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

"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.