Commutative algebra - Department of Mathematical Sciences - old ...

66 5. LOCALIZATION
(2) For a prime ideal Q ⊂ U −1 R the contracted ideal Q ∩ R is a prime ideal
and the extended (Q ∩ R)U −1 R = Q
Proof. Conclusion by 4.3.6, 4.3.7, 4.3.8.
5.1.6. Corollary. Let R be a ring and U a multiplicative subset.
(1) An ideal maximal among the ideals disjoint from U is a prime ideal.
(2) Any ideal disjoint from U is contained in a prime ideal disjoint form U.
Proof. The prime ideals disjoint from U are the prime ideals in U −1 R.
5.1.7. Proposition. The nilradical of a ring R is the intersection of all prime ideals
P .
√
0 = P
Proof. By 1.3.8 the nilradical is contained in any prime ideal. Suppose u ∈ R is
not nilpotent. Then {u n } −1 R is nonzero. Then contraction of a maximal ideal,
5.1.1, is a prime ideal in R not containing u.
5.1.8. Corollary. Let R be a ring.
(1) The radical of an ideal I is the intersection of all prime idealsP containing I
√
I = P
P
I⊂P
(2) For ideals I, J ⊂ R, √ I ∩ J = √ I ∩ √ J.
(3) If U is a multiplicative subset, then U −1√ 0 = √ 0 in U −1 R.
If R is reduced, then U −1 R is reduced.
Proof. (1) Use 5.1.7 on the factor ring R/I. (2) This follows from (1). (3) Use the
correspondence 5.1.5.
5.1.9. Definition. A prime ideal minimal for inclusion among prime ideals is a
minimal prime ideal.
5.1.10. Proposition. Any prime ideal of Q ⊂ R contains a minimal prime ideal
P ⊂ Q.
Proof. The set of prime ideals in R is ordered by inclusion. Given a decreasing
chain Pα then ∩Pα is a prime ideal. Conclusion by Zorn’s lemma.
5.1.11. Corollary. Let R ⊂ S be a subring and P ⊂ R a minimal prime ideal.
Then there is a minimal prime ideal Q ⊂ S contracting to P = Q ∩ R.
5.1.12. Exercise. (1) Let K ⊂ R be an infinite subfield and I, P1, . . . , Pn any ideals.
Show that if I ⊂ P1 ∪ · · · ∪ Pn then I ⊂ Pi for some i.
(2) Let P, P1, P2 be proper ideals. Show that if P is a maximal ideal and P n ⊂ P1 ∪ P2
then P = P1 of P = P2.