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

94 8. NOETHERIAN MODULES
Proof. Use the sequence 3.2.7.
8.1.8. Example. A module of finite length is noetherian.
8.1.9. Exercise. (1) Show that
N Z is not noetherian.
(2) Show that
N Z is not noetherian.
(3) Show that Q is not a noetherian Z-module.
8.2. Noetherian rings
8.2.1. Definition. A ring R is a noetherian ring if R is a noetherian module.
8.2.2. Proposition. The following conditions are equivalent.
(1) R is noetherian.
(2) Any ideal is finite.
(3) Any increasing sequence of ideals is stationary.
(4) Any nonempty subset of ideals of contains an ideal maximal for inclusion.
Proof. This is follows from 8.1.1.
8.2.3. Proposition. The nilradical in a noetherian ring is nilpotent, i.e. √ 0 n = 0
for some n. Some power of the radical of an ideal is contained in the ideal, i.e.
√ I n ⊂ I for some n
Proof. √ 0 = (b1, . . . , bn) such that b k i = 0. Then ( aibi) nk = 0.
8.2.4. Proposition. Given ideals I, J ⊂ R. Then the following are equivalent.
(1) R/I, R/J are noetherian
(2) R/I ∩ J is noetherian
Proof. This is a special case of 8.1.7.
8.2.5. Lemma. (1) Any factor ring of a noetherian ring is a noetherian ring.
(2) A principal ideal domain is noetherian.
(3) If M is a noetherian R-module, then R/ Ann(M) is a noetherian ring.
Proof. (1), (2) are clear. (3) Analog of the proof of 7.3.16. Let x ∈ M, then
R/ Ann(x) Rx is noetherian. If x1, . . . , xn generate M, then Ann(M) =
Ann(x1) ∩ · · · ∩ Ann(Xn). By 8.2.4 R/ Ann(M) is artinian.
8.2.6. Proposition. Let R be a noetherian ring. Any finite R-module is noetherian.
Proof. Let M be a finite R-module. There is a surjective homomorphism R n →
M → 0, 6.1.2. Conclusion by 8.2.3.
8.2.7. Corollary. Let R be a noetherian ring. Any finite R-module is finite presented.
8.2.8. Lemma. Let R be a noetherian ring and M, N noetherian modules.
(1) M ⊗R N is noetherian.
(2) HomR(M, N) is noetherian.
Proof. Conclusion by 6.1.8, 6.1.9, 8.2.6.