# Zorn's lemma in abstract algebra?

It is well konwn that Zorn's lemma implies:

Prop.1 Every commutative unital ring has a maximal ideal.

Prop.2 Every proper ideal is contained in a maximal ideal in a unital ring.

Question: Can we prove the above propositions without Zorn's lemma?

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Proposition 2 as stated is false. Take a group with no maximal subgroups, e.g., $\mathbb{Q}$, and give it $0$ multiplication; that's a ring in which no proper ideal is contained in a maximal ideal. Did you mean the ring to be unital as well? –  Arturo Magidin Jul 28 '11 at 4:22
@Arturo: Thank you, I fixed this. –  Ch Zh Jul 28 '11 at 4:32
Your one-stop shop for equivalent forms of the axiom of choice: consequences.emich.edu/CONSEQ.HTM –  Qiaochu Yuan Jul 28 '11 at 4:46
On the other hand: The Boolean Algebra Maximal Ideal Theorem (although it cannot be proved in ZF) is strictly weaker than AC. –  GEdgar Jul 28 '11 at 14:39

## 2 Answers

"Every unital ring (other than the trivial ring) has maximal ideals" is equivalent to the Axiom of Choice (and hence to Zorn's Lemma) in ZF. So one cannot prove it without Zorn's Lemma or some equivalent statement.

"In a unital ring $R$, every proper ideal is contained in a maximal ideal" follows from the first proposition by taking the quotient $R/I$ and lifting a maximal ideal using the lattice isomorphism theorem. Conversely, if in a unital ring every proper ideal is contained in a maximal ideal, then every unital ring has maximal ideals: just find a maximal ideal that contains the zero ideal. So this proposition is equivalent to the first, and hence cannot be proven without Zorn's Lemma or some equivalent statement.

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As Prof Magidin mentions, the existence of maximal ideals implies the axiom of choice. I was curious to see how this is done, and found a proof on pg. 112 of this book by Rubin and Rubin. Hodges gave the first proof in 1979 (!), and he proved something stronger, namely that the existence of maximal ideals in factorial rings is enough to imply choice.

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