Existence of prime ideals and Axiom of Choice.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.