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

7.3. ARTINIAN RINGS 87
7.2.9. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L have finite length.
(2) M/N ∩ L has finite length.
If finite length
ℓR(M/N + L) + ℓR(M/N ∩ L) = ℓR(M/N) + ℓR(M/L)
Proof. Use the sequences 3.2.8.
7.2.10. Proposition. A R-module M of finite length is finite and generated by
ℓR(M) or less elements.
Proof. There is a sequence 0 → N → M → L → 0 with L simple. Conclusion
by induction.
7.2.11. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M has finite
length. Then M has finite length as R-module and
ℓ R/I(M) = ℓR(M)
Proof. This follows from 7.1.6 and 7.2.2.
7.2.12. Example. Let K be a field. A module M is of finite length if it is a finite
dimensional vector space. Then
7.2.13. Exercise. (1) Compute
(2) Compute
ℓZ(Z/(p n1
1
ℓK(M) = rankK M
. . . pnk
k )) = n1 + · · · + nk
ℓ K[X](K[X]/((X − a1) n1 . . . (X − ak) nk )) = n1 + · · · + nk
7.3. Artinian Rings
7.3.1. Lemma. Let M be a module. The following conditions are equivalent.
(1) Any decreasing sequence Mi of submodules is stationary, Mn = Mn+1 for
n >> 0.
(2) Any nonempty subset of submodules of M contains a minimal element.
Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain a
minimal element. Then choose a non stationary sequence. (2) ⇒ (1): A descending
sequence containing a minimal element is stationary.
7.3.2. Definition. A module M which satisfies the conditions of 7.3.1 is an artinian
module.
7.3.3. Definition. A ring R is an artinian ring if R is an artinian module over
itself.
7.3.4. Proposition. Let 0 → N → M → L → 0 be an exact sequence of modules
over the ring R. Then M is artinian if and only if N and L are artinian.