Existence of prime ideals and Axiom of Choice. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:14:43Z http://mathoverflow.net/feeds/question/98549 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice Existence of prime ideals and Axiom of Choice. AliReza Olfati 2012-06-01T08:22:43Z 2012-06-01T14:16:19Z <p>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.</p> <p>Theorem: the existence of maximal ideals in a ring with unity is equivalent to Axiom of choice.</p> <p>This means that every attempt to prove the existence of maximal ideals is related to apply the Axiom of Choice.</p> <p>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 $P\cap S=\varnothing$. </p> <p>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:</p> <ul> <li><p>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?</p></li> <li><p>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?</p></li> <li><p>What kind of relation is between the Axiom of countable choice and The existence of prime ideals in a commutative ring with unity?</p></li> </ul> http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice/98558#98558 Answer by godelian for Existence of prime ideals and Axiom of Choice. godelian 2012-06-01T10:06:21Z 2012-06-01T10:06:21Z <p>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.</p> <p>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 &amp; Rubin: "Consequences of the axiom of choice".</p> <p>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.</p> http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice/98573#98573 Answer by Martin Brandenburg for Existence of prime ideals and Axiom of Choice. Martin Brandenburg 2012-06-01T14:16:19Z 2012-06-01T14:16:19Z <p>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.</p> <p>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.</p> <p>On the other hand, the Compactnass Theorem is weaker than the Axiom Choice.</p> <p>PS: Now I've realized that the whole question is a dublicate of <a href="http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma" rel="nofollow">this one</a>. See especially the answer by Chris Phan.</p>