Noetherian properties in composite generalized power series rings

Theorem 2.1

Let D ⊆ E be an extension of commutative rings with identity and ( Γ , ≤ ) a positively ordered monoid. If D + 〚 E Γ ⁎ , ≤ 〛 is a Noetherian ring, then the following statements hold.

1. D is a Noetherian ring.

2. If Γ is cancellative, then E is a finitely generated D-module and Γ is finitely generated.

3. ( Γ , ≤ ) is narrow.

Proof

1. Note that 〚 E Γ ⁎ , ≤ 〛 is an ideal of D + 〚 E Γ ⁎ , ≤ 〛 , and D ≅ D + 〚 E Γ ⁎ , ≤ 〛 / 〚 E Γ ⁎ , ≤ 〛 ; so D is a Noetherian ring.

2. Let Γ be cancellative. We first show that E is finitely generated as a D-module. Let α ∈ Γ ⁎ . Consider the ideal X α 〚 E Γ , ≤ 〛 of 〚 E Γ , ≤ 〛 generated by X α :

Proof of Claim

Suppose, by way of contradiction, that there is no such element α n + 1 . Let β ∈ Γ \ 〈 α 1 , … , α n 〉 . If X β ∉ I n , then α n + 1 ≔ β contradicts our assumption and so the claim holds. Thus, assume X β ∈ I n . Then there exist f 1 , … , f n ∈ D + 〚 E Γ ⁎ , ≤ 〛 such that X β = ∑ i = 1 n X α i ⁎ f i . Hence, we have

so β = α i 1 + γ 1 for some i 1 ∈ { 1 , … , n } and γ 1 ∈ supp ( f i 1 ) . Note that γ 1 ∉ 〈 α 1 , … , α n 〉 because β ∉ 〈 α 1 , … , α n 〉 . If X γ 1 ∉ I n , then setting α n + 1 ≔ γ 1 again contradicts our assumption because γ 1 ∉ < α 1 , … , α n > . Thus, assume X γ 1 ∈ I n . We proceed by induction. Assume m ≥ 1 and that γ 0 = β , γ 1 , … , γ m ∈ Γ and i 1 … , i m ∈ { 1 , … , n } have been chosen so that, for each 1 ≤ j ≤ m ,

Note that for β = γ 0 , γ 0 ∉ 〈 α 1 , … , α n 〉 and X γ 0 ∈ I n . Then X γ m ∈ I n implies, as in the argument above, that γ m = α i m + 1 + γ m + 1 , for some i m + 1 ∈ { 1 , 2 , … , n } and some γ m + 1 ∈ supp ( f i m + 1 ) \ 〈 α 1 , … , α n 〉 . By the contradiction hypothesis, X γ m + 1 ∈ I n . It follows that we obtain, for every m ≥ 1 ,

Since Γ is cancellative, γ m = α i m + 1 + γ m + 1 , for every m ≥ 1 . Hence, for every m ≥ 1 ,

and so we have an infinite nondecreasing chain of ideals in D + 〚 E Γ ⁎ , ≤ 〛 :

Since D + 〚 E Γ ⁎ , ≤ 〛 is Noetherian, there exists an integer N ≥ 1 such that, for every k ≥ N ,

Therefore, X γ N + 1 = X γ N ⁎ f , for some f ∈ D + 〚 E Γ ⁎ , ≤ 〛 and thus γ N + 1 = γ N + δ , for some δ ∈ supp ( f ) . Since γ N = α i N + 1 + γ N + 1 and Γ is cancellative, α i N + 1 + δ = 0 . Since 0 ≤ α for all α ∈ Γ , G ( Γ ) = { 0 } ; so α i N + 1 = 0 , which contradicts the fact that α i ∈ Γ ⁎ . This proves the claim.

By the claim, we obtain a strictly infinite chain I 1 ⊊ I 2 ⊊ ⋯ of ideals in D + 〚 E Γ ⁎ , ≤ 〛 , which is a contradiction to the fact that D + 〚 E Γ ⁎ , ≤ 〛 is a Noetherian ring.

1. Suppose to the contrary that there are infinitely many pairwise incomparable elements α 1 , α 2 , … of Γ . Consider the chain of ideals in D + 〚 E Γ ⁎ , ≤ 〛 :

Since D + 〚 E Γ ⁎ , ≤ 〛 is a Noetherian ring, there exists an integer N ≥ 1 such that, for every k ≥ N ,

Therefore, X α N + 1 = ∑ i = 1 N X α i ⁎ f i , for some f 1 , … , f N ∈ D + 〚 E Γ ⁎ , ≤ 〛 . Hence, we obtain

Thus, there exist k ∈ { 1 , … , N } and β ∈ supp ( f k ) such that α N + 1 = α k + β , which implies that α N + 1 ≥ α k , a contradiction.□

Corollary 2.2

(cf. [2, 5.2]) Let D be a commutative ring with identity and ( Γ , ≤ ) a positively ordered monoid. If 〚 D Γ , ≤ 〛 is a Noetherian ring, then the following statements hold.

1. D is a Noetherian ring.

2. If Γ is cancellative, then Γ is finitely generated.

3. ( Γ , ≤ ) is narrow.

Lemma 2.3

Let D ⊆ E be an extension of commutative rings with identity and ( Γ , ≤ ) an ordered monoid. Then an ideal of D + 〚 E Γ ⁎ , ≤ 〛 containing 〚 E Γ ⁎ , ≤ 〛 is of the form I + 〚 E Γ ⁎ , ≤ 〛 for some ideal I of D.

Proof

Let A be an ideal of D + 〚 E Γ ⁎ , ≤ 〛 containing 〚 E Γ ⁎ , ≤ 〛 and set I ≔ { f ( 0 ) | f ∈ A } . Clearly, I is an ideal of D. Choose d ∈ I . Then there exists an element f ∈ A such that d = f ( 0 ) . Since f − d ∈ 〚 E Γ ⁎ , ≤ 〛 , d ∈ A ; so I ⊆ A . Hence, I + 〚 E Γ ⁎ , ≤ 〛 ⊆ A . The reverse containment is obvious, which completes the proof.□

Lemma 2.4

[2, 3.2, 3.3] Let ( Γ , ≤ ) be an ordered monoid. Then the following statements hold:

1. If ( Γ , ≤ ) is a totally ordered monoid, then Γ is torsion-free and cancellative.

2. If ( Γ , ≤ ) is an ordered monoid, and Γ is torsion-free and cancellative, then there exists a compatible strict total order on Γ , which is finer than ≤ .

Lemma 2.5

Let D ⊆ E be an extension of commutative rings with identity, and I be a finitely generated ideal of D. Let ( Γ , ≤ ) be a positively totally ordered monoid. If E is a finitely generated D-module and Γ is finitely generated, then I + 〚 E Γ ⁎ , ≤ 〛 is finitely generated.

Proof

Note that if ( Γ , ≤ ) is a totally ordered monoid, then by Lemma 2.4, Γ is torsion-free and cancellative. Let E = e 1 D + ⋯ + e m D for some e 1 , … , e m ∈ E , and Γ = 〈 α 1 , … , α n 〉 with 0 < α 1 < α 2 < ⋯ < α n . We first claim that 〚 E Γ ⁎ , ≤ 〛 is a finitely generated ideal of D + 〚 E Γ ⁎ , ≤ 〛 . Let f ∈ 〚 E Γ ⁎ , ≤ 〛 . Define

and for i = 1 , … , n − 1 , set

Then supp ( f ) = ⋃ i = 1 n S α i and hence we write f as follows: For 1 ≤ t ≤ n ,

Note that for each α ∈ S α t ,

So for each α ∈ S α t , we have

Note that if α ∈ S α t , then α − α t ∈ Γ . Since ( Γ , ≤ ) is compatible, we have that if α , β ∈ S α t with α ≠ β , α − α t ≠ β − α t . We also note that { α − α t | α ∈ S α t } is artinian because supp ( f ) is artinian, which means that ∑ α ∈ S α t X α − α t ∈ D + 〚 E Γ ⁎ , ≤ 〛 .

Therefore, we have

and hence f = ∑ t = 1 n f t ∈ ( { e i X α t | i = 1 , … , m and t = 1 , … , n } ) . Thus, we obtain

If I = ( b 1 , … , b q ) , then we have

and thus I + 〚 E Γ ⁎ , ≤ 〛 is finitely generated as desired.□

Lemma 2.6

Let ( Γ , ≤ ) be a positively totally ordered monoid. If Γ is finitely generated, then Γ is Archimedean.

Proof

This is an immediate consequence of Lemma 2.4 and the facts that (1) S is a torsion-free, cancellative, finitely generated monoid with G ( S ) = { 0 } if and only if S is isomorphic to a submonoid of ℕ m for some integer m ≥ 1 [6, Theorem 3.11] (or [7, II. 7]); and (2) every submonoid of ℕ m is Archimedean.□

Lemma 2.7

Let D ⊆ E be an extension of commutative rings with identity and ( Γ , ≤ ) a positively totally ordered monoid. If Γ is generated by α 1 , … , α n , then the X α 1 〚 E Γ , ≤ 〛 + ⋯ + X α n 〚 E Γ , ≤ 〛 -adic topology on D + 〚 E Γ ⁎ , ≤ 〛 is Hausdorff.

Proof

Let 0 ≠ g ∈ ⋂ m ≥ 1 X α 1 〚 E Γ , ≤ 〛 + ⋯ + X α n 〚 E Γ , ≤ 〛 m . Then for each m ≥ 1 , g is a finite sum of generalized power series of the form X k m 1 α 1 + ⋯ + k m n α n ⁎ g m , where g m ∈ 〚 E Γ , ≤ 〛 and k m 1 , … , k m n are nonnegative integers with k m 1 + ⋯ + k m n = m . Recall that π ( g ) is the minimal element in supp ( g ) . Therefore, we have