TITLE: AC and Krull's theorem equivalence
QUESTION [5 upvotes]: It is well known that the axiom of choice can be used to prove Krull's theorem which states that every ring has a maximal ideal. However, i heard once that Krull's theorem is equivalent to the AC (or to Zorn's lemma). Is that true?
So, we suggest each ring (perhaps, each commutative ring) has a maximal ideal and now we need to build some ring to prove the AC (Zorn's lemma, Zermelo theorem etc). Could anybody explain how to do that?

REPLY [6 votes]: Originally the proof was given by Hodges, from 1979 [1].
Banaschewski gave a new proof of the theorem in 1994 [2].

Bibliography:

Hodges, W., Krull implies Zorn. Journal of the London Mathematical Society 2 (1979), 285-287.
Banaschewski, B., A New Proof that “Krull implies Zorn”. Mathematical Logic Quarterly 40 (1994), 478-480.