Let be a commutative ring with identity and Suppose that every prime ideal of is maximal. Prove that there exist and such that
The problem is trivial if is prime, for then it is maximal and is a field, so that always works for nonzero .
The problem is only hard when has zero divisors. We can suppose is neither a unit nor 0, those cases being trivial.
If are not all distinct then for some we have then works.
If they are all distinct then I don't know. I suspect this is impossible but I can't see why that would be. I'd love the solution!
means " is an ideal of " now because see that if is a totally ordered subset of then i.e. every chain in has a maximal element.
therefore by Zorn's lemma has a maximal element, say the claim is that is prime: so suppose is not prime. then there exist some such that
let since is a maximal element of and we have so and let
since is multiplicatively closed, we have but, since we have and thus contradiction! so is prime and therefore maximal. now,
since we must have thus because is a maximal ideal. hence for some but then we will get a contradiction:
this contradiction means that our initial assumption that is false and the proof is complete.