# Krull's theorem and AC

Lately I have been trying to prove axiom of choice based on krull theorem. The theorem states that for every ring with a unit $R$ that is not a field, there is a maximal ideal. I know it is equivalent to AC. What i have so far is: let $S$ be a set such that $\cup S \neq \phi$. Let's build a ring:  $R=P(S)$ ;$\oplus=XOR$, $\odot=\phi$ ; $\otimes=\cap$, $1=\{S\}$ It is a ring, moreover: for every $x\epsilon R$, $(-x)=x$. Also, it is not a field because for every $x\epsilon R$, $x\neq phi$ $x^{-1}$ does not exit. From this i can deduce that exits a maximal ideal.

That is all have. how do i continue??

• Possible duplicate: math.stackexchange.com/questions/317028/… – RghtHndSd Apr 29 '14 at 14:27
• @rghthndsd it does not really help me... i am trying to prove AoC based on Krull's theorem – Nir Agami Apr 29 '14 at 14:48
• In one of the answers there, there is a link to this article which proves the equivalence. From a little searching, it seems as if there is no "easy" proof that Krull implies AC, however I do not know this for certain. – RghtHndSd Apr 29 '14 at 14:56
• @rghthndsd it helps a little but it is still against my intuition. is there a more "intuitive" way to prove it- without using polynomial rings? – Nir Agami Apr 29 '14 at 15:34
• There's no "simpler" proof that I know of. One could try and come up with a proof of equivalence to one of the many other equivalents to the axiom of choice. From the existence of bases to vector spaces, to Zorn's lemma, to showing every model has an elementary extension/submodel of any cardinality. The options are unlimited, unfortunately not everything in mathematics is "simple and intuitive" and sometimes you have to work for a while to develop the right intuition for "intuitive" to be true. This is one of these cases. – Asaf Karagila Apr 29 '14 at 15:46