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

86 7. MODULES OF FINITE LENGTH
7.2.2. Lemma. Any two finite composition series have the same number of submodules.
Proof. Let l(M) be the least length of a composition series. If M is simple then
l(M) = 1. Let 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M be any composition series. For a
submodule N ⊂ M there is a filtration 0 = N ∩M0 ⊂ N ∩M1 ⊂ · · · ⊂ N ∩Mn =
N with factors N ∩Mi/N ∩Mi−1 ⊂ Mi/Mi−1 being either simple or 0. It follows
that l(N) ≤ l(M). If l(N) = l(M) then each factor is nonzero and therefore
N = M. Applying this to the composition series of M gives l(M) ≤ n ≤ l(M)
as wanted.
7.2.3. Definition. The number of nontrivial submodules in a filtration as above
will be denoted ℓR(M) and is the length of M.
7.2.4. Proposition. Given an exact sequence
0 → N → M → L → 0
Then M has a finite length if and only if both N and L have finite length. In that
case
ℓR(M) = ℓR(N) + ℓR(L)
Proof. A finite filtration with simple quotients in M induces filtrations in both N
and L, and vice versa.
The filtrations in N and L give a filtration in M with the same quotients, so the
length formula follows from 7.2.2.
7.2.5. Corollary. Let f : M → N be a homomorphism.
(1) If M has finite length if and only if Ker f, Im f have finite length. If finite
length
ℓR(M) = ℓR(Ker f) + ℓR(Im f)
(2) If N has finite length if and only if Im f, Cok f have finite length. If finite
length
ℓR(N) = ℓR(Im f) + ℓR(Cok f)
Proof. Use the sequences 3.1.8.
7.2.6. Corollary. Let f : M → M be a homomorphism on a module of finite
length. Then
ℓR(Ker f) = ℓR(Cok f)
and the following are equivalent:
(1) f is injective.
(2) f is surjective.
(3) f is an isomorphism.
Proof. Use the sequences 3.1.8.
7.2.7. Corollary. Let N ⊂ M be a submodule and suppose M has finite length.
(1) ℓR(N) ≤ ℓR(M).
(2) ℓR(N) = ℓR(M) if and only if N = M.
7.2.8. Corollary. Let N, L ⊂ M be a submodules and suppose N, L has finite
length. Then N + L, N ∩ L have finite length and
ℓR(N + L) + ℓR(N ∩ L) = ℓR(N) + ℓR(L)