Affine Algebraic Sets and Coordinate Rings

Because $S\subseteq (S)$, every point annihilating $(S)$ certainly annihilates $S$, so $V((S))\subseteq V(S)$. Conversely, if $a\in V(S)$ and $g={\sum }_{i=1}^r{h}_i{f}_i\in (S)$ with $f_i\in S$, then $g(a)={\sum }_{i=1}^r{h}_i(a)f_i(a)=0$. Hence $a\in V((S))$. Therefore $V(S)=V((S))$.

Since $(S)\subseteq \sqrt{(S)}$, we have $V(\sqrt{(S)})\subseteq V((S))$. Conversely, if $a\in V((S))$ and $g\in \sqrt{(S)}$, choose $m\ge 1$ with $g^m\in (S)$. Then $g(a{)}^m=g^m(a)=0$, and a field has no nonzero nilpotent elements, so $g(a)=0$. Thus $a\in V(\sqrt{(S)})$, proving $V((S))=V(\sqrt{(S)})$.

Combining steps 1.1 and 2.1 gives $V(S)=V((S))=V(\sqrt{(S)})$.

If $\{S_{\alpha }{\}}_{\alpha \in A}$ is any family of subsets of $k[x_1,\dots ,x_n]$, then a point lies in every $V(S_{\alpha })$ exactly when it annihilates every polynomial in every $S_{\alpha }$, that is, exactly when it lies in $V({\bigcup }_{\alpha \in A}{S}_{\alpha })$. Hence ${\bigcap }_{\alpha \in A}V(S_{\alpha })=V\text{\hspace{0.17em}⁣}\left({\bigcup }_{\alpha \in A}{S}_{\alpha }\right)\mathrm{.}$

By definition, $V(\mathrm{\varnothing })={\mathbf{A}}_k^n$ and $V(1)=\mathrm{\varnothing }$.

Let $I,J\subseteq k[x_1,\dots ,x_n]$ be ideals. If $a\in V(I)\cup V(J)$, then every product $fg$ with $f\in I$ and $g\in J$ vanishes at $a$, so $a\in V(IJ)$. Conversely, if $a\in V(IJ)$ but $a\mathrm{\notin }V(I)$ and $a\mathrm{\notin }V(J)$, choose $f\in I$ and $g\in J$ with $f(a)\mathrm{\ne }0$ and $g(a)\mathrm{\ne }0$. Then $(fg)(a)\mathrm{\ne }0$, contradicting $a\in V(IJ)$. Thus $V(I)\cup V(J)=V(IJ)$.

For arbitrary subsets $S,T$, step 1.3 and [L2] give $V(S)\cup V(T)=V((S))\cup V((T))=V((S)(T)),$ so finite unions of zero loci are zero loci. Repeating this argument proves the same for any finite union.

Steps 1.1, 1.2, and 2.1 are exactly the topology axioms for the closed subsets of ${\mathbf{A}}_k^n$, and step 1.3 gives the displayed formula $V(I)\cup V(J)=V(IJ)$.

If $X\subseteq V(J)$ and $f\in J$, then every point of $X$ lies in $V(J)$, so [L1] says $f$ vanishes at every point of $X$. Therefore $f\in I(X)$ by [L2], and hence $J\subseteq I(X)$.

Conversely, if $J\subseteq I(X)$ and $x\in X$, then every polynomial in $J$ lies in $I(X)$, so [L2] says every polynomial in $J$ vanishes at $x$. By [L1], this means $x\in V(J)$. Hence $X\subseteq V(J)$.

Applying steps 1.1 and 1.2 with $J=I(X)$ yields $X\subseteq V(I(X))$, and applying them with $X=V(J)$ yields $J\subseteq I(V(J))$.

Let $X\subseteq k^n$ be affine algebraic, say $X=V(S)$ for some set $S\subseteq k[x_1,\dots ,x_n]$. Then $X\subseteq V(I(X))$ by [L1]. The reverse inclusion holds because every element of $S$ vanishes on $X$, so $S\subseteq I(X)$ and therefore $V(I(X))\subseteq V(S)=X$ by [L1]. Hence $X=V(I(X))$.

Conversely, assume $X$ is nonempty and irreducible, and let $fg\in I(X)$. Then every point of $X$ annihilates $fg$, so $X\subseteq V(fg)=V(f)\cup V(g)$ by [L3]. Irreducibility gives $X\subseteq V(f)$ or $X\subseteq V(g)$, and then [L1] yields $f\in I(X)$ or $g\in I(X)$. Also $1\mathrm{\notin }I(X)$ because $X\mathrm{\ne }\mathrm{\varnothing }$. Therefore $I(X)$ is prime.

Let $J\subseteq k[x_1,\dots ,x_n]$ be an ideal. Then [L2] gives $I(V(J))=\sqrt{J}$. Therefore $J$ is radical exactly when $I(V(J))=J$. Together with step 1.1, this makes $X\mapsto I(X)$ and $J\mapsto V(J)$ inverse correspondences between algebraic sets and radical ideals.

Let $X$ be a nonempty algebraic set. If $I(X)$ is prime and $X=Y\cup Z$ with algebraic subsets $Y=V(J)$ and $Z=V(K)$, then [L3] gives $X=V(JK)$. Applying step 1.1 and then [L2], $I(X)=I(V(JK))=\sqrt{JK}\mathrm{.}$ Hence $JK\subseteq I(X)$. If neither $J$ nor $K$ were contained in $I(X)$, we could choose $f\in J\setminus I(X)$ and $g\in K\setminus I(X)$, and then $fg\in JK\subseteq I(X)$ would contradict primality of $I(X)$. So $J\subseteq I(X)$ or $K\subseteq I(X)$, which by [L1] means $X\subseteq Y$ or $X\subseteq Z$. Since $Y,Z\subseteq X$, one has $X=Y$ or $X=Z$. Thus $X$ is irreducible.

Steps 2.2 and 1.2 identify the irreducible algebraic sets precisely with the prime ideals under the correspondence from step 2.1.

Assume $X$ is irreducible. If $U,V\subseteq X$ are nonempty open subsets with $U\cap V=\mathrm{\varnothing }$, then $X=(X\setminus U)\cup (X\setminus V)$ is a union of two proper closed subsets, contrary to irreducibility. Therefore every two nonempty open subsets of $X$ meet.

Conversely, assume every two nonempty open subsets of $X$ meet. If $X=A\cup B$ with $A$ and $B$ proper closed subsets, then $X\setminus A$ and $X\setminus B$ are nonempty disjoint open subsets, a contradiction. Hence $X$ is irreducible.

If every nonempty open subset of $X$ is dense and $U,V$ are nonempty open subsets, then $\bar{U}=X$, so $U$ meets the nonempty open set $V$. Thus every pair of nonempty open subsets meets.

Assume every two nonempty open subsets of $X$ meet, and let $U\subseteq X$ be nonempty open. If $\bar{U}\mathrm{\ne }X$, then $X\setminus \bar{U}$ is a nonempty open set disjoint from $U$, contradiction. So $U$ is dense.

Steps 1.1 and 1.2 show that irreducibility is equivalent to pairwise intersection of nonempty opens, and steps 2.1 and 1.3 show that this is also equivalent to density of every nonempty open subset.

By [L2], the set $X$ is a classical affine variety exactly when $I(X)$ is a prime ideal of $k[x_1,\dots ,x_n]$.

By [L1] and [L3], the coordinate ring $k[X]$ is an integral domain exactly when the ideal $I(X)$ is prime.

Steps 1.1 and 1.2 prove that $X$ is a classical affine variety if and only if $k[X]$ is an integral domain.

Let $\bar{f}\in k[X]$, represented by a polynomial $f$. Define $\mathrm{\Phi }(\bar{f}):X\to k$ by $\mathrm{\Phi }(\bar{f})(x)=f(x)$. If $\bar{f}=\bar{g}$ in $k[X]$, then $f-g\in I(X)$ by [L1], so [L2] says $f(x)=g(x)$ for every $x\in X$. Thus $\mathrm{\Phi }$ is well defined.

If $\mathrm{\Phi }(\bar{f})=\mathrm{\Phi }(\bar{g})$, then $f(x)=g(x)$ for every $x\in X$, so $f-g\in I(X)$ by [L2]. Hence $\bar{f}=\bar{g}$ by [L1]. Therefore $\mathrm{\Phi }$ is injective.

Every polynomial function on $X$ is, by definition, the restriction of some polynomial $f\in k[x_1,\dots ,x_n]$, and that function is exactly $\mathrm{\Phi }(\bar{f})$. Hence $\mathrm{\Phi }$ is surjective.

Steps 1.1, 1.2, and 1.3 give the stated bijection, and step 1.2 is precisely the criterion that two polynomials define the same function on $X$ if and only if their difference lies in $I(X)$.

For any $\bar{f}\in k[X]$, the set $D_X(\bar{f})$ is open because it is the complement in $X$ of the closed subset where $\bar{f}$ vanishes. Also a point lies in $D_X(\bar{f})\cap D_X(\bar{g})$ exactly when both $\bar{f}$ and $\bar{g}$ are nonzero there, equivalently when $\bar{f}\bar{g}$ is nonzero there. Thus $D_X(\bar{f})\cap D_X(\bar{g})=D_X(\bar{f}\bar{g})$.

Let $U\subseteq X$ be Zariski-open and $x\in U$. Then $X\setminus U$ is closed in $X$, so there is a set $S$ of polynomials on the ambient affine space with $X\setminus U=X\cap V(S)$ by [L2]. Since $x\mathrm{\notin }V(S)$, choose $f\in S$ with $f(x)\mathrm{\ne }0$. Then $x\in D_X(\bar{f})\subseteq U$. Hence every open set is a union of principal opens.

Steps 1.1 and 1.2 are exactly the basis criterion.

A fraction $a\mathrm{/}{\bar{f}}^n\in k[X{]}_{\bar{f}}$ defines a function on $U$ by $x⟼\frac{a(x)}{\bar{f}(x{)}^n}\mathrm{.}$ If $a\mathrm{/}{\bar{f}}^n=b\mathrm{/}{\bar{f}}^m$, then after passing to the common denominator ${\bar{f}}^{n+m}$ their difference is $c\mathrm{/}{\bar{f}}^{n+m}$ with $c=a{\bar{f}}^m-b{\bar{f}}^n$. By [L2], some power of $\bar{f}$ annihilates $c$. Evaluating at a point of $U$, where $\bar{f}$ is nonzero, forces $c(x)=0$, so the two fractions give the same function. Thus there is a well-defined ring homomorphism $\mathrm{\Phi }:k[X{]}_{\bar{f}}\to {\mathcal{O}}_X(U)$.

If $U=\mathrm{\varnothing }$, then $\bar{f}(x)=0$ for every $x\in X$, so $\bar{f}=0$ in $k[X]$. By [L1], the localization $k[X{]}_{\bar{f}}$ is the zero ring. The only function $\mathrm{\varnothing }\to k$ is both additive and multiplicative identity, so ${\mathcal{O}}_X(U)$ is also the zero ring. Hence the statement is proved in this boundary case.

Let $\phi \in {\mathcal{O}}_X(U)$, given by a finite principal-open cover $U={\bigcup }_{i=1}^r{D}_X({\bar{h}}_i)$ and local formulas $\phi =\frac{{\bar{g}}_i}{{\bar{h}}_i}\text{on }{D}_X({\bar{h}}_i)\mathrm{.}$ Discard the indices with $D_X({\bar{h}}_i)=\mathrm{\varnothing }$; the remaining family still covers the nonempty open set $U$, and relabel it as $D_X({\bar{h}}_1),\dots ,D_X({\bar{h}}_r)$. For any $i,j$, the overlap $D_X({\bar{h}}_i)\cap D_X({\bar{h}}_j)=D_X({\bar{h}}_i{\bar{h}}_j)$ is open by [L4]. It is also nonempty because $D_X({\bar{h}}_j)$ is a nonempty open subset of $X$, while [L5] makes $D_X({\bar{h}}_i)$ dense in $X$. On that overlap the two local expressions for $\phi$ agree, so the class ${\bar{g}}_i{\bar{h}}_j-{\bar{g}}_j{\bar{h}}_i\in k[X]$ vanishes on a dense subset of $X$ and therefore on all of $X$. By [L3], ${\bar{g}}_i{\bar{h}}_j={\bar{g}}_j{\bar{h}}_i\text{in }k[X]\mathrm{.}$

Choose representatives $f,h_1,\dots ,h_r\in k[x_1,\dots ,x_n]$ of $\bar{f},{\bar{h}}_1,\dots ,{\bar{h}}_r$. Because $D_X(\bar{f})={\bigcup }_{i=1}^r{D}_X({\bar{h}}_i)$, a point of $X$ lies outside $U$ exactly when all $h_i$ vanish there. Thus $V(I(X)+(f))=V(I(X)+(h_1,\dots ,h_r))$ inside ${\mathbf{A}}_k^n$. Applying [L6] gives $f\in I(V(I(X)+(f)))=I(V(I(X)+(h_1,\dots ,h_r)))=\sqrt{I(X)+(h_1,\dots ,h_r)},$ so for some $M\ge 1$ there exist $b_1,\dots ,b_r\in k[x_1,\dots ,x_n]$ and $u\in I(X)$ with $f^M=u+b_1{h}_1+\cdots +b_r{h}_r\mathrm{.}$ Passing to $k[X]$ yields ${\bar{f}}^M={\bar{b}}_1{\bar{h}}_1+\cdots +{\bar{b}}_r{\bar{h}}_r\mathrm{.}$

Assume now that $U\mathrm{\ne }\mathrm{\varnothing }$. To prove injectivity, let $a\mathrm{/}{\bar{f}}^n\in k[X{]}_{\bar{f}}$ map to the zero function on $U$. Then $a(x)=0$ for every $x\in U$. On the complement $X\setminus U$ one has $\bar{f}(x)=0$. Therefore $(a\bar{f})(x)=0$ for every $x\in X$, so $a\bar{f}=0$ in $k[X]$. By [L2], this implies $a\mathrm{/}{\bar{f}}^n=0$ in $k[X{]}_{\bar{f}}$. Thus $\mathrm{\Phi }$ is injective.

Define $a:={\bar{b}}_1{\bar{g}}_1+\cdots +{\bar{b}}_r{\bar{g}}_r\in k[X]\mathrm{.}$ Fix $x\in D_X({\bar{h}}_j)$. Using the relations from step 1.3 and then step 1.4, we compute $a(x){\bar{h}}_j(x)={\sum }_{i=1}^r{\bar{b}}_i(x){\bar{g}}_i(x){\bar{h}}_j(x)={\bar{g}}_j(x){\sum }_{i=1}^r{\bar{b}}_i(x){\bar{h}}_i(x)={\bar{g}}_j(x)\bar{f}(x{)}^M\mathrm{.}$ Since ${\bar{h}}_j(x)\mathrm{\ne }0$, division in the field $k$ gives $\frac{a(x)}{\bar{f}(x{)}^M}=\frac{{\bar{g}}_j(x)}{{\bar{h}}_j(x)}=\phi (x)\mathrm{.}$ Because the $D_X({\bar{h}}_j)$ cover $U$, this holds for every $x\in U$. Hence $\phi =\mathrm{\Phi }(a\mathrm{/}{\bar{f}}^M)$, so $\mathrm{\Phi }$ is surjective.

Step 2.1 proves injectivity, step 2.2 proves surjectivity, and step 1.2 handles the empty-open boundary. Therefore $\mathrm{\Phi }:k[X{]}_{\bar{f}}\stackrel{\sim }{\to }{\mathcal{O}}_X(U)$ is a ring isomorphism.

The ideal $I(X)$ is radical by [L1], so [L2] yields minimal prime ideals ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ over $I(X)$ with $I(X)={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_m\mathrm{.}$ Applying [L1] again, $X=V(I(X))=V({\mathfrak{p}}_1)\cup \cdots \cup V({\mathfrak{p}}_m),$ and each $V({\mathfrak{p}}_i)$ is irreducible.

This union is irredundant. Indeed, if $V({\mathfrak{p}}_i)\subseteq V({\mathfrak{p}}_j)$, then [L1] reverses inclusions and gives ${\mathfrak{p}}_j\subseteq {\mathfrak{p}}_i$. Minimality of the ${\mathfrak{p}}_i$ over $I(X)$ forces equality, so distinct terms cannot contain one another.

Let $Y\subseteq X$ be any irreducible closed subset. Then $Y$ is an affine algebraic set, so $Y=V(\mathfrak{q})$ for some prime ideal $\mathfrak{q}$ containing $I(X)$ by [L1]. Since the ${\mathfrak{p}}_i$ are precisely the minimal primes over $I(X)$, some ${\mathfrak{p}}_i$ is contained in $\mathfrak{q}$, hence $Y=V(\mathfrak{q})\subseteq V({\mathfrak{p}}_i)$. Therefore every irreducible closed subset of $X$ lies in one of the $V({\mathfrak{p}}_i)$.

Step 2.2 shows that the maximal irreducible closed subsets of $X$ are exactly the sets $V({\mathfrak{p}}_i)$. Hence the family from step 1.1 is uniquely determined up to order and is precisely the irreducible-component decomposition of $X$.

Let $X\subseteq {\mathbf{A}}_k^n$ be an affine algebraic set. By [L1], the ring $k[X]$ is a quotient of the polynomial ring $k[x_1,\dots ,x_n]$, so [L3] shows that $k[X]$ is of finite type over $k$.

Let $A$ be a reduced affine $k$-algebra. By [L2] and [L3], there exist $n\ge 0$ and an ideal $I\subseteq k[x_1,\dots ,x_n]$ with $A\cong k[x_1,\dots ,x_n]\mathrm{/}I$. Because $A$ is reduced, if $f^m\in I$ then the class of $f$ in $A$ is nilpotent, hence zero; therefore $f\in I$. So $I$ is a radical ideal.

To see that $k[X]$ is reduced, let $\bar{f}\in k[X]$ satisfy ${\bar{f}}^m=0$ for some $m\ge 1$. Choosing a representative $f$ in the polynomial ring, this means $f^m\in I(X)$. Since $I(X)$ is radical by [L4], we have $f\in I(X)$, so $\bar{f}=0$. Thus $k[X]$ is reduced. Together with step 1.1 and [L2], this proves part (1).

By [L4], the radical ideal $I$ is the vanishing ideal of the algebraic set $X:=V(I)\subseteq {\mathbf{A}}_k^n$. Then [L1] gives $k[X]=k[x_1,\dots ,x_n]\mathrm{/}I(X)=k[x_1,\dots ,x_n]\mathrm{/}I$, so $A\cong k[X]$.

Steps 2.1 and 2.2 establish the stated object-level correspondence between affine algebraic sets and reduced affine $k$-algebras.

Fix $x\in X$. Evaluation at $x$ gives a surjective homomorphism ${\mathrm{ev}}_x:k[X]\to k$ because every constant is attained. Its kernel is exactly ${\mathfrak{m}}_x$, and the quotient $k[X]\mathrm{/}{\mathfrak{m}}_x$ is isomorphic to the field $k$. Hence [L3] shows that ${\mathfrak{m}}_x$ is maximal.

Let $\mathfrak{m}$ be a maximal ideal of $k[X]$. Under the quotient map $\pi :k[x_1,\dots ,x_n]\twoheadrightarrow k[X]$, the inverse image $M:={\pi }^{-1}(\mathfrak{m})$ is a maximal ideal of the polynomial ring. By [L2], $M=(x_1-a_1,\dots ,x_n-a_n)$ for a unique point $a\in k^n$. Because $I(X)\subseteq M$, every polynomial vanishing on $X$ vanishes at $a$, so $a\in V(I(X))=X$. Then $\mathfrak{m}={\mathfrak{m}}_a$.