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I would like to solve the beginner's standard exercise which claims that a point of $\mathrm{Spec} \ R$ is closed iff it is a maximal ideal.

The reverse implication is easy. The direct one seems much subtler. If $P = V(I)$ then, since there exist a maximal ideal $M$ containing $P$, it would follow that $\{P, M\} \subseteq V(I) = P$, whence $P=M$. In order to show the existence of $M$ I applied Zorn's lemma to the set of ideals containing $P$.

The question is the following: is there any other proof of the above implication that does not (indirectly) use the axiom of choice? My concern is that I may be using a cannon to shoot a fly. Also, it would be the first time that I see closedness of points to require the axiom of choice. If the axiom is indeed needed in general, are there "nice" classes of rings for which we could get away without it?

Eric Wofsey
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Alex M.
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    Why would try to avoid something which is completely natural, basic, and useful? Zorn's lemma is not a cannon. – Rüdiger Mar 19 '17 at 17:47
  • Seeing how "every ideal is contained in maximal ideal" is equivalent to choice, probably not. – Asaf Karagila Mar 19 '17 at 17:55
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    @Rüdiger As a lover of Choice, I think it is very instructive to find choiceless proofs when they exist (and to establish the strength of the principle in question over ZF, when they don't). It may not be a cannon, but in contexts where it's not truly necessary Choice can obscure the underlying combinatorics. So I think this is a fine question. – Noah Schweber Mar 19 '17 at 17:56
  • Alex, to prevent even more people the same answer, maybe you can edit your question to explain why this answer is not good. – Asaf Karagila Mar 19 '17 at 20:38
  • Indeed, the fact that every ideal is contained in a maximal ideal is equivalent to the axiom of choice. See e.g. http://math.stackexchange.com/questions/317028/a-confusion-about-axiom-of-choice-and-existence-of-maximal-ideals – fulges Mar 19 '17 at 23:19
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    If $P$ is not maximal, we can find $x\not\in P$ such that $P + xR\neq R$. To show ${P}$ is not closed, it is sufficient to find a prime ideal in the quotient $R/(P+xR)$. The statement that every ring has a prime ideal is equivalent to the Boolean prime ideal theorem, which is strictly weaker than AC (see http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice). – Julian Rosen Mar 20 '17 at 15:04

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The following example shows that the use of the axiom of choice cannot be avoided. Let $A$ be the ring of entire functions, and let $B$ be the localization of $A$ obtained by inverting $z-a$ for each $a\in\mathbb{C}$ (equivalently, let $B$ be the ring of meromorphic functions on $\mathbb{C}$ with only finitely many poles). The ring $A$ is a Bezout domain, and hence so is $B$. In particular, $0$ is a prime ideal of $B$.

Now suppose $P\subset B$ is a prime ideal with a nonzero element $f\in P$. Since $f$ is not a unit in $B$, it must have infinitely many zeroes. Let $Z$ be the set of zeroes of $f$. Since $Z$ is a closed discrete subset of $\mathbb{C}$, it is in bijection with $\mathbb{N}$ (exercise: you can prove this without choice). Now let $U$ be the set of subsets $S\subseteq Z$ such that there exists $g\in P$ whose vanishing set is $S$. Then $U$ must be an ultrafilter on $Z$ (the fact that $U$ is closed under binary intersections comes from the fact that $B$ is a Bezout domain and intersections of zero sets correspond to gcds of functions; the fact that if $S\cup T=Z$ then either $S\in U$ or $T\in U$ is because you can find a function $g$ vanishing on $S$ and a function $h$ vanishing on $T$ so that $gh$ is divisible by $f$, so by primeness either $g\in P$ or $h\in P$). Moreover, since any function with finitely many zeroes is a unit in $B$, $U$ must be nonprincipal.

Thus if there exists a nonzero prime ideal in $B$, there exists a nonprincipal ultrafilter on $\mathbb{N}$. Now work in a model of ZF where there exist no nonprincipal ultrafilters on $\mathbb{N}$. In this model, $0$ is the only prime of $B$, so $0$ is a closed point of $\operatorname{Spec} B$. However, $0$ is not a maximal ideal of $B$.

Eric Wofsey
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