ON THE ELASTICITY OF GENERALIZED
ARITHMETICAL CONGRUENCE MONOIDS
S. T. CHAPMAN AND DAVID STEINBERG
Abstract. An arithmetical congruence monoid (or ACM ) is a multiplicative monoid,
which consists of an arithmetic sequence and the element 1. As they are traditionally
defined, it is required that a ≤ b and a2 ≡ a (mod b) must hold to ensure closure. It is
well known that unique factorization need not occur in ACMs. In this paper, we investigate
factorization results when the requirement a ≤ b is dropped. More specifically, if M (a, b)
is an ACM, we offer results concerning the elasticity of generalized ACMs (or GACMs) of
the form Mr (a, b) = {a + kb ∈ M (a, b)|k ≥ r} ∪ {1}. We characterize when a generalized
ACM is half-factorial (i.e. lengths of irreducible factorizations are constant). Moreover,
we offer conditions, which force the elasticity to be infinite and offer a formula in the case
a 6= 1 for when it is finite.
1. Introduction
The idea of unique factorization is important in the study of number theory and algebra. Multiplicative monoids offer a rich area of research, where unique factorization need
not occur. Congruence monoids are multiplicative submonoids of integral domains which
are defined by congruences (see [8, Section 2.11], [10] and [11]). Non-principal orders in
Dedekind domains or the monoid of totally positive integers of an algebraic number field
are simple examples of congruence monoids. If the congruence monoid is regular, then it
is a Krull monoid ([8, Proposition 2.11.6]) and its arithmetic is studied with methods from
additive group theory and combinatorial number theory (see [7]) For non-regular congruence monoids there are only abstract finiteness results for the arithmetical invariants. In
the present paper we focus on (non-regular) arithmetical congruence monoids and establish
precise values for certain arithmetical invariants by number theoretic methods.
The simplest example of a non-regular congruence monoid is the well-known Hilbert
monoid,
1 + 4N0 = {1, 5, 9, 13 . . . },
which does not have unique factorization since 441 = 9 × 49 = 212 , and 9, 21, and 49 are all
irreducible. The Hilbert monoid falls into a broader class of multiplicative submonoids of
N, known as arithemetical congruence monoids, or ACMs. These are monoids of the form
M (a, b) = {n ∈ N|n ≡ a(mod b)} ∪ {1}
2000 Mathematics Subject Classification. 20M14, 20D60, 11B75.
Key words and phrases. congruence monoid, non-unique factorization, elasticity.
The second author received support from the National Science Foundation under grant DMS-0648390.
1
2
S. T. CHAPMAN AND D. STEINBERG
where a2 ≡ a(mod b) must hold to ensure multiplicative closure. Without loss of generality,
we assume henceforth that a ≤ b. We now establish some notation and definitions used
throughout this article.
Suppose M is a commutative cancelative monoid. Then, we denote the non-units of M
as M • and write A(M ) for the irreducibles, or atoms, of M . Since 1 is the only unit in any
ACM, we have
A(M (a, b)) = {m ∈ M (a, b)|m = cd with c, d ∈ M (a, b) implies c = 1 or d = 1}.
If every element of M • can be written as a product of atoms of M , then we say M is atomic.
For any m ∈ M • , suppose that
(*)
m = p1 p2 . . . ps = q1 q2 . . . qt
where each pi and qj is an atom of M . We say the monoid M is factorial if for every
m ∈ M • and factorization of the form (*), we have s = t and there exists a permutation,
σ of {1, 2, . . . , t}, so that pi and qσ(i) are associates for all 1 ≤ i ≤ t. The monoid M is
half-factorial if s = t holds under these conditions, but pi need not have an associate among
the qj . It is worth noting that ACMs are atomic and the only factorial ACM is M (1, 2) = N.
If M is not half-factorial, then it contains some element with factorizations of different
lengths. For m ∈ M • , the set of lengths of m is
LM (m) = {s ∈ N|m = p1 p2 ...ps where pi ∈ A(M )}
and the set of lengths of M is
L(M ) = {L(m) | m ∈ M • }.
When there is no confusion concerning the monoid M , we shorten LM (m) to L(m). If
L(m) = {s1 , s2 , s3 , . . . } with the si ’s listed in increasing order, then the set
4(m) = {si − si−1 |si ∈ L(m)}
and
4(M ) =
[
4(m).
m∈M •
The elasticity of an element m ∈ M • is given by
%(m) =
sup(L(m))
inf(L(m))
and we define the elasticity of M to be %(M ) = sup{%(m)|m ∈ M • } (a good survey article
on the elasticity can be found at [2]). If there is an m ∈ M • such that %(M ) = %(m), then
we say that the elasticity of M is accepted.
Our work is to generalize results from Banister, Chaika, Chapman and Meyerson [4]. In
that article, the authors were able to characterize when an ACM is half-factorial. When
it is not, they gave necessary and sufficient conditions for the elasticity to be infinite,
and offered formulas to find the elasticity when it is finite. Moreover, they found that
min4(M (a, b)) = 1, an important result since it is known that min4(M ) = gcd4(M )
ELASTICITY IN GACMS
3
for any commutative cancellative monoid, M [6, Lemma 3]. We offer similar results for
generalized ACMs, which are multiplicative monoids of the form
Mr (a, b) = {a + kb|k ∈ N0 and k ≥ r} ∪ {1}.
The argument will consider two cases. If a = 1, then M (1, b) is a Krull monoid, and its
arithmetic is closely related to the structure of its divisor class group Cl(M (1, b)) (see [8,
Section 2.5]). If a > 1, then M (a, b) is not Krull, and its arithmetic will be dependent on
gcd(a, b). When gcd(a, b) = pk for p a prime integer, then M (a, b) is called local. Some
results concerning the arithmetic of M (a, b) in the local case can be found in [3], [4] and [5].
2. Elasticity in generalized ACMs and
min4Mr (a, b).
In this section, we explore the elasticity of Mr (a, b) based on gcd(a, b). That a GACM
is atomic follows directly from the atomic property of its associated ACM. We show that if
gcd(a, b) has two distinct prime divisors, then the elasticity of M (a, b) is infinite. Moreover,
we find that if gcd(a, b) is a prime power, then %(Mr (a, b)) is finite and offer a formula for
computing it. This leaves the possibility that a and b are relatively prime, in which case
a2 ≡ a (mod b) forces a ≡ 1 (mod b) (see [4, Lemma 2.1]). Since we may assume a ≤ b this
means a = 1. We start with the case of infinite elasticity.
Proposition 2.1. If Mr (a, b) is a generalized ACM so that gcd(a, b) has at least two distinct
prime divisors, then %(Mr (a, b)) = ∞.
Proof. Let Mr (a, b) be a generalized ACM so that d = gcd(a, b) has at least two distinct
prime divisors. Say b = db1 . Since a2 ≡ a (mod b), it is not hard to show that gcd(d, b1 ) = 1
and
M (a, b) = {n ∈ N|n ≡ 1(mod b1 ) and d divides n} ∪ {1}.
Now, let [x] represent the equivalence class of an integer in Zb1 . Then, by Dirichlet’s
theorem, we may pick a prime, q ≥ a + rb so that q ∈ [d−1 ]. Thus, dq ∈ A(Mr (a, b)). By
[4], any power of dq has a factorization in M (a, b) no longer than some constant, B ∈ N.
Thus, we may write (dq)n+1 = (dq)α1 α2 . . . αr where αi ∈ A(M (a, b)) and r ≤ B.
Now, notice that for any m1 , m2 ∈ M (a, b)\Mr (a, b), we have m1 m2 ∈ (M (a, b)\Mr (a, b))∪
A(Mr (a, b)). Using this fact, for some s ≤ r we may write
α1 α2 . . . αr = α∗ β1 β2 . . . βs
where α∗ ∈ (M (a, b)\Mr (a, b)) ∪ {1} and βj ∈ A(Mr (a, b)). If α∗ = 1 we are done, so
assume otherwise. Then, we have (dq)n+1 = (dqα∗ )β1 . . . βs . But since dq is the smallest
element of M (a, b), which q divides, dqα∗ must be irreducible in Mr (a, b). Since n may be
arbitrarily large, the result follows.
In the case a = 1, the following Lemma will be of great use.
Lemma 2.2. Let b > 2 and r ≥ 1 be positive integers. Then
L(M (1, b)) ⊆ L(Mr (1, b)).
4
S. T. CHAPMAN AND D. STEINBERG
Proof. If x ∈ A(M (1, b)) and x ∈ Mr (1, b), then clearly x ∈ A(Mr (1, b)). Let y ∈ M (1, b)
and suppose that
y = α1 · · · αs = β1 · · · βt
(†)
where each αi and βj is irreducible in M (1, b). Using the Krull structure of M (1, b), write
y = p1 · · · pw
where the pi ’s are the not necessarily distinct prime divisors of y in M (1, b). By (†), there
are two partitions ω1 , . . . , ωs and θ1 , . . . , θt of {1, . . . , w} such that
Y
Y
αi =
pj and βi =
pj .
j∈ωi
j∈θi
Again, using Dirichlet’s theorem, for each 1 ≤ i ≤ w choose a prime divisor p0i such that
[pi ] = [p0i ] and p0i > 1 + rb. Set
Y
Y
p0j and βi0 =
p0j
αi0 =
j∈ωi
j∈θi
in M (1, b). It follows that each αi0 and βi0 is irreducible in M (1, b) and by construction are
each larger than 1 + rb. Hence, each is irreducible in Mr (1, b) and it follows that
z = α10 · · · αs0 = β10 · · · βt0
(‡)
is an irreducible factorization of z in M (1, b) with all irreducible factors in Mr (1, b). This
implies that
LM (1,b) (y) ⊆ LMr (1,b) (z)
By a similar argument, if
z = δ10 · · · δf0 = η10 · · · ηg0
is an irreducible factorization of z in Mr (1, b) with each δi0 and ηi0 in M (1, b), then there
exists irreducibles δ1 , . . . , δf , η1 , . . . , ηg in M (1, b) with
y = δ1 · · · δf = η1 · · · ηg .
Hence, if LM (1,b) (y) ( LMr (1,b) (z), then z must be divisible by an atom α of Mr (1, b) that
is not irreducible in M (1, b). Suppose α = γβ with γ and Q
β in M (1, b). Hence,
Q there
are disjoint subsets κ1 and κ2 of {1, . . . , w} such that γ = j∈κ1 p0j and β = j∈κ2 p0j .
By construction, all the p0i ’s are greater than 1 + rb, and hence γ and β are in Mr (1, b),
contradicting the irreducibility of α in Mr (1, b). This completes the proof.
Corollary 2.3. Let b > 2 and r ≥ 1 be positive integers.
(1) %(M (1, b)) ≤ %(Mr (1, b)).
(2) min ∆(Mr (1, b)) = min ∆(M (1, b)) = 1.
(3) Mr (1, b) is half-factorial if and only if r = 1 and b = 1, 2, 3, 4 or 6.
Proof. (1) follows directly from Lemma 2.2. For (2), since min ∆(M (1, b)) = 1 (see [4,
Proposition 2.2.5]), the assertion follows from Lemma 2.2. (3) (⇐) follows directly from [4,
Proposition 2.2.4]. (⇒)If b 6= 1, 2, 3, 4 or 6, then (1) and [4, Proposition 2.2.4] implies that
Mr (1, b) is not half-factorial for any value of r. Hence, b = 1, 2, 3, 4 or 6. If r = 1, then
ELASTICITY IN GACMS
5
M (1, b) = M1 (1, b) which is half-factorial again by [4, Proposition 2.2.4]. To complete the
argument, suppose b = 1, 2, 4 or 6 and r > 1. In each of these cases, b + 1 = p where p
is prime in Z and p ∈ M (1, b) but p 6∈ Mr (1, b). Thus, there is a smallest positive integer
k > 1 such that pk ∈ Mr (1, b) but pk−1 6∈ Mr (1, b). Clearly pk and pk+1 are both irreducible
in Mr (1, b) and (pk )k+1 = (pk+1 )k shows that Mr (1, b) is not half-factorial. If b = 3, then
1 + b = p2 where p = 2. Repeat the previous argument using (p2 )k and (p2 )k+1 where k is
again minimal so that (p2 )k ∈ Mr (1, b) but (p2 )k−1 6∈ Mr (1, b) (note here that pt 6∈ Mr (1, b)
when t is odd).
In the next two theorems, we consider %(Mr (a, b)) in the case when gcd(a, b) is 1 or a
prime power. For the case a = 1 we offer an upper bound (Theorem 2.4) and in the case
where gcd(a, b) = pk we find a precise formula (Theorem 2.7). First, we must introduce a
powerful tool in studying elasticities. For any atomic monoid, M , a function f : M → R is
a semi-length function if
(1) f (xy) = f (x) + f (y) for all x, y ∈ M and
(2) f (x) = 0 if and only if x is a unit.
Anderson and Anderson offer an important theorem concerning semi-length functions, which
can be used to bound elasticity [1]. For an atomic monoid M , which is not factorial, and a
semi-length function, f , they show that
sup{f (x)|x ∈ A(M ) and x not prime}
.
(††)
%(M ) ≤
inf{f (x)|x ∈ A(M ) and x not prime}
We begin with the case a = 1. If x ∈ A(Mr (1, b)) then write
π(x)
x=
Y
pi
i=1
where the pi are the prime divisors of x in M (1, b). Set
D(Mr (1, b)) = sup{ π(x) | x ∈ A(Mr (1, b))}.
Note that D(Mr (1, b)) ≥ D(Cl(M (1, b)) where D(Cl(M (1, b)) represents Davenport’s constant of the divisor class group of M (1, b) (see [8, Chapter 5]).
Theorem 2.4. Let b > 2 and r ≥ 1 be positive integers.
(1) There exists a positive integer N (dependent on r and b) such that π(x) ≤ N for all
atoms x of Mr (1, b). Hence, D(Mr (1, b)) is finite.
(2) %(Mr (1, b)) ≤ D(Mr2(1,b)) .
Proof. (1) Let p1 be the smallest prime integer such that gcd(p1 , b) = 1. Hence, if α
is an atom of M (1, b), then α ≥ p1 . Let k be the smallest positive integer such that
pk1 ∈ Mr (1, b). Suppose that x ∈ A(Mr (1, b)) and π(x) > 2kb. Factor x = β1 · · · βt into
atoms of M (1, b). Since π(βi ) ≤ b, it follows that t ≥ 2k. Moreover, y 0 = β1 · · · βk ≥ pk1
and hence y 0 ∈ Mr (1, b). Similarly, y 00 = βk+1 · · · βt ≥ pk1 and y 00 ∈ Mr (1, b). Thus x = y 0 y 00
reduces in Mr (1, b), a contradiction. This completes the proof of (1).
(2) Since π(x) is a semilength function, the inequality follows from (††).
6
S. T. CHAPMAN AND D. STEINBERG
The inequality in (2) above may be strict. We will demonstrate this by example but
will first require a lemma. Notice that we may partition the set A(Mr (1, b)) by writing
A(Mr (1, b)) = OA(Mr (1, b)) ∪ NA(Mr (1, b)) where
OA(Mr (1, b)) = {m ∈ A(Mr (1, b))|m ∈ A(M (1, b))}
and
NA(Mr (1, b)) = {m ∈ A(Mr (1, b))|m ∈
/ A(M (1, b))}.
We shall call these sets the old atoms of Mr (1, b) and the new atoms of Mr (1, b), respectively. Since OA(Mr (1, b)) can be determined from A(B(Z×
b )) we will focus our attention
to NA(Mr (1, b)), starting with the following lemma.
Lemma 2.5. If m ∈ NA(Mr (1, b)) then one of the following holds:
(i) m = n1 n2 n3 where n1 , n2 , n3 ∈ (M (1, b) − Mr (1, b)) ∪ {1}
(ii) m = n1 n2 where n1 ∈ M (1, b) − Mr (1, b) and n2 ∈ OA(Mr (1, b)).
Proof. We consider two cases: either m has a factor, which is an old atom, or not. Suppose
m has no old atoms as factors. Let m = a1 a2 . . . an be any factorization of m in M (1, b).
Since none of the ai are atoms of Mr (1, b) we know ai ∈ M (1, b) − Mr (1, b) for 1 ≤ i ≤ n.
If a1 a2 ∈ Mr (1, b) then in order for m to be an atom we must then have a3 a4 . . . an =
α ∈ M (1, b) − Mr (1, b). Thus we may write m = a1 a2 α as a product of three elements
from M (1, b) − Mr (1, b). If a1 a2 ∈
/ Mr (1, b) then a1 a2 = β ∈ M (1, b) − Mr (1, b) so that we
may write m = βa3 . . . an as a product of n − 1 element from M (1, b) − Mr (1, b). We may
continue this process until we have written m as a product of three or fewer elements of
M (1, b) − Mr (1, b), satisfying statement (i) above.
Now consider the second case, where m has an old atom as a factor. Then, for some
n1 ∈ OA(Mr (1, b)) and n2 ∈ M (1, b) we may write m = n1 n2 . Since m is an atom, it must
be that n2 ∈ M (1, b) − Mr (1, b), satisfying statement (ii) above.
Example 2.6. We show that ρ(M3 (1, 4)) < D(M32(1,4)) . By the previous lemma, if d =
max{π(x)|x ∈ M (1, b) − Mr (1, b)}, then D(Mr (1, b)) ≤ max{3d, d + D(Cl(M (1, b)))}. In
this case, this implies that D(M3 (1, 4)) ≤ 6. Indeed, it is not hard to see that 93 = 729 ∈
A(M3 (1, 4)) and π(729) = 6 so that D(M3 (1, 4)) = 6. Thus, ρ(M3 (1, 4)) ≤ 26 = 3. However,
M3 (1, 4) cannot have an accepted elasticity of 3. Since 729 is the unique element of length
6, in order for M3 (1, 4) to have an accepted elasticty of 6, some power of 729 would have
to factor into irreducibles all of length 2, a contradiction.
Now, suppose M3 (1, 4) has a non-accepted elasticity of 3. Let > 0 be given. Then,
there exists an x ∈ M3 (1, 4) so that
x = p1 p2 . . . pn = α1 α2 . . . αs = β1 β2 . . . βt
where each αi , βj ∈ A(M3 (1, 4)), each pk is a prime integer and 3 − < st < 3. Thus,
n
n
s > t(3 − ) implies n/s
> n/t
(3 − ). Since ns ≥ 2, we have nt > ns (3 − ) ≥ 2(3 − ) = 6 − 2.
If we let m be the number of βj so that π(βj ) = 6, then we have that
Thus,
n
(1)
m ≥ ( − 5)t ≥ (6 − 2 − 5)t = (1 − 2)t.
t
n
t
≤
6m+5(t−m)
t
=
m
t +5.
ELASTICITY IN GACMS
Similarly, notice that 3 − <
so that π(αi ) = 2 then we have
(2)
c ≥ (3 −
7
n/t
s
n
6
t implies s < 3− ≤ 3− . If
2c+3(s−c)
n
= 3 − sc . Thus,
s ≥
s
we let c be the number of αi
6
2
n
)s ≥ (3 −
)s = (1 −
)s ≥ (1 − )s.
s
3−
3−
Without loss of generality, assume that π(βj ) = 6 for 1 ≤ j ≤ m. Since 729 is the unique
element of length 6, this implies that βj = 729 for 1 ≤ j ≤ m. Now, by (1) we have
(3)
3
π(β1 β2 . . . βm ) = π(729m ) = 6m ≥ 6(1 − 2)t ≥ n(1 − 2) ≥ n.
4
Similarly, assume that π(αi ) = 2 for 1 ≤ i ≤ c. Since
(4)
n
s
< 3, by (2) we have
n
n
n
π(α1 α2 . . . αc ) = 2c ≥ 2(1 − )s = 2(1 − )
≥ 2(1 − ) ≥ .
n/s
3
2
By the Pigeon Hole Principle, (3) and (4) imply that for some u ≥ n4 , v ≥ 1 and w ≥ 1,
there exist permutations, σ of {1, 2, . . . , n} and φ of {1, 2, . . . c}, so that
pσ(1) pσ(2) . . . pσ(u) = αφ(1) αφ(2) . . . αφ(v) = 729w
Since all of the αφ(i) are of length 2, this implies that a power of 729 may be factored into
irreducibles all of length 2, a contradiction. This completes the argument.
We turn now to the case where gcd(a, b) = pk for p a prime integer.
Theorem 2.7. Let Mr (a, b) be a generalized ACM and suppose gcd(a, b) = pk for p a prime
and k a natural number.
(1) If f : M (a, b) → R sends an element to its p-adic value, and j0 = max{f (x)|x ∈
M (a, b)\Mr (a, b) and pn x ∈ A(Mr (a, b))} then
n+j
k
where n is the smallest positive integer such that pn ∈ Mr (a, b) and j = max{k −
1, j0 }.
(2) The arithmetic function g(a,b) (r) = %(Mr (a, b)) is monotonically increasing.
(3) Mr (a, b) is half-factorial if and only if r = 0 and a = p.
%(Mr (a, b)) =
Proof. Suppose the generalized ACM Mr (a, b) has gcd(a, b) = pk for p a prime and k a
natural number. Hence, we may write b = pk b1 and we leave it to the reader to confirm
gcd(p, b1 ) = 1. Then, since a2 ≡ a(mod b) we have a2 ≡ a(mod b1 ). It follows that a ≡
1(mod b1 ). This implies that m ∈ N is an element of M (a, b) if and only if m ≡ 1(mod b1 )
and pk divides m.
Now, suppose n is the smallest positive integer so that pn ∈ Mr (a, b). Such an n must
exist since (pφ(b1 ) )c ∈ Mr (a, b) for some c. Define the function f : M (a, b) → R0 to send an
element to its p-adic value. Let
j0 = max{f (x)|x ∈ M (a, b)\Mr (a, b) and pn x ∈ A(Mr (a, b))}
8
S. T. CHAPMAN AND D. STEINBERG
and define j = max{k − 1, j0 }. We define j as such since j0 may be the maximum over the
empty set. There are two cases. If j = k − 1 then the argument from [4, Theorem 2.4 (1)]
holds and we have
n+k−1
%(Mr (a, b)) =
.
k
Otherwise, we have j = j0 . Now, suppose m ∈ Mr (a, b) so that m has p-adic value at
least n + j + 1. Then, for some integer z we may write m = pn+j+1 z = (pn )(pj+1 z). We
know that either pj+1 z ∈ Mr (a, b) or pj+1 z ∈ M (a, b)\Mr (a, b). If pj+1 z is an element of
Mr (a, b), then m is not an atom. Otherwise, pj+1 z ∈ M (a, b)\Mr (a, b) and by our choice of
j, m is still not an atom. Thus, n + j is the largest p-adic value of atoms in Mr (a, b).
Now let [x] represent the equivalence class of x in Zb1 . Then, by Dirichlet’s theorem, we
may pick a prime, q ∈ [p−k ] such that q ≥ a + br. Thus, pk q ∈ Mr (a, b) and, since pk divides
all elements of Mr (a, b), k is the smallest p-adic value for atoms of Mr (a, b). Since f is a
semi-length function, it follows from (††) that %(Mr (a, b)) ≤ (n + j)/k.
Assuming j = j0 , there exists some m0 ∈ M (a, b)\Mr (a, b) so that m0 = pj z0 for some
z0 ∈ Z with gcd(z0 , p) = 1. Moreover, pn+j z0 ∈ A(Mr (a, b)). Invoking Dirichlet’s theorem
again, we may pick a sufficiently large prime, s ∈ [pj−k ], so that pk z0 s ∈ A(Mr (a, b)). Let
tc = (pk z0 s)c(n+j)+1 . Notice that since tc has p-adic value ck(n + j) + k, it can be written
as the product of at most c(n + j) + 1 atoms, and it does reach this upper bound.
Now, notice that
c(n+j−k)+1 c(n+j)+1
tc = (pn+j z0 )ck (pk z0
s
).
Thus, tc can be written as the product of ck + 1 atoms. Furthermore, since max{f (x)|x ∈
A(Mr (a, b)} = n + j, the product of ck or fewer atoms has p-adic value at most
ck(n + j) < ck(n + j) + 1.
Thus, tc cannot be written as a product of fewer than ck + 1 atoms. Hence, %(tc ) = (c(n +
j) + 1)/(ck + 1). Since limc→∞ %(tc ) = (n + j)/k, we conclude that %(Mr (a, b)) = (n + j)/k.
For (2), pick r0 ∈ N0 and suppose pn0 is the smallest power of p in Mr0 (a, b). Either
a + r0 b = pn0 or not. If not, then for any m ∈ M (a, b)\Mr0 (a, b), pn0 m ∈ A(Mr0 ) implies
pn0 m ∈ A(Mr0 +1 ). Thus,
max{f (x)|x ∈ M \Mr0 and pn0 x ∈ A(Mr0 )} ≤
max{f (x)|x ∈ M \Mr0 +1 and pn0 x ∈ A(Mr0 +1 )}
implying g(a,b) (r0 ) ≤ g(a,b) (r0 + 1). If a + r0 b = pn0 , then n0 must be the largest p-adic value
of elements in M (a, b)\Mr0 +1 (a, b). Notice, if w is the order of p in Zb1 then pn0 +w is the
smallest power of p in Mr0 +1 (a, b). Moreover, (pn0 +w )(pn0 ) ∈ A(Mr0 +1 (a, b)) since every
element of Mr0 +1 (a, b) smaller than pn0 +w has at least two distinct prime divisors. Thus,
n0 = max{f (x)|x ∈ M \Mr0 +1 and pn0 +w x ∈ A(Mr0 +1 )}
and
n0 > max{f (x)|x ∈ M \Mr0 and pn0 x ∈ A(Mr0 )},
yielding g(a,b) (r0 ) < g(a,b) (r0 + 1). Thus, g is monotonically increasing.
ELASTICITY IN GACMS
9
For (3), we have %(M0 (a, b)) = 1 if and only if a is prime and a divides b from [4]. Now,
suppose r = 1 and a is prime that divides b. Then, a is the only element in M (a, b)\M1 (a, b)
and (a2 )(a) ∈ A(M1 (a, b)). Thus, %(M1 (a, b)) = 3 by (1). The result then follows from
(2).
We conclude by showing that min4Mr (a, b) = 1 when Mr (a, b) is not half-factorial and
a 6= 1.
Theorem 2.8. If a ≥ 1 and Mr (a, b) is a generalized ACM, which is not half-factorial,
then min 4Mr (a, b) = 1.
Proof. The case a = 1 follows from Lemma 2.2. So suppose a > 1. For r = 0, the result
follows from [4, Theorem 2.7]. If r ≥ 1, then a ∈
/ Mr (a, b). Let n ≥ 2 be the smallest natural
number so that an ∈ Mr (a, b). Since an = (a)(an−1 ), any factorization of an must include an
element no larger than an−1 . Thus, an ∈ A(Mr (a, b)). Similarly, since an+1 = (a2 )(an−1 ),
2
an+1 ∈ A(Mr (a, b)) for n ≥ 3. In this case, we have an +n = (an )n+1 = (an+1 )n giving the
result. If n = 2, then a3 either is or is not an atom. If so, then a6 = (a2 )3 = (a3 )2 yielding
the result. Otherwise, a3 is the product of exactly two atoms, say a3 = m1 m2 for some
m1 , m2 ∈ A(Mr (a, b)). Then, a6 = (a2 )3 = (m1 )2 (m2 )2 and the result still holds.
References
[1] D. D. Anderson and D. F. Anderson, Elasticity of factorization in integral domains, J. Pure Appl.
Algebra 80 (1992), 217-235.
[2] D.F. Anderson, Elasticity of factorizations in integral domains: a survey, Factorization in Integral
Domains, Lecture Notes in Pure and Applied Mathematics, Marcel Dekker 189(1997), 1–30.
[3] P. Baginski, S. T. Chapman and G. Schaeffer, On the delta-set of a singular arithmetical congruence
monoid,” J. Théor. Nombres Bordeaux 20(2008), 45–59.
[4] M. Banister, J. Chaika, S. T. Chapman and W. Meyerson, On the arithmetic of arithmetical congruence
monoids, Colloq. Math. 108(2007), 105–118.
[5] M. Banister, J. Chaika, S. T. Chapman and W. Meyerson, A theorem on accepted elasticity in certain
local arithmetical congruence monoids, to appear in Abh. Math. Sem. Univ. Hamburg.
[6] A. Geroldinger, On the arithmetic of certain no integrally closed noetherian integral domains, Comm.
Algebra 19(1991), 685698.
[7] A. Geroldinger, Additive group theory and non-unique factorizations, Combinatorial Number Theory
and Additive Group Theory, Advanced Courses in Mathematics CRM Barcelona, Birkhäuser, 2009,
1–86.
[8] A. Geroldinger and F. Halter-Koch, Congruence Monoids, ACTA Arith. 112(2004), 263-296.
[9] A. Geroldinger and F. Halter-Koch, Non-unique Factorizations: Algebraic, Combinatorial and Analytic
Theory, Pure and Applied Mathematics, vol. 278, Chapman & Hall/CRC, 2006.
[10] F. Halter-Koch, Arithmetical semigroups defined by congruences, Semigroup Forum 42(1991), 59–62.
[11] R. D. James and I. Niven, Unique factorization in multiplicative systems, Proc. Amer. Math. Soc.
5(1954), 834-838.
10
S. T. CHAPMAN AND D. STEINBERG
Sam Houston State University, Department of Mathematics and Statistics, Box 2206, Huntsville,
TX 77341-2206
E-mail address: [email protected]
Trinity University, Department of Mathematics, One Trinity Place, San Antonio, TX.
78212-7200
Current address: 4828 NE 19th, Portland, OR 97211
E-mail address: [email protected]