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?

$\begingroup$ Have you tried searching the site before asking? $\endgroup$ – Asaf Karagila Apr 24 '13 at 8:39

$\begingroup$ mathoverflow.net/questions/27163/… (Also a google search for "axiom of choice"+krull returns that link in the third place, and the paper by Hodges from that answer as the second result). $\endgroup$ – Asaf Karagila Apr 24 '13 at 8:44

$\begingroup$ Yes, i have, but a bit sluggishly. Thank you! $\endgroup$ – Nikita Apr 24 '13 at 10:01

$\begingroup$ Nikita, welcome to MathOverflow. Asaf, since the question to which you link to is not the same as this question, why not just answer this question with a link to Hodge's paper? Indeed, I was say that that answer over there is more an answer to this question here than that one (although of course they are closely related). $\endgroup$ – Joel David Hamkins Apr 24 '13 at 12:20

$\begingroup$ To be honest, after i had asked my question, i found the Hodge's paper at the Journal of London mathematical society, but couldn't subscribe for the Journal (there was something like 'such page doesn't exists'). Also i hope it's not necessary to pay for it. If it's not a problem, can anybody give the link to the Hodge's paper or send it me to the email? $\endgroup$ – Nikita Apr 24 '13 at 12:53
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), 285287.
Banaschewski, B., A New Proof that “Krull implies Zorn”. Mathematical Logic Quarterly 40 (1994), 478480.