Maximal Ideals in Commutative Rings and the Axiom of Choice

(i). Let $(A_{i})_{i\in I}$ be a family of non-empty sets. Consider the set $X$ of partial choice functions for the family, i.e. the set of functions $f$ with domain $\mathrm{dom}(f)\subset I$ and $f(i)\in A_{i}$ for all $i\in\mathrm{dom}(f)$. We order $X$ by inclusion, i.e. $f\leqslant g$ iff ${\mathrm{dom}}(f)\subset{\mathrm{dom}}(g)$ and $g|_{{\mathrm{dom}}(f)}=f$.

Let $Y\subset X$ be compatible. Then for any $f,g\in Y$ and $i\in{\mathrm{dom}}(f)\cap{\mathrm{dom}}(g)$ we have $f(i)=g(i)$. Consequently $\bigcup_{f\in Y}f$ is itself a partial choice function and is an upper bound on $Y$. Assuming WZL, there is a maximal element $f\in X$. If ${\mathrm{dom}}(f)\subsetneq I$ we can take $g=f\cup\{(i,a)\}$ for some $i\in I\setminus{\mathrm{dom}}(f),\,a\in A_{i}$ and then $f<g$, contradicting maximality. Therefore ${\mathrm{dom}}(f)=I$ and $f$ is a (full) choice function for $(A_{i})_{i\in I}$. Thus AC holds.

(ii) Let $X$ be a partial order where every chain has an upper bound. Let $C$ be the set of chains in $X$, partially ordered by inclusion. We want to apply WZL to $C$, so we check that its condition holds. If $D\subset C$ is compatible then for any $c_{1},\ldots,c_{n}\in D$ we have that $\bigcup_{i=1}^{n}c_{i}\subset c\in C$ is a chain ($c$ is an upper bound on $\{c_{1},\ldots,c_{n}\}$). Therefore $e=\bigcup_{c\in D}c$ is a chain (if $x,y\in e$ then $x\in c,\,y\in d$ for some $c,d\in D$ and thus $x,y\in c\cup d\subset u\in C$ are comparable, i.e. $x\leqslant y$ or $y\leqslant x$; here $u\in C$ is an upper bound for $\{c,d\}$). Therefore $e\in C$ is a chain and is therefore an upper bound on $D$. Assuming WZL, there must be a maximal element $c\in C$ (i.e. a maximal chain in $X$). By assumption $c$ has an upper bound $x$ in $X$. If $x$ is not maximal, say $x<y$ for some $y\in C$, then $c\cup\{x,y\}\supsetneq c$ is a chain, contradicting the maximality of $c$. Thus $X$ has a maximal element as required.

∎

Clearly $1\in S$. Let $f,g\in S$ be big and let $x\in X$ be a variable. Write $f=f_{1}+f_{2},\,g=g_{1}+g_{2}$, where $f_{1}$ (resp. $g_{1}$) consists of the monomials of $f$ (resp. $g$) dominated by $x$, and $f_{2}$ (resp. $g_{2}$) consists of the monomials of $f$ (resp. $g$) not dominated by $x$. Since $f,g$ are big we must have $f_{2},g_{2}\neq 0$ and therefore $f_{2}g_{2}\not\leqslant x$ (since the monomials appearing in $f_{2},g_{2}$ do not contain variables $\leqslant x$). Since $fg=(f_{1}g_{1}+f_{1}g_{2}+f_{2}g_{1})+f_{2}g_{2}$ and $f_{1}g_{1}+f_{1}g_{2}+f_{2}g_{1}\leqslant x$ (by observation (i) above), we have $fg\not\leqslant x$ (by observation (ii) above). This is true for any $x\in X$, so $fg\in S$ is big and $S$ is multiplicative. ∎

Let $M\lhd S^{-1}A$ be a maximal ideal and consider the localization homomorphism $\mathrm{loc}:A\to S^{-1}A$ given by $\mathrm{loc}(a)=\frac{a}{1}$. Then $\mathrm{loc}^{-1}(M)\lhd A$ is maximal among the ideals of $A$ disjoint from $S$ (if $\mathrm{loc}^{-1}\subsetneq I\lhd A$ then $M\subsetneq S^{-1}I\lhd S^{-1}A$, so $1\in S^{-1}I$ and therefore $I\cap S\neq\emptyset$).∎

Apply the previous lemma to $A=R[X]$ and $S$ the set of big polynomials (which is multiplicative by Lemma 10).∎

Let $g=\sum_{i=1}^{n}g_{i}y_{i}\in(Y),\,y_{i}\in Y,\,g_{i}\in R[X]$. Since $Y$ is compatible there exists $x\in X$ such that $y_{i}\leqslant x$ for $1\leq i\leq n$. Therefore $g\leqslant x$ is small for any $g\in(Y)$ and $(Y)$ is a small ideal.∎

(i). Clearly $P\supset(Y)$, so it is enough to show that $P\subset(Y)$. Let $f\in P$ and let $m$ be a monomial appearing in $f$.

Claim. There exists $z\in Y$ which appears in $m$.

Since $f\in P$ is small we have $m\neq 1$, so let $x$ be a variable appearing in $m$. Let $g\in P$ and consider $h=x^{d}f+g\in P$, where $d>\deg g$. Since $h$ is small and the monomials appearing in $g$ and $x^{d}m$ also appear in $h$ because of our assumption on $d$ (they cannot cancel each other out), there exists $y\in X$ such that $h\leqslant y$ and therefore $m,g\leqslant y$ (note that since $x$ occurs in $m$ the condition $x^{d}m\leqslant y$ is equivalent to $m\leqslant y$). This implies in particular that $z\leqslant y$ for some $z\in X$ appearing in $m$. Consequently $qz+g\leqslant y$ is small for any $q\in R[X],\,g\in P$. Therefore the ideal $(P,z)\supset P$ is small and by the maximality of $P$ we have $z\in P\cap X=Y$, establishing the claim.

Now since every monomial appearing in $f$ contains a variable from $Y$, we have $f\in(Y)$ and the proof of (i) is complete.

(ii). Let $x_{1},\ldots x_{n}\in Y$. We assume WLOG that they are distinct. Then $x_{1}+x_{2}+\ldots+x_{n}\in P$ is small and therefore there exists an upper bound $x\in X$ on $\{x_{1},\ldots,x_{n}\}$. Hence $Y$ is compatible.

(iii) Let $Y\subset Y^{\prime}\subset X$ be compatible, $x\in Y^{\prime}$. By Lemma 15, $(Y^{\prime})\supset(Y)=P$ is a small ideal. By the maximality of $P$ we have $(Y^{\prime})=P$ and therefore $Y^{\prime}=(Y^{\prime})\cap X=P\cap X=Y$. ∎

Let $X$ be a poset such that every compatible $Y\subset X$ has an upper bound. By the assumption of Proposition 7 and Lemma 14 there exists a maximal small ideal $P\lhd R[X]$. By Proposition 16(ii-iii) the set $Y=P\cap X\subset X$ is maximal compatible and by assumption $Y$ has an upper bound $x\in X$. We claim that $x$ is a maximal element of $X$. Otherwise $x<y$ for some $y\in Y$ and $Y\cup\{x,y\}\supsetneq Y$ is compatible (since $y$ is an upper bound on the entire set), contradicting the maximality of $Y$. ∎