# Existence of prime ideals and Axiom of Choice.

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?

• 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. – François G. Dorais Jun 1 '12 at 12:31
• 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. – François G. Dorais Jun 1 '12 at 15:03

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.

• 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) $I$. – Andreas Blass Jun 1 '12 at 12:28
• 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 ...? – Ali Reza Jun 1 '12 at 12:41
• @ 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. – godelian Jun 2 '12 at 23:46

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 Compactnass Theorem is weaker than the Axiom Choice.

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

• 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. – François G. Dorais Jun 1 '12 at 14:59
• 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. – Martin Brandenburg Jun 1 '12 at 15:17
• 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? – Laurent Moret-Bailly Jun 2 '12 at 11:53
• You're right ... perhaps it doesn't work at all that way. – Martin Brandenburg Jun 2 '12 at 14:08
• @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”). – ACL Jan 25 '16 at 10:56

In light of Laurent Moret-Bailly's comment under Martin's answer, let us supplement that answer by showing how existence of prime ideals in finitely generated rings may be proven in ZF without a choice principle (not even the ultrafilter principle). In fact we will show that countable rings $R$ have prime ideals, using the weak König lemma.

(Recall the weak König lemma: if $T$ is an infinite subtree of the infinite binary tree $2^\ast$ [the nodes of $2^\ast$ are finite words in letters $0, 1$; the parent of any nonempty word is obtained by omitting its last letter], then $T$ has an infinite branch. This is easily proven in ZF; see e.g. here.)

Fix an enumeration $a_0, a_1, a_2, \ldots$ of elements of $R$. Build a tree $T$ whose nodes $w$ are labeled by ideals $P_w$ of $R$, by induction. $T_k$ will denote the set of nodes of length $k$, so $T_0$ consists of just the empty word $e$, which we label with the zero ideal. Assuming $T_{k-1}$ and the ideals $P_w$ for words $w \in T_{k-1}$ are given, define $T_k$ and the ideals there according to two cases; for this it is convenient to write $k = 2l + m$ where $m \in \{0, 1\}$ and $l$ is used to code a pair of natural numbers $(i, j)$ by the usual trick, viz. $l = \binom{i+j+1}{2} + j$.

Case $m = 0$: for each $w \in T_{k-1}$ such that $a_i a_j$ belongs to $P_w$, put $w0, w1$ in $T_k$, and define $P_{w0} := P_w + \langle a_i \rangle$, and $P_{w1} := P_w + \langle a_j \rangle$. For any other $w \in T_{k-1}$, put $w0$ in $T_k$, and define $P_{w0} = P_w$.

Case $m = 1$: for each $w \in T_{k-1}$ such that $1 \notin P_w$, put $w0$ in $T_k$, and define $P_{w0} := P_w$. If $1 \in P_w$, then put neither $w0$ nor $w1$ in $T_k$.

$T$ is infinite: it is easy to prove by induction that for each $k$ there is $w \in T_k$ such that $1 \notin P_w$. If $b$ is an infinite branch through $T$, then put $P = \bigcup_{w \in b} P_w$. Then $P$ is a proper ideal by the recipe of case $m=1$, and is prime by the recipe of case $m=0$.

• On a countable ring you can use induction to show there is a prime ideal. No need to appeal to anything. – Asaf Karagila Jan 23 '16 at 19:16
• @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. – Todd Trimble Jan 23 '16 at 19:20
• 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. – Asaf Karagila Jan 23 '16 at 19:54