Noetherian rings of composite generalized power series

Definition 2.1

[3] A monoid S is said to be

1. cancellative if s + t = s + u implies t = u for all s , t , u ∈ S ,

2. torsion-free if for any s , t ∈ S and a positive integer n , n s = n t implies s = t ,

3. finitely generated if there exists a finite subset { s 1 , … , s n } of S such that S = ⟨ s 1 , … , s n ⟩ , where ⟨ s 1 , … , s n ⟩ is the set of all elements ∑ i = 1 n k i s i with nonnegative integers k i .

Definition 2.2

([9]) A partially ordered set ( S , ≤ ) is said to be

1. artinian if every strictly descending sequence of elements of S is finite,

2. narrow if every subset of pairwise order-incomparable elements of S is finite,

3. an ordered monoid if it is an additive monoid and the order ≤ is compatible with the operation, i.e., for any s , t , u ∈ S , s ≤ t implies s + u ≤ t + u .

Definition 2.3

[9] An ordered monoid ( S , ≤ ) is said to be

1. strictly ordered if s < s ′ implies s + t < s ′ + t for all s , s ′ , t ∈ S ,

2. positively ordered if 0 ≤ s for all s ∈ S ,

3. positive strictly ordered if it is both positively ordered and strictly ordered.

Lemma 2.4

[11, Lemma 2.2] Let ( S , ≤ ) be a partially ordered set. The following conditions are equivalent:

1. ⇓ ( S , ≤ ) is artinian.

2. For every infinite sequence ( s n ) of S, there exist i < j such that s i ≤ s j .

3. ( S , ≤ ) is artinian and narrow.

Proof

(2) ⇔ (3) It follows from [13, Theorem 2.1].

(1) ⇒ (2) Let ( s n ) n ≥ 1 be an infinite sequence of S . Put S i ≔ { s j ∣ j ≥ i } for each i ≥ 1 . For each i , put I i ≔ { x ∈ S ∣ x ≤ y for y ∈ S i } . Note that I i is a lower set of S . Since S 1 ⊇ S 2 ⊇ … , we have I 1 ⊇ I 2 ⊇ … . Note that s i ∈ I i for each i . Since ⇓ ( S , ≤ ) is artinian, there exist i < j such that I i = I j . Thus, s i ∈ I j , which means that s i ≤ s k for some k ≥ j > i .

(2) ⇒ (1) Suppose that ( I n ) n ≥ 1 is an infinite strictly descending chain of lower sets of S . Consider an infinite sequence ( s n ) n ≥ 1 such that s i ∈ I i \ I i + 1 . By assumption, there exist i < j such that s i ≤ s j . Since I j is a lower set of S and s j ∈ I j , we have s i ∈ I j , which is a contradiction.□

Lemma 2.5

Let S be a strict monoid and let s 1 , … , s n be nonzero elements of S. If ∑ i = 1 n k i s i = 0 for some nonnegative integers k i , then k i = 0 for all i .

Proof

Let s 1 , … , s n be nonzero elements of S such that ∑ i = 1 n k i s i = 0 for each k i ≥ 0 . Assume that k i ≠ 0 for some 1 ≤ i ≤ n . Without loss of generality, we may assume that k 1 ≠ 0 . Then, 0 = 0 + s 1 + ( ( k 1 − 1 ) s 1 + ∑ i = 2 n k i s i ) . Since S is strict, we have s 1 = 0 , which is a contradiction.□

Lemma 2.6

Let S be a nonzero monoid. Then, the following statements hold:

1. If S is strict, then G ( S ) = { 0 } .

2. If S is cancellative with G ( S ) = { 0 } , then S is strict.

3. [11, Lemma 3.1] If ( S , ≤ ) is a positive strictly ordered monoid, then S is strict.

4. [11, Lemma 3.2] S is strict if and only if ( S , ≼ ) is a strictly ordered monoid.

5. [11, Lemma 3.3] Let S be a strict monoid. Then, S is finitely generated if and only if ⇓ ( S , ≼ ) is artinian.

Remark 2.7

1. For two strictly ordered monoids ( S , ≤ ) and ( S , ≼ ) , the fact that ( S , ≼ ) is narrow may not imply that ( S , ≤ ) is narrow. For an example, let S = N be an additive monoid. Then, N is a finitely generated strict monoid and ( N , ≼ ) is narrow. But if ≤ is the trivial order on N , then ( N , ≤ ) is not narrow.

2. Let ≤ ′ and ≤ be partial orders on a monoid S such that ≤ ′ is finer than ≤ (i.e., s ≤ t implies s   ≤ ′   t for all s , t ∈ S ). It follows from Lemma 2.4 that if ( S , ≤ ) is artinian and narrow, then ( S , ≤ ′ ) is artinian and narrow. The converse does not hold; ≼ is finer than the trivial order ≤ on N , ( N , ≼ ) is artinian and narrow, but ( N , ≤ ) is not narrow.

3. Let ( S , ≤ ) be a positive strictly ordered monoid. Then, S is strict by Lemma 2.6. So ( S , ≼ ) is a strictly ordered monoid and ≤ is finer than ≼ .

Lemma 2.8

Let ( S , ≤ ) be a positive strictly ordered monoid. If S is finitely generated, then ( S , ≤ ) is artinain and narrow.

Example 2.9

Let S = ( N × { 0 , 1 } ) \ { ( 0 , 0 ) } = { ( n , a ) ∣ n = 0 , 1 , … , a = 0 , 1 } \ { ( 0 , 0 ) } . Define an addition on S as follows: for x = ( n , a ) , y = ( m , b ) ∈ S ,

Then, ( S , + ) is a monoid with identity ( 0 , 1 ) . Since ( 2 , 0 ) + ( 1 , 0 ) = ( 3 , 0 ) = ( 2 , 0 ) + ( 1 , 1 ) , S is not cancellative. We now define an order ≤ on S as follows: for x = ( n , a ) , y = ( m , b ) ∈ S ,

Then ( S , ≤ ) is a positive strictly ordered monoid.

Theorem 3.1

Let A ⊆ B be an extension of commutative rings with identity and ( S , ≤ ) a strictly ordered monoid. Suppose that S is a strict monoid. If A + 〚 B S * , ≤ 〛 is a Noetherian ring, then the following statements hold:

1. A is a Noetherian ring and B is a finitely generated A-module.

2. S is finitely generated.

3. If ( S , ≤ ) is positively ordered, then ( S , ≤ ) is narrow.

Proof

Let R ≔ A + 〚 B S * , ≤ 〛 . We recall that the element f of R with a s = f ( s ) for each s ∈ S is written as f = ∑ s ∈ S a s X s .

(1) It is clear that 〚 B S * , ≤ 〛 is an ideal of R . So R ⁄ 〚 B S * , ≤ 〛 ≅ A is a Noetherian ring. Let s ∈ S * . Consider the ideal X s 〚 B S , ≤ 〛 of 〚 B S , ≤ 〛 generated by X s :

Note that G ( S ) = { 0 } since S is strict. Since s ∈ S * , X s g ∈ R for every g ∈ 〚 B S , ≤ 〛 . Thus X s 〚 B S , ≤ 〛 is an ideal of R . Since R is Noetherian, there exist g 1 , … , g n ∈ X s 〚 B S , ≤ 〛 such that X s 〚 B S , ≤ 〛 = ( g 1 , … , g n ) R . Since each g i = X s f i for some f i ∈ 〚 B S , ≤ 〛 , the ideal X s 〚 B S , ≤ 〛 can be written as follows:

For any b ∈ B , there are h i ∈ R such that

Since S is strict, G ( S ) = { 0 } by Lemma 2.6 (i), and s + t = s (for s , t ∈ S ) implies t = 0 . Recall that for f , g ∈ 〚 B S , ≤ 〛 and s ∈ S , X s ( f , g ) is the set { ( t , u ) ∈ S × S ∣ f ( t ) ≠ 0 , g ( u ) ≠ 0 , and t + u = s } . Hence, we have the following:

Therefore, for the element s ∈ S , we have the following:

Consequently, b ∈ ( f 1 ( 0 ) , … , f n ( 0 ) ) A , and B is a finitely generated A -module.

(2) Let ( s n ) be an infinite sequence in S . Without loss of generality, we may assume that s n ≠ 0 for all n ≥ 1 . Let I n be the ideal of R generated by X s 1 , … , X s n :

Since R is Noetherian, there exists N ≥ 1 such that X s j ∈ I N for all j > N . Hence, for j > N , X s j = X s 1 f 1 + … + X s N f N , where each f i ∈ R . Therefore,

So s j = s i + t i for some t i ∈ supp ( f i ) , where i < j . Hence, s i ≼ s j for i < j . Note that since S is strict, ( S , ≼ ) is a strictly ordered monoid. It follows from Lemmas 2.4 and 2.6 that S is finitely generated.

(3) It follows from Lemma 2.8.□

Corollary 3.2

[12, Theorem 2.1] Let A ⊆ B be an extension of commutative rings with identity and ( S , ≤ ) a positive strictly ordered monoid. If A + 〚 B S * , ≤ 〛 is a Noetherian ring, then A is Noetherian, B is a finitely generated A-module, S is finitely generated, and ( S , ≤ ) is narrow.

Corollary 3.3

Let A ⊆ B be an extension of commutative rings with identity. Let ( S , ≤ ) be strictly ordered, where S is cancellative with G ( S ) = { 0 } . If A + 〚 B S * , ≤ 〛 is a Noetherian ring, then A is Noetherian, B is a finitely generated A-module, and S is finitely generated.

Lemma 3.4

Let S be a strict monoid that is finitely generated. Then, there is a surjective strict homomorphism from ( N n , ≼ ) to ( S , ≼ ) for some n.

Proof

Let S = ⟨ s 1 , … , s n ⟩ be a finitely generated strict monoid. Note that since N n and S are strict, it follows from Lemma 2.6 that ( N n , ≼ ) and ( S , ≼ ) are strictly ordered monoids. We define a map σ from ( N n , ≼ ) to ( S , ≼ ) as follows:

Then, σ is a surjective monoid homomorphism by [14, Theorem 1.2]. It follows from Lemma 2.5 that if 0 ≠ x ∈ N n , then 0 ≠ σ ( x ) ∈ S . If a , b ∈ N n with a ≺ b , then b = a + c for some 0 ≠ c ∈ N n . Since σ is a monoid homomorphism, σ ( b ) = σ ( a ) + σ ( c ) and 0 ≠ σ ( c ) ∈ S . Hence, σ ( a ) ≺ σ ( b ) in S . Therefore, σ is strict.□

Lemma 3.5

Let R be a commutative ring with identity, and let σ : ( S , ≤ ) → ( S ′ , ≤ ′ ) be a strict homomorphism of strictly ordered monoids. If ( S , ≤ ) is narrow and σ is surjective, then the induced ring homomorphism σ ¯ : 〚 R S , ≤ 〛 → 〚 R S ′ , ≤ ′ 〛 is surjective.

Proof

It follows from [8, p. 78] that the induced map σ ¯ : 〚 R S , ≤ 〛 → 〚 R S ′ , ≤ ′ 〛 is a ring homomorphism. Let f ′ = ∑ s ′ ∈ S ′ a s ′ X s ′ ∈ 〚 R S ′ , ≤ ′ 〛 . Then, supp ( f ′ ) is artinian and narrow in S ′ . Since σ is surjective, there is a map τ from S ′ to S such that σ ∘ τ is the identity map on S ′ . Since σ is strict and ( S , ≤ ) is narrow, τ ( supp ( f ′ ) ) is artinian and narrow in S . Put f ≔ ∑ s ∈ τ ( supp ( f ′ ) ) a σ ( s ) X s . Then, f ∈ 〚 R S , ≤ 〛 and σ ¯ ( f ) = f ′ . Hence, σ ¯ is surjective.□

Remark 3.6

(1) If ( S , ≤ ) and ( S , ≤ ′ ) are strictly ordered monoids, and ≤ ′ is finer than ≤ , then 〚 R S , ≤ 〛 is a subring of 〚 R S , ≤ ′ 〛 . In particular, if ( S , ≤ ) is narrow, then 〚 R S , ≤ 〛 = 〚 R S , ≤ ′ 〛 (see also [8, 1.18]).

(2) In Lemma 3.5, the condition that ( S , ≤ ) is narrow is necessary. We give an example such that the induced map σ ¯ is not surjective when the map σ is surjective. Let K be a field. Let ≤ and ≤ ′ be the trivial order and usual order on N , respectively. The identity map σ : N → N is a surjective strict homomorphism of strictly ordered monoids from ( N , ≤ ) to ( N , ≤ ′ ) . Note that 〚 K N , ≤ 〛 = K [ X ] and 〚 K N , ≤ ′ 〛 = K 〚 X 〛 . Then, there does not exist surjective ring homomorphism from K [ X ] to K 〚 X 〛 . Hence, the induced map σ ¯ is not surjective.

Theorem 3.7

If B is a finitely generated A-module over a Noetherian ring A and ( S , ≤ ) is a positive strictly ordered monoid that is finitely generated, then A + 〚 B S * , ≤ 〛 is a Noetherian ring.

Proof

Since ( S , ≤ ) is a positive strictly ordered monoid and S is finitely generated, it follows from Lemmas 2.6 and 3.4 that there is a surjective strict monoid homomorphism τ : ( N n , ≼ ) → ( S , ≼ ) for some n . Since ( S , ≤ ) is positive strictly ordered, we have s ≼ t ⇒ s ≤ t for all s , t ∈ S . Thus, the identity map i : ( S , ≼ ) → ( S , ≤ ) is a surjective strict monoid homomorphism. Hence, the composite map σ = i ∘ τ : ( N n , ≼ ) → ( S , ≤ ) is a surjective strict monoid homomorphism. Note that ( N n , ≼ ) is narrow by Lemmas 2.4 and 2.6. It follows from Lemma 3.5 that σ induces the surjective ring homomorphism σ ¯ : 〚 B N n , ≼ 〛 → 〚 B S , ≤ 〛 , which is defined as follows: for each f = ∑ u ∈ N n a u X u ∈ 〚 B N n , ≼ 〛 ,

Since σ is a surjective strict monoid homomorphism and ( N n , ≼ ) is a positive strictly ordered monoid, we have that σ ( 0 ) = 0 and σ ( N n * ) = S * . For each f ¯ = ∑ s ∈ S a s X s ∈ 〚 B S * , ≤ 〛 , supp ( f ) is artinian and narrow. Since ( N n , ≼ ) is artinian and narrow, σ − 1 ( supp ( f ) ) is an artinian and narrow subset of N n . For each s ∈ supp ( f ) , choose u s ∈ N n such that σ ( u s ) = s . Put X ≔ { u s ∈ N n ∣ s ∈ supp ( f ) } ⊆ N n * . Note that X is artinian and narrow. Consider f = ∑ u s ∈ X a s X u s . Then, f ∈ 〚 B N n , ≼ 〛 and σ ¯ ( f ) = f ¯ . Hence, we have that σ ¯ ( a ) = a for a ∈ B and σ ¯ ( 〚 B N n * , ≼ 〛 ) = 〚 B S * , ≤ 〛 .

We also note that there is a ring isomorphism ρ : 〚 B N n , ≼ 〛 → B 〚 X 1 , … , X n 〛 , which is defined as follows: for each f = ∑ u = ( k 1 , … , k n ) ∈ N n a u X u ∈ 〚 B N n , ≼ 〛 ,

We note that ρ ( a ) = a for a ∈ B and ρ ( 〚 B N n * , ≼ 〛 ) = ( X 1 , … , X n ) B 〚 X 1 , … , X n 〛 . Now, we have the following (commutative) diagram:

Here, the ring homomorphism ϕ : B 〚 X 1 , … , X n 〛 → 〚 B S , ≤ 〛 is surjective, which is defined as follows: for each f ′ ∈ B 〚 X 1 , … , X n 〛 ,

Note that A + 〚 B N n * , ≼ 〛 is a subring of 〚 B N n , ≼ 〛 . Hence, we have the following:

It follows from [2, Proposition 2.1] that A + ( X 1 , … , X n ) B 〚 X 1 , … , X n 〛 is a Noetherian ring. Since ρ is an isomorphism, A + 〚 B N n * , ≼ 〛 is a Noetherian ring. Hence, A + 〚 B S * , ≤ 〛 is a Noetherian ring.□