Some strange behaviors of the power series ring R [ [ X ] ]

Let R be a commutative ring with identity. Let R[X] and R[[X]] be the polynomial ring and the power series ring respectively over R. Being the completion of R[X] (under the X-adic topology), R[[X]] does not always share the same property with R[X]. In this paper, we present some known strange behaviors of R[[X]] compared to those of R[X].


Introduction
In this paper, a ring means a commutative ring with identity.Let R be a ring.Let R[X] and R[[X]] be the polynomial ring and the power series ring respectively over R. Being the completion of R[X] (under the X-adic topology), R[[X]] does not always share the same property with R[X].In this paper, we present some strange behaviors of R[[X]] compared to those of R [X].Aspects of R[[X]] that will be considered include Krull dimension, transcendence degree, and Noetherian property.
Let R be a ring.If there exists a chain P 0 ⊂ P 1 ⊂ • • • ⊂ P n of n + 1 prime ideals of R, but no such chain of n + 2 prime ideals, then we say that the Krull dimension of R is n (or R is n-dimensional) and write Krull-dim R = n.Otherwise, we say that the Krull dimension of R is infinite (or R is infinite-dimensional) and write Krull-dim R = ∞.For a cardinal number α, we say that dim R = α if R has a chain of prime ideals with length α but no longer chains.(The length of a chain P of prime ideals of R is defined by |P| − 1, where |P| denotes the cardinality of P. For two chains P and Q of prime ideals, we say that P is longer than Q if |P| > |Q|).We also say that dim R ≥ α if there is a chain of prime ideals of R with length ≥ α and that dim R ≤ α if every chain of prime ideals of R has length ≤ α.Hence, dim R = α if and only if dim R ≥ α and dim R ≤ α.We note that if Krull-dimR = n < ∞, then dim R = Krull-dim R. Furthermore, dim R ≥ ℵ 0 implies Krull-dimR = ∞ but not vice versa.
The Krull dimension of the polynomial ring R[X] is fairly well-known for a finitedimensional ring R. For example, dim R[X 1 , . . ., X n ] = dim R + n if R is a Noetherian ring or a Prüfer domain [16,17].In general, dim R + 1 ≤ dim R[X] ≤ 2 dim R + 1 [16].For the power series He defines an ideal I of a ring R to be an SFT ideal if there exist a finitely generated ideal J ⊆ I and k ∈ N such that a k ∈ J for each a ∈ I and he calls a ring R an SFT ring if every ideal of R is an SFT ideal.Typical examples of non-SFT domains are finite-dimensional nondiscrete valuation domains and non-Noetherian almost Dedekind domains.Kang and Park [9] showed that dim 01002 (2018) https://doi.org/10.1051/itmconf/20182001002ICM 2018 valuation domain V. Kang and Toan [12] proved that dim V[[X]] ≥ 2 ℵ 1 for a 1-dimensional nondiscrete valuation domain V. Loper and Lucas [14,15] also achieved the result that dim D[[X]] ≥ 2 ℵ 1 for a non-Noetherian almost Dedekind domain or a 1-dimensional nondiscrete valuation domain D. Recently, Toan and Kang [11] generalize the above results by In Section 2, we are going to introduce their technique in proving this result.
Let D be an integral domain with quotient field K.It is easy to see that the polynomial rings D[X] and K[X] have the same quotient field.However, as mentioned in [3], it is rare that D[[X]] and K[[X]] have the same quotient field.Except for the trivial case when D is a field, the only example showing that D[[X]] and K[[X]] have the same quotient field is given by Gilmer in [7].For two integral domains D 1 ⊆ D 2 , denote by tr.d.(D 2 /D 1 ) the transcendence degree of the quotient field of D 2 over that of D 1 .Hence, for a cardinal number α, tr.d.(D 2 /D 1 ) ≥ α if there exists a subset of the quotient field of D 2 with cardinality at least α that is algebraically independent over the quotient field of D 1 .Sheldon showed in [18] In [2], Arnold and Boyd made a great improvement of this result by showing that if We present here the technique in [8], where the authors showed that whenever K Note that in general the bound ℵ 1 is the greatest lower bound that one can obtain since under the continuum hypothesis the cardinality of the quotient field of Let α be an infinite cardinal number (e.g., α = ℵ 0 , ℵ 1 , . ..).An ideal I of a ring R is called an α-generated ideal if I can be generated by a set with cardinality ≤ α.R is called an αgenerated ring if every ideal of R is an α-generated ideal.By definition, an ℵ 0 -generated ring is a ring whose ideals are countably generated.Trivial examples of ℵ 0 -generated rings are those that have only countably many elements (so that each ideal has itself as a countable generating set).Every Noetherian ring is obviously an ℵ 0 -generated ring.However, the converse does not hold.Polynomial rings R[X 1 , X 2 , . . ., X n , . ..] in countably infinite indeterminates over countable rings R, the ring O of algebraic integers, the ring Int(Z) of integer-valued polynomials on Z, and 1-dimensional nondiscrete valuation domains are good examples of ℵ 0 -generated rings that are not Noetherian rings.
Even though the class of ℵ 0 -generated rings is strictly larger than the class of Noetherian rings, it is shown in [10] that when restricted to power series rings, they are actually the same.In other words, the concepts "ℵ 0 -generated ring" and "Noetherian ring" are the same for the power series ring R [[X]].This shows a strange behavior of the power series ring R[[X]] compared to that of the polynomial ring R[X].Indeed, for any infinite cardinal number α, R is an α-generated ring if and only if R[X] is an α-generated ring, which is an analogue of Hilbert Basis Theorem stating that R is a Noetherian ring if and only if R[X] is a Noetherian ring.

Krull dimension of R[X]
The aim of this section is to show that R is a non-SFT domain, which contrasts with the fact that the Krull dimension of the polynomial ring R[X] is always finite provided that R is.The following result is from [16,Theorems 2 and 9].

Construction of an η 1 -set A
We first construct an η 1 -set A. This η 1 -set will be the index set of an infinite chain of prime ideals in D[[X]] when D is a non-SFT domain.
• Let N = {1, 2, 3, . ..} be the set of positive integers and let U be the set of all subsets U of N such that U = {n, n + 1, . ..} for some n ∈ N.
• For two sequences of positive integers s = {s n }, t = {t n }, define s t (or t s) if for each positive integer k, there is a set U ∈ U (depending on k) such that s n > kt n for each n ∈ U, i.e., s n > kt n for all large n.
• Let S be the collection of all A such that A has the following properties.
-A is a nonempty collection of strictly increasing sequences s = {s n } of positive integers.
-If s ∈ A, then s b, where b = {b n } is the sequence defined by b n := n for all n.
-If s, t ∈ A and s t, then s t or t s.
If u is the sequence defined by u n := b 2 n for each n, then it is easy to see that u b.It follows that the set S is nonempty.Order S by set-theoretic inclusion.By Zorn's Lemma, there exists a maximal element in S. Let A be a maximal element in S.This choice of A will be fixed through the rest of this section.For s, t ∈ A, we define s t if and only if s = t or s t.Then (A, ) becomes a totally order set.

Chains of prime ideals in D[[X]]
Suppose that D is a non-SFT domain.Then there is an ideal I of D which is not SFT.As in [1, p. 300], there exists a sequence a 0 , a 1 , . . ., a n , . . . of for each m ≥ 1. Fix this sequence a 0 , a 1 , . . ., a n , . ... For an element s = {s n } ∈ A, let ] generated by all power series f s such that s v i for each i = 1, 2, . . ., n, ITM Web of Conferences 20, 01002 (2018) https://doi.org/10.1051/itmconf/20182001002ICM 2018 Then S is a subset of the quotient field of D[[X]].For elements v 1 , . . ., v n (not necessarily distinct) in A, let B = {v 1 , . . ., v n } be a multiset.We also let B := n.Define Then the set S can be rewritten as where the union is taken over all multisets B consisting of elements in A.

Let D[[X]][S ] be the ring generated by S over D[[X]]. Denote by S D[[X]][S ] the ideal of D[[X]]
[S ] generated by S .Then this ideal is not the unit ideal ([11, Lemma 3.4]).
Theorem 2.5 ([6, Theorem 19.6])Let P be a prime ideal of an integral domain R.There exists a valuation overring W of R with maximal ideal Q such that Q ∩ R = P.
. Since S R is not the unit ideal of R, one can choose a prime ideal P of R such that P contains S R. By Theorem 2.5, there exists a valuation overring W of R with maximal ideal Q such that Q ∩ R = P.For each α ∈ A, let I α W be the ideal of W generated by I α and let Since W is a valuation domain, √ I α W is a prime ideal of W. Hence, P α is a prime ideal of D [[X]].For α, β ∈ A, it can be shown that α β if and only if P α ⊃ P β .Hence, the set A totally ordered set A is Dedekind-complete provided that every nonempty subset of A that has an upper bound has a supremum.Finally we can now present the main result of [11].
3 Transcendence degree in power series rings Proof.This is obvious.

Construction of an upper fathomless set B
In this subsection, we construct an uncountable set B. In the next subsection, we will construct a set {a f } f ∈B of power series in K [[x]] that is algebraically independent over the quotient field of D [[x]] when K [[x]] and D [[x]] have different quotient fields.
Let N = {1, 2, . ..} be the set of positive integers.Set S be the collection of all functions f : N → N satisfying the following three conditions (see [18]).
• f (i) > i for all i ∈ N.
• For each k ≥ 1, there exists an I ≥ 1 (depending on k) such that The function f (i) = 2 2 i satisfies these three conditions.Hence, the set S is nonempty.For a function f in S, denote im( Definition 3.2 For two functions f and g in S, we define g f (we also write f g) if there exists an integer I ≥ 1 such that the following hold.
• im(g| ≥I ) ⊆ im( f ) and g(i) > f (i) for all i ≥ I.
• If s 1 < s 2 are both in im(g| ≥I ), then they are not adjacent in im( f ), i.e., there is no i such that s 1 = f (i) and s 2 = f (i + 1).
Remark 3.3 For f, g, h ∈ S, if f g and g h, then f h.
Let C(S) denote the collection of all nonempty subsets B of S satisfying: for any two different functions f and g in B, either f g or g f .We order C(S) by set-theoretic inclusion.By Zorn's Lemma, there exists a maximal element in C(S).Let B be a maximal element in C(S).This choice of B will be fixed through the rest of the section.For f, g ∈ B, we define f g if and only if f = g or f g.Then (B, ) becomes a totally ordered set.

Transcendence degree in power series rings
Suppose that D is an integral domain with quotient field K such that the quotient field of

K[[X]] properly contains the quotient field of D[[X]].
Then by [7,Theorem 1], there exists a sequence {a i } ∞ i=1 of nonzero elements of D such that ∩ ∞ i=1 (a i ) = (0).Replacing {(a i )} ∞ i=1 by a (suitable) strictly descending subsequence if necessary, we may assume that the sequence ITM Web of Conferences 20, 01002 (2018) https://doi.org/10.1051/itmconf/20182001002ICM 2018 Therefore, if i 1 ≥ i 2 then (a i 1 ) ⊆ (a i 2 ) and hence a i 1 is a multiple of a i 2 .Let B be the upper fathomless set constructed in Subsection 3.2.For each f ∈ B, define a power series We will give sketch of proof of the following main result of the section.For a power series p in K[[X]], the support of p, denoted by supp(p), is the set of all nonnegative integer s for which the coefficient of X s in p is nonzero.

Theorem 3.6 ([8, Theorem 12]) Suppose that D is an integral domain with quotient field K such that D[[X]] and K[[X]] have different quotient fields. Then the quotient field of K[[X]] has uncountable transcendence degree over the quotient field of D[[X]], i.e, t.r.(K[[
Sketch of proof.We show that the set {a f } f ∈B is algebraically independent over the quotient field of D [[X]].Suppose on the contrary that there are some Let P(t 1 , . . ., t J ) be a nonzero polynomial with coefficients in the quotient field of Multiplying both sides of this equation by a suitable element in D[[X]], we may assume that the coefficients of P(t 1 , . . ., t J ) are in D [[X]].Let M be the set of all power series of the form a e(1) f 1 . . .a e(J) f J corresponding to those monomials t e(1) 1 . . .t e(J) J which occur in P(t 1 , . . ., t J ) with nonzero coefficient.Then the relation P(a f 1 , . . ., a f J ) = 0 is actually a linear combination of the elements of M with coefficients in D [[X]].Order the elements of M lexicographically, using the exponents of a f 1 , then a f 2 , and so forth.Write M = {m 0 , m 1 , . . ., m R }, where m 0 is the largest term in this ordering.Then P(a f 1 , . . ., a f J ) = 0 has the form ]] (r = 0, 1, . . ., R), d 0 0. Denote by k the index of the first nonzero coefficient d 0,k of d 0 .The proof is finished by showing a contradiction that 0 d 0,k ∈ ∩ ∞ i=1 (a i ), which is technically complicated and is omitted.

ℵ 0 -generated power series rings
The purpose of this section is to show that the power series ring R [[X]] is an ℵ 0 -generated ring if and only if R[[X]] is a Noetherian ring (and hence) if and only if R is a Noetherian ring.In order to prove this result, we only need to show that if R is a non-Noetherian ring, then the power series ring R[[X]] is not an ℵ 0 -generated ring since if R is a Noetherian ring, then R[[X]] is also a Noetherian ring (for example, see [13,Theorem 71]) and hence an ℵ 0 -generated ring.Suppose that R is a non-Noetherian ring.Our task is to construct an ideal J of R[[X]] that cannot be generated by any countable subset of J. Using the η 1 -set (A, ) in Section 2, we construct generators for the J of R[[X]] as follows.
Let R be a non-Noetherian ring.Then there exists a sequence a 0 , a 1 , . . ., a m , . . . of elements in R such that a m (a 0 , a 1 , . . ., a m−1 ) https://doi.org/10.1051/itmconf/20182001002ICM 2018 for each m ≥ 1.For each integer m ≥ 0, we let I m := (a 0 , a 1 , . . ., a m ).Then a m I m−1 for each m ≥ 1.For each sequence s = {s n } ∈ A, we define We let J be the ideal of R [[X]] generated by all f s with s ∈ A. The following is the main result of [10].( (3) R is a Noetherian ring.
Proof.We only need to prove that (1) implies (3).Suppose that R is not a non-Noetherian ring.We show that R[[X]] is not an ℵ 0 -generated ring.It suffices to show that the ideal J constructed above is not a countably generated ideal.Suppose on the contrary that J is countably generated.Then there exists a countable subset B of A such that J is generated by where h(s) ∈ R[[X]] and s ∈ B. Since v B, by taking a finite intersection of members of U, we can find a set U ∈ U such that v m < s m for each m ∈ U and for each s appearing in the finite sum (1).Choose any number m ∈ U. Since v m < s m , the coefficient of f s at X j belongs to I m−1 for all j ≤ v m .It follows that the coefficient of h(s) f s at X v m belongs to I m−1 .This holds for every s appearing in the finite sum (1).Therefore, the coefficient of s h(s) f s at X v m belongs to I m−1 .This is a contradiction since the coefficient of Corollary 4.2 If R is an ℵ 0 -generated ring, then the power series ring R[[X]] is not necessarily an ℵ 0 -generated ring.
Proof.Let R be any ℵ 0 -generated ring which is not a Noetherian ring (for example, let R = Q[X 1 , X 2 , . . ., X n , . ..], the polynomial ring in countably infinite indeterminates over the field of rational numbers Q).Then R[[X]] is not an ℵ 0 -generated ring by Theorem 4.1.
The above corollary contrasts with the following famous result for polynomial rings.In comparing with Theorem 4.1, on has the following result, whose its proof follows the standard one of Hilbert Basis Theorem.• The concepts "Noetherian ring" and "ℵ 0 -generated ring" are the same for the power series ring R[[X]] while they are different for the polynomial ring R[X].In fact, the polynomial ring R[X] is an α-generated ring if and only if R is an α-generated ring, which is an analogue of Hilbert Basis Theorem stating that the polynomial ring R[X] is a Noetherian ring if and only if R is a Noetherian ring.
Techniques involved in the proofs of the above results are quite complicated.By exploring these, we hope that similar techniques can be further developed to have a better understand of the power series ring R[[X]].

Definition 2 . 3
Let (Y, ) be a totally ordered set and B, C be subsets of Y.We say that B C if b c for each b ∈ B and each c ∈ C. A totally ordered set (Y, ) is an η 1 -set if for any two countable subsets B, C of Y such that B C, there exists an element a ∈ Y such that B a C, i.e., b a c for each b ∈ B and each c ∈ C.

3. 1
Transcendence degree in polynomial rings Theorem 3.1 Let D be an integral domain with quotient field K. Then D[X] and K[X] have the same quotient field.Hence, tr.d.(K[X]/D[X]) = 1.

Definition 3 . 4 ATheorem 3 . 5
totally ordered set (Y, ) is called an upper fathomless set if for every nonempty countable subset C of Y, there exists an element y ∈ Y such that y c for all c ∈ C. By definition, every upper fathomless set is an uncountable set.The following is [8, Theorem 7].The set (B, ) is an upper fathomless set and hence is uncountable.

Theorem 4 . 3 (
Hilbert Basis Theorem.)If R is a Noetherian ring, then so is the polynomial ring R[X].

Theorem 4 . 4 ( 5 Conclusion
[10, Theorem 22]) For any infinite cardinal number α, a ring R is an αgenerated ring if and only if the polynomial ring R[X] is an α-generated ring.ITM Web of Conferences 20,01002 (2018)   https://doi.org/10.1051/itmconf/20182001002ICM 2018Being the completion of the polynomial ring R[X], the power series ring R[[X]] however has very strange behaviors.The paper presents the following three examples.•If dim R is finite, then so is dim R[X].However, dim R[[X]] can be infinite, in fact ≥ 2 ℵ 1 , even if dim R is finite.• If D isan integral domain with quotient field K, then D[X] and K[X] share the same quotient field.However, the quotient field of K[[X]] is much larger than that of D[[X]].In fact, the quotient field of K[[X]] is often has uncountable transcendence degree over that of D[[X]].