Classical Affine Varieties: Coordinate Rings, Morphisms, and Rational Maps

If $a\in V(S)$, then every $q={\sum }_i{r}_i{s}_i\in (S)$ satisfies $q(a)={\sum }_i{r}_i(a)s_i(a)=0$. Conversely $S\subseteq (S)$, so vanishing on $(S)$ implies vanishing on $S$. Thus $V(S)=V((S))$.

Write $J=(S)$. If $a\in V(J)$ and $q^m\in J$ for $m\ge 1$, then $q(a{)}^m=0$, hence $q(a)=0$ in the field $k$. Thus $V(J)\subseteq V(\sqrt{J})$. The reverse inclusion follows from $J\subseteq \sqrt{J}$.

For any indexed family $(S_{\lambda })$, a tuple vanishes on ${\bigcup }_{\lambda }{S}_{\lambda }$ exactly when it vanishes on every $S_{\lambda }$. Hence ${\bigcap }_{\lambda }V(S_{\lambda })=V({\bigcup }_{\lambda }{S}_{\lambda })$; the empty intersection is $k^n$.

If $a\in V(I)\cup V(J)$, every product of an element of $I$ with one of $J$ vanishes at $a$, hence every element of $IJ$ does. If $a$ belongs to neither locus, there are $f\in I,g\in J$ with $f(a),g(a)\mathrm{\ne }0$; then $(fg)(a)\mathrm{\ne }0$, so $a\mathrm{\notin }V(IJ)$. This proves both inclusions.

Replace arbitrary equation sets by generated ideals and iterate the two-set union identity. The zero-set identities for 0 and 1 supply the empty union and the full space. Intersecting these identities with $X$ verifies the induced closed-set axioms.

$E\subseteq V(J)$ says that for every $a\in E$ and every $f\in J$, $f(a)=0$. Interchanging these two universal quantifiers says precisely $J\subseteq I(E)$. Enlarging $E$ imposes more conditions on $I(E)$; enlarging $J$ imposes more equations on $V(J)$. This proves both reversals.

Every point of $E$ lies in $V(I(E))$. If a closed set $V(J)$ contains $E$, step 1.1 gives $J\subseteq I(E)$ and therefore $V(I(E))\subseteq V(J)$. Thus $V(I(E))$ is the smallest closed set containing $E$. Applying this to the already closed set $E=V(J)$ gives the last identity.

The ring is a finite-variable polynomial ring over algebraically closed $k$, so the strong Nullstellensatz applies with the assumed AC and gives $I(V(J))=\sqrt{J}$. The other composite is the identity on closed sets by F2. For radical $J$ the first composite is also the identity, proving the stated inverse bijections and their inclusion reversal.

If $X$ is nonempty irreducible and $fg\in I(X)$, then $X=(X\cap V(f))\cup (X\cap V(g))$. Irreducibility forces one of these closed sets to be $X$, hence $f\in I(X)$ or $g\in I(X)$. Also $1\mathrm{\notin }I(X)$ because $X$ has a point. Thus $I(X)$ is prime.

Conversely suppose $P=I(X)$ is prime. Then $X$ is nonempty, since $I(\mathrm{\varnothing })=R$. If $X=C\cup D$ with $C,D$ proper closed subsets, choose $f\in I(C)\setminus I(X)$ and $g\in I(D)\setminus I(X)$; these exist by the injective reversing correspondence in step 1.1. The product vanishes on $C\cup D=X$, contradicting primality. Thus $X$ is irreducible. For any prime $P$, the elementary implication $f^r\in P\Rightarrow f\in P$ makes $P$ radical, so step 1.1 realizes it by such an $X$.

For a point $a$, evaluation onto $k$ has kernel $I(\{a\})$. A proper ideal strictly containing this kernel would contain an $f$ with $f(a)\mathrm{\ne }0$; subtracting $f-f(a)$ in the kernel puts a nonzero constant in that ideal, hence 1. Thus the kernel is maximal. Conversely F5 writes each maximal ideal as $(x_i-a_i{)}_i$, whose locus is exactly $\{a\}$. The unit ideal has empty locus, and the empty set has vanishing ideal $R$.

Let $\pi :R\to A$ be the quotient and $H⊴A$. Its inverse image contains $I(X)$, so its zero locus lies in $X$ and equals $V_X(H)$. Polynomial vanishing upstairs gives $I(V_X(H))=\sqrt{{\pi }^{-1}H}$. Passing to the quotient using F6 and F7 yields $I_X(V_X(H))=\sqrt{H}$. This also proves the relative closed-set correspondence.

Two disjoint nonempty opens $U,V$ give the proper closed cover $X=(X\setminus U)\cup (X\setminus V)$. Conversely a proper closed cover $X=C\cup D$ gives disjoint nonempty opens $X\setminus C,X\setminus D$. Taking complements proves both directions of the criterion.

If $U$ is nonempty open in irreducible $X$, each nonempty open $W$ of $X$ meets $U$ by step 1.1. Thus no proper closed subset of $X$ contains $U$, which says $\bar{U}=X$. If $V_1,V_2$ are nonempty opens of $U$, they are opens of $X$ because $U$ is open, so they intersect. The criterion makes $U$ irreducible.

The density assertion is F1 with the irreducible nonempty space $X$. For a finite list $U_1,\dots ,U_r$, start with $W_0=X$ and set $W_j=W_{j-1}\cap U_j$. If $W_{j-1}$ is nonempty open, F1 gives $W_j\mathrm{\ne }\mathrm{\varnothing }$; intersection preserves openness. Finite induction gives $W_r$ nonempty open.

F1 now makes $W_r$ dense. This includes $r=0$, since $W_0=X$ is nonempty and its closure is itself.

If $X$ is a variety, F1 makes $I(X)$ prime. The polynomial ring is commutative, so F3 applied to $I(X)$ and the quotient in F2 says $k[X]$ is a nonzero domain.

If $k[X]$ is a nonzero domain, F3 makes $I(X)$ prime, and F1 makes $X$ nonempty irreducible. In particular the zero ring $k[\mathrm{\varnothing }]$ is excluded on both sides.

Give all functions $X\to k$ pointwise operations. Evaluating a polynomial at each point defines a unital $k$-algebra homomorphism $e:R\to k^X$, by iterated polynomial evaluation, since sums and products evaluate to sums and products. Its image is exactly the polynomial functions. Its kernel consists exactly of the polynomials vanishing on every point, namely $I(X)$.

F3 gives $R\mathrm{/}I(X)\cong \mathrm{im}e$ by $[f]\mapsto (a\mapsto f(a))$, and constants show it is an isomorphism over $k$. When $X$ is empty there is one function, its function algebra is the zero ring, and $\ker e=R$; the same quotient identification applies.

For $f,g\in k[X]$, $f(x)g(x)\mathrm{\ne }0$ in the field $k$ exactly when both factors are nonzero. This proves the intersection identity, including $f=0$ or $g=0$ and $f=1$.

If $U=X\setminus (X\cap V(S))$ is open, then $x\in U$ exactly when some $s\in S$ does not vanish at $x$. Hence $U={\bigcup }_{s\in S}{D}_X(\bar{s})$. Each member is open and contained in $U$. This gives a principal neighbourhood of each point of $U$, and the empty $S$ gives the empty union.

Near a given point, intersect the neighbourhoods on which $s=g\mathrm{/}h$ and $t=a\mathrm{/}b$ with $h,b$ nowhere zero. Then $s+t=(gb+ah)\mathrm{/}(hb)$, $st=ga\mathrm{/}(hb)$, and $-s=-g\mathrm{/}h$ on this intersection; the denominators are nonzero there. Constants are $c\mathrm{/}1$. Pointwise ring laws therefore give a $k$-algebra, including the zero function algebra on the empty open.

Restricting a quotient expression to its intersection with a smaller open preserves regularity. Pointwise sums, products and constants restrict to themselves, so restriction is a unital algebra homomorphism; iterated restrictions agree.

Given $U={\bigcup }_i{U}_i$ and regular $s_i$ with equal restrictions on every overlap, the union of their function graphs is a function $s:U\to k$: for a fixed point all available values coincide. On $U_i$ it equals $s_i$, hence near each point it has the quotient expression of that section. Thus it is regular. A function with these restrictions must have that same value at every point, proving uniqueness. For the empty cover of the empty open its graph is empty.

Restriction $A\to {\mathcal{O}}_X(D(f))$ is a unital algebra map. The function $1\mathrm{/}f$ is regular on $D(f)$, so F6 extends restriction to the displayed map. If $a\mathrm{/}{f}^r$ maps to zero, $a$ vanishes on $D(f)$; hence $fa$ vanishes everywhere on $X$, since $f=0$ outside $D(f)$. F2 gives $fa=0$ in $A$, and F7 makes the fraction zero. This proves injectivity without cancellation in $A$.

Fix $s\in {\mathcal{O}}_X(D(f))$. At each point take a local expression $s=g\mathrm{/}h$ and refine its neighbourhood to $D(a)\subseteq D(f)\cap D(h)$ by F3. The inclusion $V_X(h)\subseteq V_X(a)$ gives $a\in \sqrt{(h)}$ by F1. Thus $a^e=hb$ for some $e\ge 1,b\in A$. On $D(a)$, set $t=a^e$ and $c=gb$; then $D(t)=D(a)$ and $s=c\mathrm{/}t$.

Consider the set of all pairs $(t,c)$ obtained in step 1.2; their opens $D(t)$ cover $D(f)$ and are contained in it. Thus $f$ vanishes on the simultaneous zero locus of the $t^2$. By F1, $f\in \sqrt{(t^2:(t,c)\text{ as above})}$. F8 supplies finitely many of these pairs $(t_i,c_i)$ and $u_i\in A$ with $f^N={\sum }_{i=1}^m{u}_i{t}_i^2$, for some $N\ge 1$. No compactness theorem or simultaneous choice of neighbourhoods is needed.

For $p\in D(f)$ and each selected pair, if $t_i(p)\mathrm{\ne }0$ then $c_i(p)=s(p)t_i(p)$; if $t_i(p)=0$ then both $c_i(p)t_i(p)$ and $s(p)t_i(p{)}^2$ are zero. Hence ${\sum }_i{u}_i(p)c_i(p)t_i(p)=s(p){\sum }_i{u}_i(p)t_i(p{)}^2=s(p)f(p{)}^N$. Division by the nonzero scalar $f(p{)}^N$ shows that $({\sum }_i{u}_i{c}_i{t}_i)\mathrm{/}{f}^N$ maps to $s$. This proves surjectivity.

If $D(f)$ is empty, $f$ is the zero function on $X$, hence zero in $A$ by F2. Localizing at 0 is the zero ring by F7; the empty domain has one function and its algebra is zero. If $f=1$, the same construction gives global sections and $A_1=A$. Together with injectivity and surjectivity this proves all cases.

Every ideal of the field $k$ is either 0 or $k$: a nonzero member is invertible, putting 1 in the ideal. These ideals are generated by 0 and 1, respectively, so F2 makes $k$ Noetherian and F1 makes $R=k[x_1,\dots ,x_n]$ Noetherian, including $n=0$. Apply F3 to the radical ideal $I(X)$, with its stipulated DC cost, to obtain its distinct minimal primes $P_1,\dots ,P_m$ with $I(X)={\bigcap }_i{P}_i$.

Put $X_i=V(P_i)$. F4 makes each nonempty irreducible. A point outside every $X_i$ admits $f_i\in P_i$ nonzero there; the finite product belongs to every $P_i$ and is nonzero at the point. Thus $V({\bigcap }_i{P}_i)\subseteq {\bigcup }_i{X}_i$; the reverse inclusion follows because every element of the intersection vanishes on each $X_i$. So $X={\bigcup }_i{X}_i$. For $m=0$, the intersection is $R$ and the union empty. Distinct minimal primes are incomparable; F4 therefore makes the $X_i$ incomparable.

An irreducible nonempty closed set $Z$ covered by finitely many closed sets must be contained in one: repeatedly split the cover as the first member and the union of the others; if $Z$ is not contained in the first, irreducibility puts it in the remaining union. For a one-member cover this ends immediately; a zero-member cover cannot cover nonempty $Z$. Thus every irreducible closed subset of $X$ is contained in some $X_i$, and incomparability makes the $X_i$ precisely the maximal ones. It also prevents removal of an $X_i$ from the cover.

If $X={\bigcup }_j{Y}_j$ is another finite irredundant irreducible closed decomposition, step 3.1 puts each $Y_j$ inside an $X_i$ and that $X_i$ inside some $Y_h$. Irredundancy forces $Y_j=Y_h$ (otherwise $Y_j$ could be removed), so $Y_j=X_i$. Reversing the two covers shows their members coincide. After removal of duplicate indexing, this is uniqueness up to permutation.

At $x\in X$, evaluation on $R$ kills $I(X)$ and hence defines evaluation on $A$; constants make it surjective. Its kernel is maximal by the same argument as F2, or by correspondence with the maximal evaluation ideal upstairs. Two points with equal kernels have equal inverse images upstairs, and F4 makes the points equal.

If $M$ is maximal in $A$, its inverse image in $R$ is maximal by F3. F4 identifies it with the evaluation ideal at a unique $x$. Since it contains $I(X)$, $x\in V(I(X))=X$. Thus $M={\mathfrak{m}}_x$. Evaluation induces a bijection $A\mathrm{/}{\mathfrak{m}}_x\to k$: equality of values is exactly equality modulo the kernel, and every constant is attained. It preserves all operations. For $X=\mathrm{\varnothing }$, $A=0$ has no proper, hence no maximal, ideals.

Apply F1 to $f=1$. Then $D(1)=X$, so the displayed isomorphism is $A_1\cong {\mathcal{O}}_X(X)$. The maps $A\to A_1$, $a\mapsto a\mathrm{/}1$, and $A_1\to A$, $a\mathrm{/}{1}^r\mapsto a$, are mutually inverse algebra maps.

The composite sends $a$ to its function on $X$, which is exactly the canonical map in the statement. If $X$ is empty both algebras are the zero ring by the empty case of F1.

If $r$ is regular on an open $U$ and $r(x)\mathrm{\ne }0$, write $r=a\mathrm{/}b$ near $x$ with $b$ nowhere zero. Intersect that neighbourhood with $D(a)$; it is a neighbourhood of $x$ where $r$ is nowhere zero. Thus the nonvanishing set of $r$ is open, and its zero set is relatively closed in $U$.

Write $C=Y\cap V(S)$. Each coordinate polynomial restricted to $Y$ is globally regular (its denominator is 1), so F3 makes its pullback regular. Therefore ${\varphi }^{-1}(C)$ is the intersection of their closed zero sets by step 1.1. This proves continuity, including empty and full closed sets.

For $x\in {\varphi }^{-1}(W)$, take a neighbourhood $W_0\subseteq W$ of $\varphi (x)$ where $s=g\mathrm{/}h$ with $h$ nowhere zero. On the open ${\varphi }^{-1}(W_0)$, $g\circ \varphi ,h\circ \varphi$ are regular and the latter is nowhere zero. Near $x$, write them as $a\mathrm{/}b,c\mathrm{/}d$ with $b,d$ nonzero. Shrink further to where $c\mathrm{\ne }0$ using step 1.1. Then $s\circ \varphi =ad\mathrm{/}(bc)$ is a valid local quotient, proving the local pullback property.

If $\varphi :X\to Y$ is a morphism, F1 and F2 identify $s\mapsto s\circ \varphi$ as a map $k[Y]\to k[X]$. Pointwise addition, multiplication and constants show it is a unital $k$-algebra homomorphism.

Conversely let $\alpha :k[Y]\to k[X]$ be a unital $k$-algebra map, with $Y\subseteq k^m$ and coordinate classes $y_j$. Define $\varphi (x)=(\alpha (y_1)(x),\dots ,\alpha (y_m)(x))$. If $P\in I(Y)$ then polynomial evaluation and the homomorphism laws give $P(\varphi (x))=\alpha (P(y_1,\dots ,y_m))(x)=0$. Hence $\varphi (x)\in V(I(Y))=Y$.

Every global regular function of $Y$ is a polynomial in the $y_j$ by F1 and F3. Substitution in step 1.2 gives $s\circ \varphi =\alpha (s)$, a polynomial and hence regular function on $X$. Thus $\varphi$ is a morphism and its pullback is $\alpha$. If one starts with $\varphi$, its pulled-back coordinate values reconstruct exactly $\varphi (x)$, so the constructions are inverses.

For composable morphisms, $s\circ (\psi \circ \varphi )=(s\circ \psi )\circ \varphi$, so $(\psi \circ \varphi {)}^{\ast }={\varphi }^{\ast }\circ {\psi }^{\ast }$; F8 ensures the composites are morphisms, and the identity pulls each function to itself. This also gives naturality of the bijection. If $X$ is empty there is one map to any $Y$ and one unital homomorphism to $k[X]=0$. If $Y$ is empty and $X$ nonempty there is neither a set map nor a unital map $0\to k[X]$, since $0=1$ would force $k[X]=0$. Both empty gives one on each side.

Let $B$ be a nonzero finite-type domain. By F7 choose a surjection $k[T_1,\dots ,T_r]\to B$ with kernel $P$. It is proper since $B\mathrm{\ne }0$, and $uv\in P$ forces one image to be zero since $B$ is a domain; thus $P$ is prime. F6 gives a variety $V(P)$ with $I(V(P))=P$. Its coordinate ring is the presentation quotient $B$. Together with the bijection and composition law this proves the stated antiequivalence on domains.

For every algebraic set $X$, F1 places $k[X]$ in the proposed algebra category, and F4 makes pullback a contravariant functor that is bijective on every hom-set. Thus it is fully faithful, including the empty cases already verified there.

Given a reduced finite-type algebra $B$, choose a surjective presentation $R=k[T_1,\dots ,T_m]\to B$ with kernel $J$. If $h^r\in J$ for $r\ge 1$, the image of $h$ is nilpotent and hence zero by reducedness, so $h\in J$. Therefore $J$ is radical. F3 gives $I(V(J))=J$, and the presentation identifies $k[V(J)]=R\mathrm{/}J$ with $B$. For $B=0$, $J=R$ and $V(J)=\mathrm{\varnothing }$.

The object realization can be made intrinsic: use the maximal ideals of $B$, with the topology and regular functions transported from any presentation. F5 identifies the points, and the algebra isomorphism between two presentations gives inverse morphisms by F4. Their pullbacks are the prescribed identities on $B$, so full faithfulness forces all such comparison maps and their composites to agree. Thus the realization is independent up to canonical isomorphism, and step 1.1 with step 1.2 proves the antiequivalence.