1. Noetherian Module
Let (Σ, ≤) be a partially ordered set. An increasing sequence {xi } of elements of Σ is
stationary if there is N > 0 so that xi = xN for all i ≥ N.
Lemma 1.1. T.F.A.E.
(1) Every increasing sequence x1 ≤ x2 ≤ · · · in Σ is stationary.
(2) Every nonempty subset of Σ has a maximal element.
Proof. (1) implies (2):
Suppose that (2) is false. Since Σ is nonempty, we can choose x1 ∈ Σ. Since Σ has no
maximal element, Σ \ {x1 } is nonempty; otherwise x1 would be the maximal element of
Σ. Then choose x2 ∈ Σ \ {x1 } so that x1 ≤ x2 . Inductively, we can chose an increasing
sequence {xn } of elements of Σ so that xn ∈ Σ \ {x1 , · · · , xn−1 } for each n ≥ 1. Since Σ has
no maximal element, {xn } is not stationary.
Let A be a commutative ring with identity. An A-module M is Noetherian if every
increasing sequence of A-submodules {Mn } of M is stationary. Here increasing means
Mn ⊆ Mn+1 for all n ≥ 1. A ring A is a Noetherian ring if A is a Noetherian A-module.
Proposition 1.1. M is a Noetherian A-module if and only if every submodule of A is
finitely generated.
Proof. Assume that M is Noetherian. Let N be an A-submodule of M. We want to show
that N is finitely generated.
Let Σ be the set of all finitely generated A-submodule of N. Then 0 ∈ Σ and hence Σ is
nonempty. On Σ, define N1 ≤ N2 by N1 ⊂ N2 . Then (Σ, ≤) is a partially ordered set. Since
M is Noetherian, every increasing sequence in Σ is stationary (submodules of N are also
submodules of M ). By Lemma 1.2, every nonempty subset of Σ has a maximal element. Let
us take the subset of Σ to be Σ itself. Let N0 be a maximal element of Σ. Claim N0 = N.
Suppose not. Choose x ∈ N \ N0 . Then Ax + N0 is again a finitely generated A-submodule
of N ; hence Ax + N0 ∈ Σ. Since x 6∈ N, N0 ( Ax + N0 in Σ, which leads to a contradiction
to the assumption that N0 is a maximal element of Σ. This proves that N = N0 is a finitely
generated A-submodule of M.
Conversely, assume that every submodule of A is finitelySgenerated. Let {Mn } be an
increasing sequence of A-submodules of M. Denote M 0 = n Mn . Then M 0 is again an
A-submodule of M. Since M 0 is finitely generated over A, choose β = {x1 , · · · , xn } a set
of generators of M, Then β is contained in MN for some N. Hence M is contained in Mn .
This shows that M = Mn . Therefore Mk = Mn for all k ≥ n. By definition, {Mn } is
stationary.
Lemma 1.2. Let M 0 be a submodule of M and π : M → M/M 0 be the quotient map. If
N1 ⊂ N2 and N1 ∩ M = N2 ∩ M, and π(N1 ) = π(N2 ), then N1 = N2 .
Proof. Let x ∈ N2 . Then π(x) ∈ π(N2 ) = π(N1 ). There is y ∈ N1 so that π(x) = π(y).
Then x − y ∈ ker π = M 0 . Since N2 contains both x, y, x − y ∈ N. Thus x − y ∈ N ∩ M 0 .
Proposition 1.2. Let 0 → M 0 → M → M 00 → 0 be an exact sequence of A-modules. Then
M is Noetherian if and only if M 0 , M 00 are Noetherian.
1
2
Proof. Let f : M 0 → M and g : M → M 00 . Since the sequence is exact, we identity M 0 with
a submodule of M and M 00 with the quotient M/M 0 . Assume that M is Noetherian.
Let {Jk } and {Ks } be increasing sequence of A-modules in M 0 and M 00 . Then g −1 (Ks )
are A-submodules containing M 0 and {Jk } are submodules of M 0 ⊂ M. Since M is Noetherian, {Jk } and {g −1 (Ks )} are stationary. Since {Jk } is an arbitrary increasing sequence
of stationary submodules of M 0 , M 0 is Noetherian. On the other hand, since {g −1 (Ks )} is
stationary, there is N so that g −1 (Ks ) = g −1 (KN ) for all s ≥ N. We have already known
Ks ⊂ KN for all s ≥ N. Suppose x ∈ KN . By subjectivity of g, there is z ∈ g −1 (KN ) so
that g(z) = x. Since g −1 (KN ) = g −1 (Ks ) for s ≥ N, z ∈ g −1 (Ks ). Hence x = g(z) ∈ Ks .
We obtain KN ⊂ Ks for s ≥ N/ Therefore Ks = KN for all s ≥ N, i.e. {Ks } is stationary.
Since every increasing sequence of submodules of M 00 is stationary, M 00 is also Noetherian.
Conversely, assume that M 0 , M 00 are Noetherian. Let {Mn } be an increasing sequence of
submodules of M. Then {f −1 (Mn )} and {g(Mn )} are increasing sequences of submodules
of M 0 and M 00 respectively. Then there exists N so that for any n ≥ N,
Mn ∩ M = f −1 (Mn ) = f −1 (MN ) = MN ∩ M
and g(Mn ) = g(MN ). By Lemma 1.2, Mn = MN for all n ≥ N.
Proposition 1.3. If Mi , 1 ≤ i ≤ n are Noetherian A-modules, so is
Ln
i=1 Mi .
Proof. This can be proved by induction. The statement is true for n = 1. Suppose the
statement is true for n − 1. Then consider the exact sequence of A-modules
(1.1)
0→
n−1
M
i=1
Mi →
n
M
Mi → Mn → 0.
i=1
Ln−1
Mn is Noetherian.
By induction hypothesis,
i=1 Mi is Noetherian and by assumption,
L
Using Proposition 1.2 and the exact sequence (1.1), we find ni=1 Mi is Noetherian.
Corollary 1.1. Let A be a Noetherian ring. If M is a finitely generated A-modules, then
M is a Noetherian A-module.
Proof. Let {x1 , · · · , xn } be a set of generators of M. Define
n
X
n
ψ : A → M, (a1 , · · · , an ) 7→
ai xi .
i=1
Then ψ is an A-module homomorphism. Since {x1 , · · · , xn } generates M, ψ is surjective.
This gives us an exact sequence of A-modules
(1.2)
0 → K → An → M → 0,
where K = ker ψ. Since A is a Noetherian ring, A is a Noetherian A-module. By Proposition
1.2, An is a Noetherian A-module. By Proposition 1.2 and the exact sequence (1.2), M is
a Noetherian A-module.
Proposition 1.4. Let A be a Noetherian ring and I an ideal of A. Then the quotient ring
A/I is a Noetherian ring.
Proof. By Proposition 1.2, A/I is a Noetherian A-module.
Let {Ik } be an increasing sequence of ideals of A/I. Then {Ik } is an increasing sequence of A-modules. Since A/I is a Noetherian A-module, every increasing sequence of
A-submodules of A/I is stationary. This shows that {Ik } is stationary. Hence A/I is a
Noetherian ring.