Skip to main content
edited title
Link
Nikita
  • 171
  • 2

AC and Krull's theorem equilvalenceequivalence

Source Link
Nikita
  • 171
  • 2

AC and Krull's theorem equilvalence

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?