NOETHERIAN MODULES AND SHORT EXACT SEQUENCES
COMMUTATIVE ALGEBRA
α
β
Proposition. Let 0 −→ L −→ M −→ N −→ 0 be a short exact sequence of
A-modules. Then,
M is Noetherian ⇐⇒ L and N are Noetherian
Proof. (⇒) Given an ascending chain of submodules {Li }∞
i=1 in L, we get ascending
in
M
.
Since
M
is
Noetherian,
there exists a positive
chain of submodules {α(Li )}∞
i=1
integer n such that α(Ln ) = α(Ln+1 ) = · · · . Applying α−1 to both sides
α−1 (α(Ln )) = α−1 (α(Ln+1 )) = α−1 (α(Ln+2 )) = · · ·
Since α is injective, α−1 (α(Li )) = Li for each i. So we obtain
Ln = Ln+1 = Ln+2 = · · ·
showing that L is Noetherian.
Similarly, given an ascending chain of submodules {Ni }∞
i=1 in N , we get ascending
chain of submodules {β −1 (Li )}∞
in
M
.
Since
M
is
Noetherian, there exists a
i=1
positive integer p such that β −1 (Np ) = β −1 (Np+1 ) = · · · . Applying β to both
sides,
β(β −1 (Np )) = β(β −1 (Np+1 )) = β(β −1 (Np+2 )) = · · ·
Since β is surjective, β(β −1 (Ni )) = Ni for each i. So we obtain
Np = Np+1 = Np+2 = · · ·
showing that N is Noetherian.
(⇐) Suppose {Mi }∞
i=1 is an ascending chain of submodules of M , then identifying
α(L) with L (which can be done, since α is injective), and taking intersections, we
get a chain of the form
L ∩ M1 ⊆ L ∩ M2 ⊆ · · ·
of submodules in L. Similarly, applying β gives a chain
β(M1 ) ⊆ β(M2 ) ⊆ · · ·
of submodules in N . Since L and N are Noetherian, each of these chains terminate.
To prove that M is Noetherian, it suffices to prove the following lemma:
Lemma. For submodules, M1 ⊆ M2 ⊆ M ,
α(L) ∩ M1 = α(L) ∩ M2
and β(M1 ) = β(M2 ) ⇒ M1 = M2
Proof. Suppose m ∈ M1 . Then, β(m) ∈ β(M1 ) = β(M2 ), so that there is n ∈ M2
such that β(m) = β(n). Then, β(m − n) = 0, so that m − n ∈ M1 ∩ ker(β) =
M1 ∩ α(L) = M2 ∩ α(L). It follows that m − n ∈ M2 , so that m ∈ M2 . This shows
that M1 ⊆ M2 . Similarly, we can prove M2 ⊆ M1 . Thus, M1 = M2 , as desired.
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