One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem. Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple proof of the following statement.

Theorem: the existence of maximal ideals in a ring with unity is equivalent to Axiom of choice.

This means that every attempt to prove the existence of maximal ideals is related to apply the Axiom of Choice.

Another important theorem in commutative algebra is Cohen's theorem, which tells us that if $R$ is a commutative ring with unity and $I$ is an ideal of $R$ disjoint from a multiplicative closed subset $S\subset R$, then there exists a prime ideal $P$ so that $I \subset P$ and $P\cap S=\varnothing$.

Cohen's theorem implies that In a commutative ring with unity there exists a prime ideal. Notice that this prime ideal need not be a maximal ideal but we need to apply Zorn's Lemma to show the existence of it. Now Here are my Questions:

  • Is it true that For Showing the existence of prime ideal in a commutative ring with unity we need the Axiom of choice or we can show the existence of it without applying this Axiom?

  • If the Answer of above Question is negative, what kind of Axiom weaker than Axiom of choice is needed to show the existence of prime ideals in a commutative ring?

  • What kind of relation is between the Axiom of countable choice and The existence of prime ideals in a commutative ring with unity?

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    $\begingroup$ Many of these questions can be answered by looking at Consequences of the Axiom of Choice by Howard and Rubin. There is also a website where one can query relationships between different consequences of AC -- consequences.emich.edu/conseq.htm -- the form you're interested in is 14AO. $\endgroup$ Jun 1, 2012 at 12:31
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    $\begingroup$ As Martin pointed out, this is nearly a duplicate of mathoverflow.net/questions/27163. In particular, Chris Phan's answer there directly addresses the question asked here. $\endgroup$ Jun 1, 2012 at 15:03

3 Answers 3


The existence of prime ideals in commutative rings with unity is equivalent in $ZF$ to the Boolean prime ideal ($BPI$) theorem, which is strictly weaker than the axiom of choice. The first reference for this is D. Scott: "Prime ideal theorems for rings, lattices and Boolean algebras", Bulletin of the American Mathematical Society (60) pp. 390.

As for the relation between BPI and the axiom of countable choice, neither of them implies the other, since there are models of $ZF$ where one holds while the other fails. You can find these in the usual reference, Howard & Rubin: "Consequences of the axiom of choice".

The theorem you mention which implies the existence of prime ideals but seems a bit stronger, is actually equivalent to $BPI$ as well. That it implies $BPI$ is trivial, and the other implication is theorem 4.1 of Rav, Y.: "Variants of Rado's selection lemma and their applications" Mathematische Nachrichten (79) 1, pp. 145.

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    $\begingroup$ The apparently stronger form, where $P$ includes $I$ and is disjoint from $S$, follows from the weaker form (mere existence of prime ideals) by applying the weaker form to the ring you get by localizing $R$ at $S$ (i.e., inverting all elements of $S$) and dividing by (the image of) $I$. $\endgroup$ Jun 1, 2012 at 12:28
  • $\begingroup$ Thank you very much for your references and sketch of proofs. But when I read your claims another Question came in my mind to. Is it true that the structure of maximal ideals are more algebraic than prime ideals? I mean that When I consider the prime ideals in commutative rings I could translate it to the language of boolean algebra and lattice theory, But maximal ideals are not wide enough to conduct it to another structure theory such as boolean algebra ...? $\endgroup$
    – Ali Reza
    Jun 1, 2012 at 12:41
  • $\begingroup$ @ Andreas: Nice! I didn't notice this direct deduction. @AliReza: Sorry, I wouldn't know how to answer your question since I'm not sure what exactly you have in mind or if I understand what exactly you mean. It all seems very subjective and relative. $\endgroup$
    – godelian
    Jun 2, 2012 at 23:46
  • $\begingroup$ Note that the D. Scott reference is an abstract. $\endgroup$
    – Kapil
    Mar 2 at 2:32

Although godelian has already answered, let me give a more direct answer (with more proofs instead of references; perhaps they coincide). First, notice that the existence of prime ideals in $R$ disjoint from multiplicative subsets $S$ and containing a given ideal $I$ is actually (quantified over $R$) equivalent to the existence of prime ideals in $R$ (quantified over $R$, of course $\neq 0$). The non-trivial direction just uses $S^{-1} (R/I)$. So we actually have only one statement, the existence of prime ideals.

I claim that the Compactness Theorem (for propositional logic) implies the existence of prime ideals: For each $a \in R$ let $p_a$ be a new variable. Consider the theory whose axioms are $p_0$, $\neg p_1$, $p_a \wedge p_b \longrightarrow p_{a+b}$, $p_a \longrightarrow p_{ab}$, $p_{ab} \longrightarrow p_{a} \vee p_{b}$ for all $a,b \in A$. A model of this theory is precisely a prime ideal of $R$. Since finitely generated rings are noetherian (Hilbert) and noetherian rings have maximal and therefore prime ideals, the theory is finitely consistent. Hence, it is consistent.

On the other hand, the Compactness Theorem is weaker than the Axiom Choice.

PS: Now I've realized that the whole question is a duplicate of this one. See especially the answer by Chris Phan.

  • $\begingroup$ You can get the result as stated without localization and quotient by adding the axioms $p_a$ for $a \in I$ and $\lnot p_b$ for $b \in S$. It looks like this does not affect the argument for finite consistency. $\endgroup$ Jun 1, 2012 at 14:59
  • $\begingroup$ Yes, but that's not really clever. The equivalence simplifies the problem right ahead. We shouldn't care about "Cohen's Theorem" (as the OP called it) at all. $\endgroup$ Jun 1, 2012 at 15:17
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    $\begingroup$ One should be careful that the Dependent Choice Axiom is needed to prove that ``every increasing sequence terminates'' implies the existence of maximal elements. So, can the proof of Hilbert's theorem be arranged to take care of this? $\endgroup$ Jun 2, 2012 at 11:53
  • $\begingroup$ You're right ... perhaps it doesn't work at all that way. $\endgroup$ Jun 2, 2012 at 14:08
  • $\begingroup$ @LaurentMoret-Bailly: The PhD Thesis of Hervé Perdry, hlombardi.free.fr/liens/TheseHervePerdry.pdf, took care of the question of constructive noetherianity by rewriting the definition of noetherian as “every increasing sequence pauses” (meaning “is not strictly increasing”). $\endgroup$
    – ACL
    Jan 25, 2016 at 10:56

In light of Laurent Moret-Bailly's comment under Martin's answer, let's add the simple argument that finitely generated rings have prime and even maximal ideals, without appealing to anything like a choice principle.

If $R$ is finitely generated (a quotient of some $\mathbb{Z}[x_1, \ldots, x_n]$), then it is countable. Enumerate its elements: $a_1, a_2, \ldots$. Form ideals $P_n$ by recursion: define $P_0 = \{0\}$, and given $P_{n-1}$, define $P_n = P_{n-1} + \langle a_n\rangle$ if this doesn't contain $1$, otherwise define $P_n = P_{n-1}$. The union of the $P_n$ is then a maximal ideal: it is proper because $1$ doesn't belong to any $P_n$. And if $a \notin P$ for some $a = a_n$, then $a_n \notin P_n$, meaning that $1 \in P_{n-1} + \langle a_n\rangle$ according to how $P_n$ was defined, and then $1 \in P + \langle a\rangle$. This proves maximality.

  • $\begingroup$ On a countable ring you can use induction to show there is a prime ideal. No need to appeal to anything. $\endgroup$
    – Asaf Karagila
    Jan 23, 2016 at 19:16
  • $\begingroup$ @AsafKaragila Isn't the argument I wrote down just an induction, really? The proof of yours of WKL that I cited is an induction argument. $\endgroup$
    – Todd Trimble
    Jan 23, 2016 at 19:20
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    $\begingroup$ Yes, of course, but I meant that you can enumerate $R$ and just construct by hand a prime ideal, adding one element after another. There's no need to appeal to WKL, just to the theorem about recursive definitions. $\endgroup$
    – Asaf Karagila
    Jan 23, 2016 at 19:54
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    $\begingroup$ @Kapil Coming back to this after 8 years, I would tell my past self to discard this absurdly contorted argument (and I may do just that: delete the post, or completely rewrite it). Asaf was right. Just enumerate the elements $a_1, a_2, \ldots$, and build a maximal ideal $P$ by taking the union of ideals $P_n$ where $P_0 = \{0\}$, defining $P_n = P_{n-1} + \langle a_n\rangle$ if this is proper, else defining $P_n = P_{n-1}$. $\endgroup$
    – Todd Trimble
    Mar 2 at 14:23
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    $\begingroup$ FWIW the original WKL-based argument isn't silly at all if you're paying attention to computability-theoretic complexity. Indeed, it's basically the key behind the observation that while the existence of principal ideals is equivalent to $\mathsf{ACA}_0$, the existence of prime ideals is strictly weaker; see Downey/Lempp/Mileti, as well as the followup by them + Hirschfeldt/Kach/Montalban. Pardon the nostalgia trip: the DLM paper was the first paper I read! $\endgroup$ Mar 2 at 20:37

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