-1

Given a commutative ring A. Prove that the set of prime ideals in A has minimal element with respect to inclusion.

I try to apply Zorn's lemma to prove above problem, but I can't find the way to construct a lower bound for any chain in prime ideals set. Please give me an idea to construct this. Thanks in advance!

VN_nmd
  • 55
  • 1
    Did you consider an intersection over all the ideals in the chain? Also, there is a similar argument that every commutative ring has a maximal ideal; it also uses the Zorn lemma, but with the union over all the ideals in the chain as an upper bound – G. Chiusole Feb 21 '20 at 08:51
  • @G. Chiusole I try to apply the same way as you have said, but the intersection of prime ideals is not a prime ideal (radical of $A$ as example), so i'm stuck there – VN_nmd Feb 21 '20 at 08:54
  • 2
    The intersection of arbitrary prime ideals is not necessarily prime, that is true. But you aren't taking an arbitrary set of prime ideals. A chain is very special, just argue from first principles. – RghtHndSd Feb 21 '20 at 08:56
  • I added a 3rd dupe link - which includes an excerpt of a nice expostion by Kaplansky. – Bill Dubuque Feb 21 '20 at 15:54

1 Answers1

2

Hint:

Consider the intersection of the prime ideals in a chain (the set of all prime ideals is not a chain, in general) and prove the contrapositive: if $x$ and $y$ do not belong to the intersection, neither does $xy$.

Bernard
  • 179,256