanishing on each proper piece whose product vanishes on all of $X$.

A morphism $f:X\to Y$ of affine varieties is dominant when its image is Zariski dense in $Y$. It is dominant exactly when the pullback homomorphismis injective. When $X$ and $Y$ are irreducible, this injection extends to their function fields and gives $\dim X\ge \dim Y$.

This is the dense-image case of a morphism of algebraic varieties.

For ideals $I,J\subseteq k[x_1,\dots ,x_n]$,The second equality also follows from $V(IJ)=V(I)\cup V(J)$ and

The algebraic sethas the two irreducible componentsEach is irreducible because the linear change $u=x-z$, $v=x+z$ identifies its coordinate ring withan integral domain.