(1) Every increasing sequence x1≤ x2

1. Noetherian Module

Let (Σ, ≤) be a partially ordered set. An increasing sequence {x_i} 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 x_n∈ Σ \ {x₁, · · · , xn−1} for each n ≥ 1. Since Σ has no maximal element, {x_n} is not stationary.

 Let A be a commutative ring with identity. An A-module M is Noetherian if every increasing sequence of A-submodules {M_n} of M is stationary. Here increasing means Mn⊆ M_n+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 finitely generated. Let {M_n} be an increasing sequence of A-submodules of M. Denote M⁰ = S

nMn. Then M⁰ is again an A-submodule of M. Since M⁰ is finitely generated over A, choose β = {x₁, · · · , x_n} a set of generators of M, Then β is contained in M_N for some N. Hence M is contained in M_n. This shows that M = Mn. Therefore M_k = Mn for all k ≥ n. By definition, {Mn} is

stationary. 

Lemma 1.2. Let M⁰ be a submodule of M and π : M → M/M⁰ 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 = M⁰. Since N2 contains both x, y, x − y ∈ N. Thus x − y ∈ N ∩ M⁰.

 Proposition 1.2. Let 0 → M⁰ → M → M⁰⁰→ 0 be an exact sequence of A-modules. Then M is Noetherian if and only if M⁰, M⁰⁰ are Noetherian.

1

2

Proof. Let f : M⁰→ M and g : M → M⁰⁰. Since the sequence is exact, we identity M⁰ with a submodule of M and M⁰⁰ with the quotient M/M⁰. Assume that M is Noetherian.

Let {Jk} and {K_s} be increasing sequence of A-modules in M⁰ and M⁰⁰. Then g⁻¹(Ks) are A-submodules containing M⁰ and {J_k} are submodules of M⁰ ⊂ M. Since M is Noe- therian, {Jk} and {g⁻¹(Ks)} are stationary. Since {Jk} is an arbitrary increasing sequence of stationary submodules of M⁰, M⁰ is Noetherian. On the other hand, since {g⁻¹(K_s)} is stationary, there is N so that g⁻¹(K_s) = g⁻¹(K_N) for all s ≥ N. We have already known Ks ⊂ K_N 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(K_N) = g⁻¹(K_s) for s ≥ N, z ∈ g⁻¹(K_s). Hence x = g(z) ∈ K_s. We obtain KN ⊂ K_s for s ≥ N/ Therefore Ks = KN for all s ≥ N, i.e. {Ks} is stationary.

Since every increasing sequence of submodules of M⁰⁰ is stationary, M⁰⁰ is also Noetherian.

Conversely, assume that M⁰, M⁰⁰are Noetherian. Let {Mn} be an increasing sequence of submodules of M. Then {f⁻¹(M_n)} and {g(M_n)} are increasing sequences of submodules of M⁰ and M⁰⁰ 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 M_i, 1 ≤ i ≤ n are Noetherian A-modules, so isLn

i=1M_i.

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

M_i →

n

M

i=1

M_i → M_n→ 0.

By induction hypothesis, Ln−1

i=1 Mi is Noetherian and by assumption, Mn is Noetherian.

Using Proposition 1.2 and the exact sequence (1.1), we find Ln

i=1M_i 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, · · · , x_n} be a set of generators of M. Define ψ : Aⁿ→ M, (a1, · · · , an) 7→

n

X

i=1

aixi.

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 {I_k} be an increasing sequence of ideals of A/I. Then {I_k} is an increasing se- quence 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.