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

8
Noetherian modules
8.1. Modules and submodules
8.1.1. Lemma. Let M be a module. The following conditions are equivalent.
(1) Any increasing sequence Mn of submodules is stationary, Mn = Mn+1 for
n >> 0.
(2) Any nonempty subset of submodules of M contains a maximal element.
(3) Any submodule of M is finite.
Proof. (1) ⇔ (2): See the proof of 7.3.1. (2) ⇒ (3): Let N be a submodule of
M and choose a maximal element N ′ in the set of finite submodules of N. For
y ∈ N, the module N ′ + Ry is finite, so N ′ = N ′ + Ry gives N = N ′ finite.
(3) ⇒ (1): The union ∪Mn is a submodule, generated by x1, . . . , xm ∈ Mk, so
Mn = ∪Mn, n > k.
8.1.2. Definition. A module M which satisfies the conditions of 8.1.1 is a noetherian
module.
8.1.3. Proposition. Let 0 → N → M → L → 0 be an exact sequence of modules
over the ring R. Then M is noetherian if and only if N and L are noetherian.
Proof. See the proof of 7.3.4.
8.1.4. Corollary. Let f : M → N be a homomorphism.
(1) M is noetherian if and only if Ker f, Im f are noetherian.
(2) N is noetherian if and only if Im f, Cok f are noetherian.
Proof. Use the sequences 3.1.8.
8.1.5. Proposition. Let f : M → M be a homomorphism on a noetherian module.
Then the following are equivalent
(1) f is surjective
(2) f is an isomorphism
Proof. Analog of the proof of 7.3.4. There is a number n such that Ker f ◦2n =
Ker f ◦n . For x ∈ Ker f ◦n there is y such that x = f ◦n (y). Then f ◦2n (y) =
f ◦n (x) = 0, so y ∈ Ker f ◦2n = Ker f ◦n . Then x = f ◦n (y) = 0 and f is
injective.
8.1.6. Proposition. A finite direct sum of noetherian modules is noetherian.
8.1.7. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L are noetherian
(2) M/N ∩ L is noetherian
93