multiplicative sets in rings and prime ideals
Let be a chain in (i.e. for any either or ). Consider . Obviously is an ideal. Furthermore, if , then there is such that . Thus , so . Lastely each , which completes this part of proof.
By Zorn’s Lemma there is a maximal element . We will show that this ideal is prime. Let be such that . Assume that neither nor . Then and and these inclusions are proper. Therefore both and do not belong to (because is maximal). This implies that there exist and . Thus
where and . Note that . We calculate
Of course , because and by our assumption that . This shows, that . But and . Contradiction.
Corollary. Let be a commutative ring, an ideal in and a multiplicative subset such that . Then there exists prime ideal in such that and .
Proof. Let be the projection. Then is a multiplicative subset in such that (because ). Thus, by proposition, there exists a prime ideal in such that . Of course the preimage of a prime ideal is again a prime ideal. Furthermore . Finaly , because . This completes the proof.
|Title||multiplicative sets in rings and prime ideals|
|Date of creation||2013-03-22 19:03:58|
|Last modified on||2013-03-22 19:03:58|
|Last modified by||joking (16130)|