Weakly Integrally Closed Domains and Forbidden Patterns
by
Mary E. Hopkins
A Dissertation Submitted to the Faculty of The Charles E. Schmidt College of Science in Partial Ful…llment of the Requirements for the Degree of Doctor of Philosophy
Florida Atlantic University Boca Raton, Florida May 2009
Acknowledgements
I am deeply grateful to my advisor, Dr. Fred Richman, for his patience and gentle guidance, and for helping me to understand algebra on a deeper level. I would also like to thank my family and Dr. Lee Klingler for supporting me personally through this challenging process. Many thanks are given to Dr. Timothy Ford and Dr. Jorge Viola-Prioli for their insightful remarks which proved to be very helpful in writing my dissertation. Lastly, I will be forever grateful to Dr. James Brewer for taking me under his wing, opening my eyes to the beautiful world of algebra, and treating me like a daughter. I dedicate this dissertation to my grandmother and Dr. Brewer.
iii Abstract
Author: Mary E. Hopkins
Title: Weakly Integrally Closed Domains and Forbidden Patterns
Institution: Florida Atlantic University
Dissertation advisor: Dr. Fred Richman
Degree: Doctor of Philosophy
Year: 2009
An integral domain D is weakly integrally closed if whenever there is an element x in the quotient …eld of D and a nonzero …nitely generated ideal J of D such that xJ J 2, then x is in D. We de…ne weakly integrally closed numerical monoids similarly. If a monoid algebra is weakly integrally closed, then so is the monoid. A pattern F of …nitely many 0’s and 1’s is forbidden if whenever the characteristic binary string of a numerical monoid M contains F , then M is not weakly integrally closed. Any stretch of the pattern 11011 is forbidden. A numerical monoid M is weakly integrally closed if and only if it has a forbidden pattern. For every …nite set S of forbidden patterns, there exists a monoid that is not weakly integrally closed and that contains no stretch of a pattern in S. It is shown that particular monoid algebras are weakly integrally closed.
iv Contents
1 Introduction ...... 1
2 Background and Preliminary Results ...... 4
3 Forbidden Patterns ...... 9 3.1 Terminology and Preliminary Results ...... 9 3.2 Examples of Forbidden Patterns ...... 12 3.3 A Characterization of Forbidden Patterns ...... 15
4. A Test to Determine if a Monoid is Weakly Integrally Closed ...... 20 4.1 An Algorithm to Determine if a Monoid has a Bad Zero ...... 20 4.2 A Characterization of Weakly Integrally Closed Monoids...... 23
5 Question 1 Generalized ...... 25
6 Weakly Integrally Closed Domains ...... 30
7 Open Problems...... 37
References ...... 39
v 1 Introduction
Noether introduced the concept of integrality in the 1920s. Let D D0 be integral n n 1 domains. An element x D0 is said to be integral over D if x + dn 1x + + 2 d1x + d0 = 0 for some natural number n and di D. Kaplansky cites an equivalent 2 de…nition in [6]: x D0 is integral over D if there is a …nitely generated D-submodule 2 J of D0 such that xJ J. An element x D0 is almost integral over D if there is 2 n a nonzero d D such that dx D for every natural number n. Any element x D0 2 2 2 that is integral over D is also almost integral over D. We say that the domain D is
(completely) integrally closed in D0 if whenever an element x D0 is (almost) 2 integral over D, then x is in D. In [3], Brown gives an example of a 3-by-3 matrix over the ring D = K[[t3; t4]], where K is a …eld and t an indeterminate, whose minimum polynomial, X2 t5X, has coe¢ cients which do not lie in D. Brewer and Richman, intrigued by Brown’s example, ask when must the coe¢ cients of the minimum polynomial of a n-by-n matrix over a domain lie in that domain. In [2], this question led to the notion of a weakly integrally closed domain. An element x D0 is said to be strongly integral 2 over D if there is a nonzero …nitely generated ideal J of D such that xJ J 2. If we use Kaplansky’s de…nition of an integral element, then it is easy to see that any element x D0 that is strongly integral over D is also integral over D. We say that D 2 is weakly integrally closed in D0 if every element x D0 that is strongly integral 2 over D is in D. If D is weakly integrally closed in it’squotient …eld, then we simply say that D is weakly integrally closed. Let M be a numerical monoid (i.e. an additive submonoid of the natural numbers)
1 m1 m2 and let K[M] = k0 + k1t + k2t + ::: : ki K and mi M , where K is a …eld. f 2 2 g We represent M by its in…nite characteristic binary string. For example, the string 10111 ::: represents the monoid generated by 2 and 3. Brewer and Richman show the following important results:
Theorem 1 ([2], Theorems 1 and 3) If the binary string of a monoid M contains a stretch of the pattern 11011, then K[M] is not weakly integrally closed.
For example, the monoids M = 0; 3; 4; 6; 7; 8 ::: and M 0 = 0; 6; 8; 12; 14; 15; 16;::: f g f g have the following in…nite binary string representations:
M : 10011011111 ::: and (1)
M 0 : 100000101000101111 ::: . (2)
Since the binary strings of M and M 0 contain the patterns 11011 and 1 1 0 1 1, respectively, then the rings K [M] and K [M 0] are not weakly integrally closed. Thus, Brown’sring K[t3; t4] is not weakly integrally closed.
Theorem 2 ([2], Theorem 5) Let i and m be a pair of natural numbers such that m 2i and let M = 0; i; m; m + 1;::: . Then K[M] is weakly integrally closed. f g
For example, the ring K[t4; t7; t8;:::] is weakly integrally closed. The motivation for this dissertation is to answer the following two questions asked by Brewer and Richman in [2].
Question 1 Are weakly integrally closed monoid rings K [M] characterized by the property that the binary string of the monoid M contains no stretch of the pattern 11011?
Question 2 Is the ring K[t4; t5; t11] weakly integrally closed?
2 When investigating a particular property of monoid rings, it is common for math- ematicians to study the analogous property in the monoid itself. In Gilmer’s book, Commutative Semigroup Rings, he extends the concept of integrality to monoids and studies the relationship between integrally closed/completely integrally closed monoids and their corresponding monoid rings. Similarly, Geroldinger has exten- sively researched completely integrally closed monoids and their rings in papers such as The Complete Integral Closure of Monoids and Domains [4]. We will do the same by de…ning strongly integral elements for monoids. In doing so, we will negatively answer Question 1 and a¢ rmatively answer Question 2. In Chapter 2, the concept of being weakly integrally closed is extended to monoids and it is shown that if a monoid ring is weakly integrally closed, then so is the monoid. In Chapter 3, we give examples of patterns of …nitely many 0’sand 1’ssuch that when found in the characteristic binary string of a numerical monoid, then the monoid ring is not weakly integrally closed; we call such patterns forbidden. Question 1 is answered in Example 31. In Chapter 4, we discuss a JavaScript program that we wrote which determines if a pattern is forbidden. We prove that this program determines whether or not a numerical monoid is weakly integrally closed. Chapter 5 is devoted to showing that there is no …nite set S of forbidden patterns such that whenever a monoid M contains no stretch of a pattern in S, then M is weakly integrally closed. In Chapter 6, we show that particular monoid rings are weakly integrally closed. We answer Question 2 and show that the ring K tl; tl+1; tl+s; tl+s+1;::: is weakly integrally closed for all 3 < s l.
3 2 Background and Preliminary Results
We list some relevant notation, de…nitions and results used throughout the rest of this dissertation. Throughout this dissertation, all rings are commutative and contain 1.
Let D D0 be domains.
De…nition 3 ([2]) An element x in D0 is said to be strongly integral over D if there exists a nonzero …nitely generated ideal J of D such that xJ J 2.
De…nition 4 ([2]) The domain D is weakly integrally closed in D0 if every ele-
ment x D0 that is strongly integral over D is an element of D. 2
Theorem 5 ([2], Theorem 2)
1. (Intersections) If each member in a family of subrings Di of D is weakly inte-
grally closed in D, then iDi is weakly integrally closed in D. \
2. (Transitivity) If D is weakly integrally closed in D0, and D0 is weakly integrally
closed in D00, then D is weakly integrally closed in D00.
3. (Weakness) If D is integrally closed in D0, then D is weakly integrally closed in
D0.
De…nition 6 A monoid is a nonempty set closed under an associative binary oper- ation that has an identity element.
For the rest of this dissertation, we assume that all monoids are commutative and we denote the monoid operation by +. Also, 0 is contained in the natural numbers.
Then the natural numbers N form a monoid under addition.
4 De…nition 7 A monoid is numerical if it is an additive submonoid of the natural numbers.
We extend the concept of a strongly integral element to monoids. Let U be an arbitrary monoid and let S be a submonoid of U.
De…nition 8 ([5]) An ideal I of S is a nonempty subset of S such that I = I +S = i + s : i I and s S . f 2 2 g
De…nition 9 ([5]) An ideal I of S is …nitely generated if there is a …nite subset B of S such that I = B + S.
De…nition 10 An element u U is strongly integral over S if there exists a 2 …nitely generated ideal J of S such that u + J J + J.
De…nition 11 ([5]) An element u U is integral over S if nu S for some n > 0. 2 2
De…nition 12 A monoid M is cancellative if whenever there are m; a; b M such 2 that m + a = m + b, then a = b.
Clearly, any numerical monoid is cancellative.
Theorem 13 Let U be a cancellative monoid and let S be a submonoid of U. If u U is strongly integral over S, then u is integral over S. 2
Proof. Let u U be strongly integral over S. Then there is a …nitely generated ideal 2 J of S such that u + J J + J J. We will proceed by induction on the number of generators of any …nitely generated ideal J 0 of S such that u + J 0 J 0. Let F = f1; : : : ; fn be a set of generators of J and let N = 1; : : : ; n . Then f g f g there is a map : N N such that for each i N, ! 2
u + fi = f(i) + si
5 for some si S. Let F 0 = f(i) : i N . If F 0 = F , then i N fi = i N f(i) so 2 f 2 g 2 2
nu = i N si since U is cancellative. Thus, nu S: P P 2 2 WeP might as well assume F 0 is properly contained in F . Then F 0 has fewer than n
elements. Let J 0 = F 0 + S. Then u + J 0 J 0. Thus, u is integral over S by induction.
Let M be a numerical monoid. If d = gcd (M), then M=d has …nite complement in the natural numbers and is isomorphic to M [5]. Thus, we may assume that M has …nite complement in the natural numbers. Let D a domain and t an indeterminate.
m1 mn Then D [M] = d0 + d1t + + dnt : n N, di D, mi M and D [[M]] = f 2 2 2 g m1 m2 d0 + d1t + d2t : di D, mi M . In the case where D is a …eld, then D[t] f 2 2 g and D[[t]] are integrally closed, so they are weakly integrally closed by Theorem 5.
Lemma 14 Let K be a …eld and M a numerical monoid. Then:
1. K[[M]] is WIC if and only if K[[M]] is WIC in K[[t]]
2. K[M] is WIC if and only if K[M] is WIC in K[t].
Proof. We will prove (1); the proof of (2) is identical. Let Q be the quotient …eld of K[[M]] and let F be the quotient …eld of K[[t]]. Assume that K[[M]] is WIC. Suppose there is x K[[t]] and a nonzero …nitely 2 generated ideal J of K[[M]] such that xJ J 2. Let j be a nonzero element of J. 2 Then xj = j1 for some j1 J since J J. Thus, x = j1=j Q so x K[[M]] by 2 2 2 assumption. Hence, K[[M]] is WIC in K[[t]]. Assume K[[M]] is WIC in K[[t]]. But K[[t]] is integrally closed in F so is WIC in F . Thus, K[[M]] is WIC in F by Theorem 5. Let x Q be strongly integral over 2 K[[M]]. Then x F since Q F so x K[[M]]. Hence, K[[M]] is WIC. 2 2 Therefore, when deciding whether or not the monoid rings K[M] or K[[M]] are weakly integrally closed, it su¢ ces to consider only the strongly integral elements of
6 K[[t]].
De…nition 15 A numerical monoid M is weakly integrally closed (or WIC) if every natural number x that is strongly integral over M is an element of M.
Theorem 16 Let K be a …eld and M a numerical monoid. Then the following is true: K [[M]] is WIC = K [M] is WIC = M is WIC: ) )
Proof. Assume K [[M]] is WIC. Then K[[M]] is WIC in K[[t]] by Lemma 14. Let x K [t] and let J be a nonzero …nitely generated ideal of K [M] such that xJ J 2 2 for some . Let F be a …nite set of generators for J. Then xf J 2 for every f F . 2 2 Let L be the ideal generated by F in K [[M]]. Then J L K [[M]] since K [M] K [[M]]. Thus, L is a nonzero …nitely generated ideal of K [[M]] for which xf L2 2 for every f F so xL L2. Moreover, x K [[t]] since K [t] K [[t]]. Therefore, 2 2 x K [[M]] since K [[M]] is WIC in K[[t]]. Consequently, x K [t] K [[M]] = K [M] 2 2 \ so K [M] is WIC in K[t].
Assume K [M] is WIC. Then K[M] is WIC in K[t] by Lemma 14. Let x N and 2 let J be a nonempty ideal of M such that x + J J + J. Let F be a …nite set of generators for J. For every f F there exist kf ; lf F and mf M such that 2 2 2
x + f = kf + lf + mf .
Let L be the ideal generated by tf : f F in K[M]. Then txtf = tkf +lf +mf L2 f 2 g 2 for every f F . Moreover, L is nonzero since F is nonempty. Thus, tx K [M] since 2 2 K [M] is weakly integrally closed in K[t]. Hence, x M so M is weakly integrally 2 closed. Thus, the monoid M = 0; i; m; m+1;::: is weakly integrally closed by Theorem f g 2. Neither of the converses to Theorem 16 are known to be true. Unfortunately, we
7 have no counterexamples.
Theorem 17 Let M be a numerical monoid. If the binary string of M contains a stretch of the pattern 11011, then M is not weakly integrally closed.
Proof. If M contains a stretch of the pattern 11011, then there is k 0 and n > 0 such that k; k + n; k + 3n; and k + 4n are elements of M and such that k + 2n is not an element of M. Let x = k + 2n and let F = k; k + n; k + 3n; k + 4n . Then f g
x + k = (k + n) + (k + n) ;
x + (k + n) = k + (k + 3n) ;
x + (k + 3n) = (k + n) + (k + 4n) ; and
x + (k + 4n) = (k + 3n) + (k + 3n) .
Thus, x + F F + F , so x + J J + J. Hence, M is not weakly integrally closed since x M. 62 Thus, we have another proof of Theorem 1. If the binary string of M contains the pattern 11011, then both K[M] and K[[M]] are not weakly integrally closed by Theorem 16. In the next chapter, we investigate patterns such as 11011 that necessarily lead to monoids that are not weakly integrally closed. We will see that there are in…nitely such patterns. Then, by Theorem 16, their corresponding monoid rings are not weakly integrally closed.
8 3 Forbidden Patterns
3.1 Terminology and Preliminaries
Are there patterns other than the pattern 11011 such that whenever a monoid contains it, then it is necessarily weakly integrally closed? We will see that the answer is yes in the next section. But …rst, we list de…nitions and results that are used throughout the rest of this dissertation.
De…nition 18 A pattern F is a pair of disjoint …nite subsets, (F0;F1), of the natural numbers whose union is either empty or contains 0. Let UF = F0 F1. If UF is [ nonempty, then 0 = min (UF ).
We identify a pattern F with a string, where elements of F0 are represented by zeros, elements of F1 are represented by ones, and elements y N UF such 2 that min (UF ) < y < max (UF ) are represented by dots. For example, the string 1 010 01111 represents the pair ( 2; 4; 6 ; 0; 3; 7; 8; 9; 10 ). We will use the term f g f g pattern to refer either to the pair or its string.
De…nition 19 The length lF of a pattern F is de…ned as follows:
max(UF ) + 1 ; if UF is nonempty lF = 8 <> 0 ; if UF is empty.
:> De…nition 20 Let F and L be patterns. Then L is a stretch (or m-stretch) of F if
L0 = mF0 and L1 = mF1 for some m > 0, in which case, we write L = mF .
9 If UF is nonempty, then the length of mF is m (lF 1) + 1. Let F be the pattern 11011 and L the pattern 1 1 0 1 1. Then (F0;F1) = ( 2 ; 0; 1; 3; 4 ) and f g f g (L0;L1) = ( 6 ; 0; 3; 9; 12 ) so L is a 3-stretch of F . Since lF = 5, then lL = f g f g 3 4 + 1 = 13.
De…nition 21 Let F and L be patterns. Then L contains F (by z) if z + F0 L0 and z + F1 L1 for some integer z.
If F is the pattern 10 1 and L is the pattern 1101011, then L contains F by z = 3.
De…nition 22 Let M be a numerical monoid. Then we say that M contains F (by z) if (z+ F0) M is empty and z + F1 M for some integer z. \
Let M = 0; 4; 6; 8; 11; 12; 13;::: and let F be the pattern 1 100111. Then f g (F0;F1) = ( 3; 4 ; 0; 4; 5; 6; 7 ) so M contains F by z = 6. f g f g
De…nition 23 Let F be a pattern such that F1 is nonempty. Let F 0 = x F0 : f 2 T min (F1) < x < max (F1) and let = min (F 0 F1). De…ne the sets F = F 0 g [ 0 T T T T and F = F1 . Then the trimmed pattern F of F is the pair F ;F . 1 0 1 Suppose that (F0;F1) = ( 0; 4; 10 ; 2; 3; 5; 6; 8 ). Then F 0 = 4 and = 2 so f g f g f g F T ;F T = ( 2 ; 0; 1; 3; 4; 6 ). 0 1 f g f g De…nition 24 We say that F is forbidden if whenever a monoid M contains F , then M is not weakly integrally closed.
The pattern 11011 is forbidden by Theorem 17. It is easy to see that any pattern that contains a forbidden pattern is forbidden.
De…nition 25 If x F0 and C is a nonempty subset of F1 such that x+C C +C, 2 then we say that x is a bad zero of F .
10 Observe that we used the fact that the pattern 11011 has a bad zero to show that it is forbidden in Theorem 17.
Theorem 26 Any pattern that has a bad zero is forbidden.
Proof. Let F be a pattern that has a bad zero and let M be a monoid that contains F .
Then there exist an x N M and a …nite subset F 0 M such that x+F 0 F 0 +F 0. 2 Let J be the ideal of M generated by F 0. Then clearly x + J J + J so M is not weakly integrally closed. Hence, F is forbidden. The next result follows immediately from the de…nition of a forbidden pattern and Theorem 16.
Theorem 27 Let K be a …eld. If the binary string of a monoid M contains a for- bidden pattern, then K[M] is not weakly integrally closed, so neither is K[[M]].
De…nition 28 A pattern F is a block if UF is nonempty and if y UF for every 2 0 < y < max (UF ) (i.e. its string representation has no dots).
For example, the patterns 11011 and 1101011 are blocks.
De…nition 29 The block set of F is the set of all blocks H such that H contains F
and lH = lF . We refer to an element of the block set of a pattern F as a block of F . The block set of the pattern 011 01 is 011001; 011101 . f g
3.2 Examples of Forbidden Patterns
When convenient, we will use exponents l 0 in the string representation of a pattern. For example, 1021 2 014 is shorthand for the pattern 1001 01111. The next theorem generalizes the forbidden pattern 11011 to an in…nite family of forbidden patterns.
11 Theorem 30 Let j 1. The pattern 11 (01)j 1 has a bad zero.
Proof. Let F be the pattern 11 (01)j 1. Then
F0 = 2i : 1 i j and f g
F1 = 0 (2i + 1) : 0 i j 2 (j + 1) . f g [ f g [ f g
Let x = 2i for some 1 i j. Then
x + 0 = 1 + (2i 1) and x + 2 (j + 1) = (2i + 1) + (2j + 1) .
Let f = (2k + 1) for some 0 k j. Then
0 + [2(i + k) + 1] ; if j i + k x + f = 8 > 2 (j + 1) + [2 (i + k j) 1] ; if j < i + k. < :> Thus, x + f F1 + F1 for all f F1, so every element of F0 is a bad zero. Hence, 2 2 F has a bad zero. So, if a monoid M contains the pattern 11 (01)j 1, then M is not weakly integrally closed and neither is its monoid algebra K [M] by Theorem 16. This next example negatively answers Question 1. We will show that the exclusion of the pattern 11011 and all of its stretches from a monoid is not enough to guarantee that its monoid algebra will be weakly integrally closed.
Example 31 Let M = 0; 5; 6; 8; 10; 11; 12;::: . Then the monoid M contains no f g stretch of the pattern 11011 but its monoid algebra, K [t5; t6; t8; t14], is not weakly integrally closed.
12 Proof. The binary string of M,
1000011010111 ::: ,
contains the pattern 1101011. Thus, M is not weakly integrally closed by Theorem 30 so K [M] is not weakly integrally closed by Theorem 16. Clearly, M does not contain the pattern 11011. Suppose M contains an n-stretch of the pattern 11011 for some n 2. Then k; k + n; k + 3n; k + 4n M and k + 2n M for some k 0. 2 62 If k = 0, then k + n 5 so k + 2n 10. So we might as well assume that k 5. Thus, k + n 8 since k + n M so k + 2n 10. Therefore, M contains no stretch 2 of the pattern 11011. Hence, K [t5; t6; t8; t14] is a monoid algebra that is not weakly integrally closed but whose corresponding monoid contains no stretch of the pattern 11011. The next theorem is another generalization of the forbidden pattern 11011.
Theorem 32 Let l 0. The pattern 1 l 1 l 01l+2 has a bad zero. Proof. Let F be the pattern 1 l 1 l 01l+2. Then
F0 = 2(l + 1) and f g
F1 = 0 l + 1 2l + i : 3 i l + 4 . f g [ f g [ f g
Let x = 2 (l + 1). Then
x + 0 = (l + 1) + (l + 1) ,
x + (l + 1)n = 0 + 3 (l + 1) , and
x + (2l + 3) n = (l + 1) + (3l + 4).
Let f = 2l + i for some 4 i l + 4. Then x + f = (2l + 3) + (2l + i 1). Hence, 13 x + f F1 + F1 for every f F1, so F has a bad zero. 2 2 For example, the pattern 10100111 has a bad zero. Therefore, the monoid M = 5; 7; 10; 11; 12;::: , as well as its monoid algebra K [t5; t7; t11], are not weakly inte- f g grally closed. We will see that a forbidden pattern need not have a bad zero. But …rst we list an important lemma that will be used repeatedly throughout this dissertation.
Lemma 33 If C is a nonempty …nite subset in N and x is an element of N C such that x + C C + C, then there are a; b; c; d C such that a < b < x < c < d. 2
Proof. Let a = min (C) and d = max (C). Then x + a = c1 + c2 and x + d = c3 + c4 for some c1; c2; c3; c4 C such that a c1; c2 and c3; c4 d. 2 Let i; j 1; 2 such that i = j . If ci = a, then cj = x which contradicts x C. 2 f g 6 62 Thus, a < c1; c2. If x < ci, then cj = x + a ci < x + a x = a which contradicts the minimality of a. Thus, c1; c2 < x. Therefore, there exists some b C such that 2 a < b < x.
Let k; l 3; 4 such that k = l. If ck = d, then cl = x which contradicts x C. 2 f g 6 62 Thus, c3; c4 < d. If ck < x, then cl = x + d ck > x + d x = d which contradicts the maximality of d. Thus, x < c3; c4. Therefore, there exists some c C such that 2 x < c < d. Hence, there exist a; b; c; d in C such that a < b < x < c < d. So, for example, the pattern 100 ::: 00111 ::: 1 contains no bad zeros.
Theorem 34 The pattern 1101 111 has no bad zeros, but each of its blocks has a bad zero, so 1101 111 is forbidden.
Proof. Let F be the pattern 1101 111. Then F0 = 2 and F1 = 0; 1; 3; 6; 7; 8 . f g f g Let C be a nonempty subset of F1 such that 2 + C C + C. Then 1 C by Lemma 2
14 33. Thus, 2 + 1 C + C so 0; 3 C. Then 2 + 3 C + C. But 5 F1 + F1 so 2 2 2 62 5 C + C. Hence, F has no bad zeros. 62 We will show that every block of F has a bad zero. The following is a list of all the blocks of F :
H1 = 110100111;H2 = 110101111;H3 = 110110111;H4 = 110111111
It is easy to see that H1 contains 1 1 0111, H2 contains 1101011, H3 contains 11011, and H4 contains 11011. Let 1 i 4. Then Hi contains a pattern with a bad zero by Theorems 30 and 32. Thus, Hi itself has a bad zero. Let M be a monoid that contains the pattern F . Then M contains some block H of F , so contains a pattern that has a bad zero. Hence, M is not weakly integrally closed, so F is forbidden.
3.3 A Characterization of Forbidden Patterns
In Lemma 26, we saw that any pattern with a bad zero is forbidden. The converse need not be true, as we saw with the forbidden pattern 1101 111. We generalize the results of Lemma 26 to characterized an arbitrary forbidden pattern. But …rst we make a brief remark.
Remark 35 Let M be a numerical monoid, let e be the smallest positive integer in M and let g be the largest integer not in M. We know that g exists since M has …nite complement in N. Let x N and let J be an ideal of M such that x + J J + J. 2 Let n = min (J). Then x + n = j1 + j2 for some j1; j2 J. We will show that when 2 trying to prove that M is weakly integrally closed, then one might as well assume that
e n < j1; j2 < x g:
15 Proof. If n = 0, then x = x + 0 = j1 + j2 J + J M. If n = j1 or n = j2, then 2 x J M. Thus, we might as well assume that 0 < n < j1; j2 or else x M. If 2 2 j1 x or j2 x, then either x J M (and we are done) or x+n = j1 +j2 > x+n, 2 a contradiction. Thus, we might as well assume that j1; j2 < x. If x > g, then x M. 2 Also, n e since n > 0. Hence, we might as well assume that e n < j1; j2 < x g.
The results from this remark are used in the proofs of Lemma 36 and Theorem 47.
Lemma 36 Let H be a block of length d and let
2 2 2 2 M = 0 d + H1 2d ; 2d + 1; 2d + 2;::: . f g [ [ f g If H has no bad zeros, then M is weakly integrally closed.
Proof. It is clear that M is a monoid that contains H. Assume that H has no bad zeros. Let x N and let J be an ideal of M such that x + J J + J. We will show 2 that x M. 2 Let n = min (J). Then x+n = j1 +j2 for some j1; j2 J. By Remark 35, we might 2 2 2 2 as well assume that d n < j1; j2 < x < 2d or else x M. Then n; j1; j2 d +H1, 2 2 2 2 2 so d > 1 and d n; j1; j2 d + d 1. Thus, x + n = j1 + j2 2 (d + d 1) which implies that x d2 + 2d 2. 2 Let I = J (d + H1). Suppose that there is i I such that x + i I + I. \ 2 62 But i J, so x + i J + J. Thus, x + i = j3 + j4 for some j3; j4 J. Without 2 2 2 2 loss of generality, we may assume that j3 I. Then j3 2d since 0 J. Thus, 62 62 2 2 2 2 x + i = j3 + j4 3d , so x 2d d + 1. Then 2d d + 1 x d + 2d 2, so d2 3d+3 0. But d2 3d+3 3=4 for all real numbers d, so we have a contradiction. 2 Thus, x + I I + I. Then min (I) x max (I), so x d + (H0 H1) since H is a 2 [ 16 2 block. But H has no bad zeros, so x d + H1 M. Hence, M is weakly integrally 2 closed. Therefore, a block is forbidden if and only if it has a bad zero. Now, we are ready for our characterization of forbidden patterns.
Theorem 37 A pattern is forbidden if and only if each of its blocks has a bad zero.
Proof. Let F be a forbidden pattern and let H be a block of F . Let M be the monoid of Lemma 36, so M contains H. Then M contains F , a forbidden pattern, since H contains F . Thus, M is not weakly integrally closed. Hence, H has a bad zero by Lemma 36. Let F be an arbitrary pattern and assume that each of its blocks has a bad zero. Let M be a monoid containing F . Then M contains a block of F . But each block of F has a bad zero, so is forbidden. Thus, M is not weakly integrally closed. The next theorem implies that we can generalize Lemma 36.
Theorem 38 Let N be a …nite set of blocks that contain no bad zeros. Then there is a block that contains every block in N and that contains no bad zeros.
Proof. We will …rst show that the theorem is true when N contains exactly two elements. Let S = G; H . Then G and H are blocks that contain no bad zeros. Let f g n = 2lG + lH and let L be the pair (L0;L1), where
L0 = G0 lG; lG + 1; : : : ; n 1 (n + H0) and [ f g [
L1 = G1 (n + H1) . [
It is clear that L contains the blocks G and H. The pattern L is a block since G and
H are blocks. Let x be an element of L0 L1 and S a nonempty subset of L1 such [ that x + S S + S. We will show that L has no bad zeros by showing x L1. 2 17 Let G0 = S G1 and let H0 = S (n + H1). Suppose that x + G0 G0 + G0. \ \ Then min (G1) x max (G1) by Lemma 33. Thus, x is contained in G0 G1 since [ G is a block. So x G1 L1 since G has no bad zeros. The same argument shows 2 that if x + H0 H0 + H0, then x (n + H1) L1. Thus, we might as well assume 2 that there is g G0 and h H0 such that x + g G0 + G0 and x + h H0 + H0. 2 2 62 62 Since g lG 1, then x + g (n + lH 1) + (lG 1) < 2n and since h n, then x + h n. Note the following observations. If f is an element of G0+G0, then f 2 (lG 1) < n. If f is an element of H0 + H0, then f 2n. If f is an element of G0 + H0, then n f (lG 1) + (n + lH 1) = n + lG + lH 2. By the above observations, x + g H0 + H0 since x + g < 2n and x + h G0 + G0 62 62 since x + h n. Then x + g and x + h are elements of G0 + H0. Therefore, x + g n and x + h n + lG + lH 2, so x n g n (lG 1) = lG + lH + 1 and x n+lG +lH 2 h n+lG +lH 2 n = lG +lH 2. Thus, lG +lH < x < lG +lH , a contradiction. Therefore, x is an element of L1, so L has no bad zeros. Hence, the theorem is true by induction.
De…nition 39 A pattern F is said to be safe if it is not forbidden.
Then a pattern F is safe if and only if a block of F has no bad zeros by Theorem 37.
Theorem 40 For every …nite set S of safe patterns, there is a weakly integrally closed monoid that contains every pattern in S.
Proof. Let S be a …nite set of safe patterns. Each pattern in S has a block that has no bad zeros. For each pattern F of S, …x HF to be a block of F that has no bad zeros. Let N = HF : F is an element of S . Then, by Theorem 38, there is some f g block H such that H contains every block in N and such that H contains no bad
18 zeros. Let M be the monoid from Lemma 36. Then M is weakly integrally closed since H has no bad zeros. Moreover, M contains every block in N, so M contains every element of S.
19 4 An easy test to determine if a monoid is weakly
integrally closed
4.1 Algorithm to determine if a pattern has a bad zero
We have developed a JavaScript program that …nds any bad zeros of a pattern. Before discussing the algorithm that the program is based on, we list some relevant de…nitions and lemmas.
De…nition 41 If a pattern F has a bad zero x F0 by C F1, then we say that the 2 pattern ( x ;C), denoted Cx, is a closed pattern of F (by x). f g
De…nition 42 Let F be a pattern and let U x be a closed pattern of F . Then U x is said to be the maximum pattern of F by x if L U for every closed pattern Lx of F .
For example, if F is the pattern 1011011, then it is not di¢ cult to see that x = 4 is
the only bad zero of F and that (4;F1) and (4; 2; 3; 5; 6 ) are the only closed patterns f g of F . Then (4;F1) is the maximum pattern of F by x = 4.
Lemma 43 If F is a pattern with a bad zero x F0, then there is a maximum pattern 2 x U of F for every element x F0. 2
Proof. Let x F0 be a bad zero of F . Let x = C F1 : C is nonempty, 2 f x + C C + C and let U = C. Then U F since C F for every C . C x 1 1 x g 2 2 Let u U. Then u C for someS C x. Thus, x + u C + C U + U, so (x; U) is 2 2 2 2
20 a closed pattern of F by x. Then it is clear that the pattern U x is maximum pattern
of F by x since C U for every C x. 2 Let F be a pattern and let x be an element of F0. Let = S F1 : S is f nonempty . De…ne the map : as g x !
(S) = s S : x + s S + S x f 2 2 g
for each S . If maps S onto itself, then (x; S) is a closed pattern of F by x. 2 x Thus, if (F1) = F1, then (x; F1) is the maximum closed pattern of F by x F0 x 2
Lemma 44 Let F be a pattern and x an element of F0. Let L0 = x (F1) and
x Li+1 = (Li) for i 0. If C is a closed pattern of F , then C Li for all i 0. x
x Proof. Let C be a closed pattern of F . Then C F1 and x + C C + C, so
x + C F1 + F1. Thus, C L0 = x (F1) = f F1 : x + f F1 + F1 . If C Li 1 f 2 g for some i > 0, then x + C C + C Li 1 + Li 1. Thus, C l Li 1 : x + l f 2 2
Li 1 + Li 1 = x (Li 1) = Li. Therefore, C Li for all i 0 by induction. g The following algorithm is the basis of our JavaScript program.
Algorithm 45 Let F be a pattern and x an element of F0. Let L0 = x (F1) and
Li+1 = (Li) for i 0. x
1. If Li is empty for some i 0, then x is not a bad zero.
2. If Li is nonempty and Li+1 = Li for some i 0, then x is a bad zero and (x; Li) is the maximum pattern of F by x.
Proof. It is clear that 1 follows immediately from Lemma 44. To show 2, suppose that Lk = Lk+1. Then Lk = (Lk) = l Lk : x+l Lk +Lk , so x+Lk Lk +Lk. x f 2 2 g Note that
F1 L0 L1 ::: Lk = Lk+1. 21 x Thus, Lk F1, so (x; Lk) is a closed pattern of F . Let U be the maximum pattern x of F by x. Then Lk U. Since U is a closed pattern of F , then U Lk by 44. Therefore, U = Lk, so (x; Lk) is the maximum pattern of F by x.
If the pattern (x; Li) is the maximum pattern of F , then for each element l Li, 2 i we will apply the map x recursively to Gl = Li l . We give an example to f g demonstrate how the program works.
Example 46 Let F be the pattern 101001111. Then we use the algorithm to show that F contains one bad zero x F0 and three closed patterns. 2
Proof. If F is the pattern 101001111, then F0 = 1; 3; 4 and F1 = 0; 2; 5; 6; 7; 8 . f g f g By Lemma 33, x = 1 cannot be a bad zero of F . If x = 3, then L0 = 2; 5; 6; 7; 8 , f g L1 = 5; 6; 7; 8 , L3 = 7; 8 and L4 is empty. Thus, x = 3 cannot be a bad zero by f g f g 1 of Algorithm 45.
Let x = 4. Then L0 = F1. Thus, the pattern (4;F1), i.e. 1 1 01111, is the i maximum pattern of F by x = 4. Let Gl = Li l . By Lemma 33, it su¢ ces to f g check Gl for 4 < f < 9. If L0 = (G5), then L1 = 0; 2; 6; 8 = L2. Thus, the x f g pattern 1 1 0 1 1 is a closed pattern of F by x = 4. Note that by Lemma 33, the pattern 1 1 0 1 1 contains no closed patterns other than itself. If L0 = (G6), x then L1 = 0; 5; 8 , so (x; L1) is not a bad zero by Lemma 33. If L0 = (G7), then f g x L1 = 0; 6; 8 , so (x; L1) is not a bad zero by Lemma 33. If L0 = (G8), then f g x L1 = 0; 2; 5; 6; 7 = L2, so the pattern 1 1 0111 is a closed pattern. We repeat f g the algorithm on L1 l for each l L1. Let Gl = L1 l . Note that it su¢ ces to f g 2 f g check only G5, G6 and G7 by Lemma 33. We leave it to the reader to see that the pattern 1 1 0111 contains no closed patterns other than itself. Hence, the pattern
22 101001111 contains exactly the following closed patterns:
1 1 0 1 1; 1 1 0111; and 1 1 01111.
Next, we will characterize weakly integrally closed monoids in a way that allows us to use Algorithm 45 to easily determine if a numerical monoid is weakly integrally closed.
4.2 A characterization of weakly integrally closed monoids
Let M be a monoid, let e be the smallest positive integer in M and let g be the largest integer not in M. Let
H0 = x N M : e < x g and f 2 g
H1 = m M : e m 2g + 1 e . f 2 g
Then the test block H of M is the pattern (H0;H1). Clearly, M contains its test block H. We will show that it su¢ ces to consider only the test block of a monoid M when determining if M is weakly integrally closed.
Theorem 47 A monoid is weakly integrally closed if and only if its test block contains no bad zeros.
Proof. Let M be a monoid and H its test block. If H has a bad zero, then H is forbidden by Theorem 26, so M is not weakly integrally closed.
23 Assume that H has no bad zeros. Let x N and let J be an ideal of M such 2 that x + J J + J. Let n = min (J). Then x + n = j1 + j2 for some j1; j2 J. By 2 Remark 35, we might as well assume that e n < j1; j2 < x g or else x M and 2 we are done. Then x H0 H1. Let 2 [
I = j J : j < g g + 1; g + 2;:::; 2g + 1 e . f 2 g [ f g
Then I is not empty since n I. Let i I. We will show that x + i I + I. 2 2 2 Suppose that i < g. Then i J, so x + i J + J. Thus, there are j3; j4 J such 2 2 2 that j3 + j4 = x + i < 2g. Then j3 < 2g j4 2g e and j4 < 2g j3 2g e, so j3; j4 I. Thus, if i < g, then x + i I + I. 2 2 Suppose that i > g. Observe that
x + i = n + (i + x n) = (g + 1) + (i + x (g + 1)) .
Clearly, n; g + 1 I. We will show that either i + x n I or i + x (g + 1) I. 2 2 2 Assume that i + x n I. But i + x n > g since i > g and x > n. Thus, 62 i + x n > 2g + 1 e since i + x n I. Then i + x (g + 1) > g since n e. But 62 also, i + x (g + 1) < i since x < g + 1. Thus, i + x (g + 1) I. Then x + i I + I, 2 2 so x + I I + I. But I H1 and H has no bad zeros. Hence, x H1 M, so M is 2 weakly integrally closed. So, if the binary string of a monoid M is 10001001111 :::, then it su¢ ces to check the pattern 100111 for any bad zeros to determine whether or not M is weakly integrally closed. It is easy to see that this pattern has no bad zeros by Lemma 33, so M is weakly integrally closed by Theorem 47.
24 5 Question 1 Generalized
We saw that excluding the pattern 11011 and its stretches from a monoid does not guarantee that the monoid is weakly integrally closed. Does there exist a …nite set of forbidden patterns that su¢ ces? Speci…cally,
Is there a …nite set of patterns such that whenever a monoid M con- tains no stretch of any of these patterns, then K[M] is necessarily weakly integrally closed?
In Example 31, we used the fact that the forbidden pattern 1101011 contains no stretch of the pattern 11011. Similarly, we will show that for every …nite set of forbidden patterns S, there is j > 0 such that the pattern 11(01)j1 contains stretch of an element of S. Then we construct a monoid containing the pattern 11(01)j1 but containing no stretch of a pattern in S. First, some technical lemmas:
Lemma 48 Any forbidden pattern that 11(01)j1 contains has length equal to 2j + 3.
Proof. Suppose 11(01)j1 contains a forbidden pattern L. Then there is a block H
j j of L such that 11(01) 1 contains H. Thus, lL = lH 2j + 3, the length of 11(01) 1, and there is m 0 such that
m + H0 2i : 1 i j and f g
m + H1 0 2i + 1 : 0 i j 2 (j + 1) . f g [ f g [ f g
Furthermore, H has a bad zero by Theorem 37 since L is forbidden. Thus, there is x m+H0 and nonempty H0 m+H1 such that x+H0 H0 +H0. Let = min (H0) 2 25 and = max (H0). Then 0 < 2 (j + 1). We will show that +1 = 2j +3. Observe that x+ = h1 +h2 and x+ = h3 +h4 for some h1; h2; h3; h4 H0 such 2 1 that < h1; h2; h3; h4 < . Thus, h1; h2; h3; h4 B so x + ; x + 2i : i 1 . 2 2 f g As a result, ; 2k : k 0 since x A. Therefore, = 0 and = 2 (j + 1) so 2 f g 2 + 1 = 2j + 3. Then 2j + 3 lH since H0 m + H1. Hence, lL = lH = 2j + 3. For example, the pattern 1101010101011 has length 13 and
1101 1 1 1 11; 11 101 1 1 11; 11 101 11; 11 1 101 1 11; 11 1 1 101 11; 11 1 1 1 1011 and 1 1 0 1 1 lists every closed pattern of F , all of which have length 13.
Lemma 49 Any stretch of a forbidden pattern is forbidden.
Proof. Let F be a forbidden pattern and let m > 0. Then mF is a stretch of F . Let G be in the block set of mF . Then G contains mH for some block H of F . And H has a bad zero by Theorem 37. Thus, there is x H0 such that x + H0 H0 + H0 2 for some nonempty H0 H1. Let m > 0, y = mx and L = mH0. Then y mH0, 2 L mH1 and y + L L + L, so mH has a bad zero. Thus, G has a bad zero since it contains mH. Hence, mF is forbidden by Theorem 37.
Lemma 50 For every forbidden pattern F , the set F1 has at least 4 elements.
26 Proof. Let F be a forbidden pattern. Let V be the set of all y (F0 F1) such 62 [ that 0 < y < max (F0 F1). Let G0 = F0 V and G1 = F1. Then G is a block of [ [ F . Thus, G has a bad zero by Theorem 37 since F is forbidden. Then there exist
a; b; c; d G1 = F1 such that a < b < c < d by Lemma 33, so F1 has at least 4 2 elements. The pattern 11011, with length 5, has the smallest length of all forbidden patterns.
Lemma 51 A pattern is forbidden if and only if its trimmed pattern is forbidden.
Proof. Let F be a pattern. Suppose its trimmed pattern F T is forbidden. Then F is forbidden since F contains F T . Assume that F is forbidden. Let G be a block of F T . Then G = HT for some
block H of F and H has a bad zero since F is forbidden. Thus, there is x H0 and 2 a nonempty …nite subset H0 H1 such that x + H0 H0 + H0. Then min (H1) < T T x < max (H1) by Lemma 33, so x H . Thus, H has a bad zero. Hence, G has a 2 0 bad zero for every block G of F T , so F T is forbidden.
Theorem 52 Let S be a …nite set of forbidden patterns. Then there exists j > 0 such that 2j + 3 is greater than the length of every pattern in S and 11(01)j1 contains no stretch of an element in S.
Proof. Fix j = F S (lF 1). Let F S. Then 2j + 3 2 (lF 1) + 3 > lF . Let 2 2
m > 0 and K = mFQ . Suppose lK = 2j + 3. Then m (lF 1) = 2 F S (lF 1) + 2 so 2 lF 1 divides 2. But lF 1 4 by Lemma 50, so lF 1 does not dividesQ 2. Therefore, j lK = 2j + 3. Hence, 11(01) 1 contains no stretch of any element of S by Lemma 48. 6
27 Lemma 53 Let L be a pattern. There is a monoid M containing L such that for any m > 0 and any trimmed forbidden pattern F for which lF < lL,
if M contains mF , then L contains mF .
2 2 2 2 Proof. Let d = lL and M = 0 (d + L1) 2d ; 2d + 1; 2d + 2;::: . Then M f g [ [ f g is a monoid that contains L. Let K = mF . Then K is a trimmed forbidden pattern by Lemma 49. Assume that M contains K. Then z + K1 M and (z + K0) M is \ 2 empty for some z 0. Let K0 = z + K0, K0 = z + K1 and I = (d + L1) K0 . Let 0 1 \ 1 = z + lK 1. We will show that I contains exactly one element.
Suppose that I is empty. Then 100 ::: 012d2 11 ::: 1 is a block of K, so it has a bad zero by Theorem 37 since K is forbidden. But this contradicts Lemma 33. Thus, I is not empty. We might as well assume that z = 0 or 2d2 or else we are done. 2 2 2 2 Then lK = z + 1 2d (d + d 1) = d d + 1. Thus, m (lF 1) d d, so m > d since lF < lL = d. Then I contains exactly one element since K is an m-stretch.
If k = z, then 100 ::: 012d2 z11 ::: 1 z is a block of K, a contradiction to Lemma
33. If k = , then K1 contains only two elements, a contradiction to Lemma 50.
2 Thus, z = 0 and 2d , so 100 ::: 01k00 ::: 01 2 11 ::: 1 is a block, call H, of K. 2d Then there is x H0 and H0 H1 such that x + H0 H0 + H0 since K is forbidden. 2 2 2 Also, 0 H0 and k < x < 2d by Lemma 33. Thus, x + 0 = k + k = 2k 2d , a 2 2 contradiction. Hence, z, (d + L1) so L contains K. 2
Theorem 54 For every …nite set S of forbidden patterns, there is a monoid that is not weakly integrally closed and whose binary string contains no stretch of an element of S.
28 Proof. Let S be a …nite set of forbidden patterns and let ST be the set of the trims of elements of S. By Theorem 52, there is j > 0 such that 11(01)j1 contains no stretch of an element in ST . Let M be the monoid from Lemma 53, where 11(01)j1 is the block L. Then M contains the forbidden pattern 11(01)j1, so M is not WIC. By Lemma 53, M contains no stretch of an element of ST since 11(01)j1 doesn’t. Thus, M contains no stretch of an element of S. Thus, there is no …nite set S of forbidden patterns such that whenever a monoid contains no stretch of an element in S, then its monoid algebra is necessarily WIC.
29 6 Weakly integrally closed domains
Let K be a …eld. Recall that the ring K [ti; tm; tm+1;:::] is weakly integrally closed by Theorem 2. The proof to show that the power series ring is also weakly integrally closed is identical to Brewer and Richman’sproof of Theorem 2. We restate Theorem 2 for formal power series rings.
Theorem 55 Let i; m 0. If Mi;m = 0; i; m; m+1;::: is a monoid, then K [[Mi;m]] f g is weakly integrally closed.
1 u Let S be a ring, t an indeterminate and let R = S[t]. Let a = a0 +a1t +:::+aut be an element of R. If a is nonzero, then de…ne the value vt(a) to be the least j such that aj = 0. If a = 0, then de…ne vt (a) = 0. Let J be any nonzero ideal of R and 6 2 de…ne the value of J as vt(J) = min vt(e) : 0 = e J . Let n = vt(J). If xJ J , f 6 2 g 1 v then vt(xe) 2n, so vt (x) n. Note that if b = b0 + b1t + ::: + bvt is an element of R, then (ab)i = akbi k. 0 k i Recall, in QuestionP 2, that Brewer and Richman ask if the ring K[t4; t5; t11] is weakly integrally closed. Consider the monoid M = 0; i; i + 1; m; m + 1;::: , where f g 0 m 2i. If i = 4 and m = 8, then K[M] is the ring K[t4; t5; t11]. Suppose that m = i + 3. Then the characteristic binary string of M, 10 ::: 01i1i+101m111 :::, contains the pattern 11011. Thus, M is not weakly integrally closed by Theorem 17, so K[M] and K[[M]] are not weakly integrally closed by Theorem 16. Therefore, we will require that 2 < i + 3 < m 2i and prove that the formal power series ring K[[ti; ti+1; tm; tm+1;:::]] is weakly integrally closed.
30 Lemma 56 Let i 0. If M = 0; i; i+1; i+4; i+5; i+6;::: , then K[[M]] is weakly f g integrally closed.
i i+1 i+2 Proof. Let R = k[[M]] and R0 = k[[t ; t ; t ;:::]]. Then R0 is weakly integrally closed by Theorem 55. we will show that R is weakly integrally closed in R0.
Let x be an element of R0 and J a nonzero …nitely generated ideal of R such that
2 xJ J . Let n = vt (J). Then vt (x) n. If n i + 4, then xi+2 = xi+3 = 0 so x R and we are done. Thus, we might as well assume that n < i + 4, so n = 0, 2 n = i or n = i + 1.
Before continuing, we insert a remark. Let a R and b R0 such that vt (a) ; vt (b) 2 2 n. Then
(ab)n+i+2 = an+kbi+2 k and 0 k i+2 n X (ab)n+i+3 = an+kbi+3 k. 0 k i+3 n X Thus, if n = 0 or n = i + 1, then
(ab)n+i+2 = anbi+2 and
(ab)n+i+3 = anbi+3,
and if n = i, then
(ab)n+i+3 = aibi+3 + ai+1bi+2
since ak = 0 for all 0 < k < i and all i + 1 < k < i + 4, and bk = 0 for all 0 < k < i.
Suppose that e; e0 J. If n = 0 or n = i + 1, then (ee0) = ene0 = 0 and 2 n+i+2 i+2
(ee0)n+i+3 = enei0+3 = 0. If n = i, then (ee0)n+i+3 = eiei0+3 + ei+1ei0+2 = 0. So for
2 any element f of J , if n = 0; i + 1, then fn+i+2 = fn+i+3 = 0 and if n = i, then
31 fn+i+3 = 0. Therefore, the following equations are true for any element e of J:
0 = enxi+2 = enxi+3 ; if n = 0; i + 1
0 = enxi+3 + en+1xi+2 ; if n = i
Let c be an element of J such that vt (c) = n. Then cn = 0. If n = 0 or n = i + 1, 6 then 0 = cnxi+2 = cnxi+3 by the above remarks. Thus, xi+2 = xi+3 = 0, so x R. 2 Suppose that n = i. Then 0 = cixi+3 +ci+1xi+2 by the above remarks. If xi+2 = 0, then 0 = cixi+3. Thus, xi+3 = 0 since ci = 0, so x R. Therefore, assume that 6 2 xi+2 = 0. Let x0 = x (xn=cn) c. Then vt(x0) i + 1, so vt (x0c) 2i + 1. 6 We need to pause for another remark. If e J, then ei+1 = eici+1=ci since xi+2 = 0 2 6 and ci = 0. Thus, e = c + b for some K and b R such that vt (b) i + 4. 6 2 2 Therefore, if f is an element of J 2, then
f = c2 + d
for some k and d R0 such that vt(d) 2i + 4. 2 2 2 2 2 Since xc J , then x0c J . Thus,by the above remark, x0c = c + d for some 2 2 k and d R such that vt(d) 2i+4. If = 0, then vt (x0c) = 2i, a contradiction. 2 2 6 Thus, = 0, so vt (d) 2i+4. Therefore, vt(x0c) 2i+4, implying that vt(x0) i+4. Then x0 R, so x R. Thus, R is weakly integrally closed in R0. Since R0 is weakly 2 2 integrally closed, then so is R by Theorem 5.
Theorem 57 Let 2 < i + 3 < m 2i. If Mi;m = 0; i; i + 1; m; m + 1;::: , then f g Ri;m = K[[Mi;m]] is weakly integrally closed.
Proof. The theorem is true for m = i + 4 by Lemma 56. Fix i 0 and let Mi;m = Mm and Ri;m = Rm. We will show that Rm is weakly integrally closed in
Rm 1, for m > i + 4. 32 Let x be an element of Rm 1 and J a nonzero …nitely generated ideal of Rm such 2 that xJ J . Let n = vt (J). Then vt (x) n. We might as well assume that n < m or else vt (x) m so x Rm. Thus, n = 0, n = i or n = i + 1. 2 If a Rm and b Rm 1 such that vt (a) ; vt (b) n, then 2 2
(ab)n+m 1 = anbm 1 + an+kbm 1 k. 1 k m 1 n X
Suppose that there is 1 k m 1 n such that an+k = 0. We will show that 6 bm 1 k = 0. Since an+k = 0, then n + k < m, so i n + k i + 1 . Thus, n i, so 6 n = 0 or n = i. Moreover, i 1 n k i implies that m i 2 m n k 1 m i 1. Thus, n + 2 < m k 1 i + n 1, since i + 4 < m 2i. If n = 0, then 0 < m k 1 < i, so bm k 1 = 0. Suppose that n = i. Then i + 2 < m k 1. Also, m k 1 < m 1 since k 1. Therefore, i + 2 < m k 1 < m 1, so bm k 1 = 0. Thus,
(ab)n+m 1 = anbm 1.
If e; e0 J, then (ee0)n+m 1 = enem0 1 = 0. Thus, fn+m 1 = 0 for every element f of 2 J 2. Let e be an element of J such that v (e) = n. Then
0 = (xe)n+m 1 = enxm 1
2 since xe J . Thus, xm 1 = 0 since en = 0. Therefore, x Rm so Rm is weakly 2 6 2 integrally closed in Rm 1. If Rm 1 is weakly integrally closed, then Rm is weakly
integrally closed by Theorem 5. Hence, Rm is weakly integrally closed by induction.
Therefore the answer to Question 2 is yes, so Brewer and Richman’sring K[t4; t5; t11] is weakly integrally closed..
33 Theorem 58 Let i; m; p 0 and let
Mi;p;m = 0 i + p : 0 m; m + 1;::: f g [ f g [ f g
be a numerical monoid. Then K[[Mi;p;m]] is weakly integrally closed.
Proof. If m < 2, then Mi;p;m is the set of all natural numbers, so K[[Mi;p;m]] = K[[t]]
m m+1 and we are done. If m i, then Rm = K[[t ; t ;:::]] is weakly integrally closed by Theorem 55. Thus, we might as well assume that m > i. Fix i; p 0. Let Mm = Ml;p;m and Rm = K[[Mi;p;m]] for m > 1. We will show that Rm is weakly
integrally closed in Rm 1. Note that there is j 0 such that i+jp < m i+(j + 1) p. Fix j. If m = i + jp + 1, then Rm = Rm 1 and we are done. Thus, we might as well assume i + jp + 1 < m < i + (j + 1) p + 1.
Let x be an element of Rm 1 and J a nonzero …nitely generated ideal of Rm such 2 that xJ J . We will show that xl+m 1 = 0. Let n = vt (J). Then vt (x) n. We may assume that n < m or else we are done. Thus, n = 0 or n = i + p for some 0 j. We will insert a remark before continuing. Let a Rm and b Rm 1 such that 2 2 vt (a) ; vt (b) n. Then
(ab)n+m 1 = anbm 1 + an+kbm 1 k. 1 k m 1 n X
Let 1 k m 1 n. Suppose that an+k = 0. We will show that bm 1 k = 0. Since 6 m 1 k < m 1, then it su¢ ces to show that
i + p < m 1 k < i + ( + 1) p for some 0 (3)
34 Since i+jp+1 < m < i+(j + 1) p+1, then i+jp k < m 1 k < i+(j + 1) p k. Since 0 < n + k < m, then n + k = i + p for some 0 j, so k = i + p n. Thus, (j ) p + n < m 1 k < (j + 1) p + n
Therefore, if n = i + p for some 0 j, then (3) is true and we are done. Therefore, we might as well assume that n = 0. Suppose that p does not divide i.
m m+1 Then 2i m since Mm is a monoid, so i m. Therefore, Rm = K[[t ; t ;:::]], so Rm is weakly integrally closed by Theorem 55. Thus, we assume that p divides i. Since n = 0, then (j ) p < m 1 k < (j + 1) p. Therefore, (3) is satis…ed, so bl+m 1 k = 0. Thus, if an+k = 0, then bl+m 1 k = 0, so an+kbl+m 1 k = 0 for all 6
1 k m 1 n. Consequently, (ab)n+m 1 = anbm 1.
Let e; e0 J. Then (ee0)n+m 1 = enem0 1 = 0 by the above observation. Thus, if 2 2 f is any element of J , then fn+m 1 = 0. Assume e J has value n. Recall that 2
vt (x) n. Then 0 = (xe)n+m 1 = enxm 1, so xm 1 = 0 since en = 0. Therefore, 6 x Rm implying that Rm is weakly integrally closed in Rm 1. Hence, Rm is weakly 2 integrally closed by induction. For example, the ring K[[t10; t13; t16; t19; t20;:::]] is weakly integrally closed. The next theorem requires the following lemma.
Lemma 59 Let A B be integral domains and let Q be the quotient …eld of A. If Q B = A and B is weakly integrally closed, then so is A. \
Proof. Let Q (A) and Q (B) be the quotient …elds of A and B. We will show that Q (A) is weakly integrally closed in Q (B). Let x be an element of Q (A) and J a nonzero …nitely generated ideal of Q (B) such that xJ J 2. Since Q (B) is a …eld, then J = Q (B), so 1 J. Thus, x = x 1 J 2 = Q (A), so Q (A) is weakly integrally 2 2 closed in Q (B). Therefore, if Q (A) B = A and B is weakly integrally closed in \
35 Q (B), then so is A by Theorem 5. The above lemma is also true for B, an integrally closed domain.
De…nition 60 A monoid M is said to be torsion-free if whenever na = nb for some natural number n > 0 and a; b M, then a = b. 2
Theorem 61 Let D be an integral domain and M a torsion-free and cancellative monoid. If D[M] is weakly integrally closed, then so is D.
Proof. The ring D[M] is an integral domain since M is torsion-free and cancellative [5]. Let Q be the quotient …eld of D. We will show that Q D[M] = D. It is \ clear that D Q D[M]. Let y be an element of Q D[M]. Then y = u0=v0 = \ \ m1 mn d0 + d1t + ::: + dnt , where uo; v0; dk D, v0 = 0, and mk M 0 . Thus, 2 6 2 f g m1 mn v0 (d1t + ::: + dnt ) = 0, so dk = 0 for all 0 < k n since v0 is nonzero. There- fore, y = d0 D, so Q D[M] = D. Hence, if D[M] is weakly integrally closed, 2 \ then so is D. It is not di¢ cult to see that Theorem 61 is also true for the power series ring D[[M]].
36 7 Open Problems
Let K be a …eld and 0 3a < m i i. Let Ma;i;m = 0; i; i+1; : : : ; i+a; m; m+1;::: f g and Ra;i;m = K[[Ma;i;m]]. Then R0;i;m and R1;i;m are weakly integrally closed by
Theorems 55 and 57. Thus, the monoids M0;i;m and M1;i;m are weakly integrally
closed by Theorem 16. We will show that the monoid Ma;i;m is weakly integrally
closed for every a 0. Observe that Ma;i;m has the following characteristic binary string:
Ma;i;m : 100 ::: 01i1 ::: 1i+a0 ::: 01m1111 ::: .
Theorem 62 Let 0 3a < m i i. Then Ma;i;m = 0; i; i + 1; : : : ; i + a; m; m + f 1;::: is a weakly integrally closed monoid. g
Proof. Let M = Ma;i;m. Since m i i, then m 2i, so M is a monoid. Let H0 = x N : i + a < x < m and let H1 = l M : i l 2m i 1 . f 2 g f 2 g Then (H0;H1) is the test block H of M. It su¢ ces to show that H has no bad zeros
by Theorem 47. Let x be an element of H0 and F H1. Then we will show that x + F F + F . 6 Let n = min (F ). Assume that x + F F + F . Then x + n = f1 + f2 for some f1; f2 F . If say f2 m, then x + n = f1 + f2 n + m > n + x since x < m. 2 Thus, f1; f2 < m, so f1; f2 i + a, so x + n = f1 + f2 2 (i + a). Therefore, x 2 (i + a) n 2 (i + a) i = i + 2a, so x i + 2a. Let J = i; i + 1; : : : ; i + a and let L = J F . Since n J, then J is not f g \ 2 empty. Since x > i + a, then x + J J + J by Lemma 33. Let j be an element of 6 J such that x + j J + J. But x + j F + F since x + F F + F and j F . 62 2 2
37 Thus, x + j = f3 + f4 for some f3; f4 F . We may assume that f4 J. Therefore, 2 62 f4 m, so f3 + f4 i + m. But x + j (i + 2a) + (i + a) = 2i + 3a < i + m. Thus, i + m < i + m, a contradiction. Hence, x + F F + F , so M is weakly integrally 6 closed.
Question 1 Let 0 3a < m i i. Is the ring K[[Ma;i;m]] weakly integrally closed?
Question 2 If a monoid M is weakly integrally closed, then must K[M] be weakly integrally closed?
38 References
[1] J. Brewer, Power Series Over Commutative Rings, Lecture Notes in Pure and Applied Mathematics 64, 1981.
[2] J. Brewer and F. Richman, Weakly integrally closed domains: minimum polyno- mials over matrices, Communications in Algebra 28 (2000), 4735 4748.
[3] W. C. Brown, Null ideals and spanning ranks of matrices, Communications in Algebra 26 (1998), 2401 2417.
[4] A. Geroldinger, F. Halter-Koch, and G. Lettl, The complete integral closure of monoids and domains, Rend. Mat. Appl., VII. Ser., 15 : 281 292; 1995.
[5] R. Gilmer, Commutative Semigroup Rings, University of Chicago Press, Chicago, 1984.
[6] I. Kaplansky, Commutative Rings, Allyn and Bacon, Inc., Boston, 1970.
39