Arithmetic of Semigroup Semirings
We define semigroup semirings by analogy with group rings and semigroup rings. We study the arithmetic properties and determine sufficient conditions under which a semigroup semiring is atomic, has finite factorization, or has bounded factorization. We also present a semigroup-semiring analog (altho...
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irk-123456789-1654652020-02-14T01:28:08Z Arithmetic of Semigroup Semirings Ponomarenko, V. Статті We define semigroup semirings by analogy with group rings and semigroup rings. We study the arithmetic properties and determine sufficient conditions under which a semigroup semiring is atomic, has finite factorization, or has bounded factorization. We also present a semigroup-semiring analog (although not a generalization) of the Gauss lemma on primitive polynomials. Напiвгруповi напiвкiльця визначаються по аналогiї з груповими кiльцями та напiвгруповими кiльцями. Вивчено арифметичнi властивостi та отримано достатнi умови, за яких напiвгрупове напiвкiльце є атомним, має скiнченну факторизацiю або має обмежену факторизацiю. Також наведено напiвгрупово-напiвкiльцевий аналог (хоча i не узагальнення) гауссiвської леми про примiтивнi полiноми. 2015 Article Arithmetic of Semigroup Semirings / V. Ponomarenko // Український математичний журнал. — 2015. — Т. 67, № 2. — С. 213–229. — Бібліогр.: 15 назв. — рос. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/165465 512.5 ru Український математичний журнал Інститут математики НАН України |
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We define semigroup semirings by analogy with group rings and semigroup rings. We study the arithmetic properties and determine sufficient conditions under which a semigroup semiring is atomic, has finite factorization, or has bounded factorization. We also present a semigroup-semiring analog (although not a generalization) of the Gauss lemma on primitive polynomials. |
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Ponomarenko, V. |
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Arithmetic of Semigroup Semirings |
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Arithmetic of Semigroup Semirings |
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Arithmetic of Semigroup Semirings |
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Arithmetic of Semigroup Semirings |
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Arithmetic of Semigroup Semirings |
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arithmetic of semigroup semirings |
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Інститут математики НАН України |
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2015 |
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Arithmetic of Semigroup Semirings / V. Ponomarenko // Український математичний журнал. — 2015. — Т. 67, № 2. — С. 213–229. — Бібліогр.: 15 назв. — рос. |
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Український математичний журнал |
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AT ponomarenkov arithmeticofsemigroupsemirings |
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2025-07-14T18:37:35Z |
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UDC 512.5
V. Ponomarenko (San Diego State Univ., USA)
ARITHMETIC OF SEMIGROUP SEMIRINGS
АРИФМЕТИКА НАПIВГРУПОВИХ НАПIВКIЛЕЦЬ
We define semigroup semirings by analogy with group rings and semigroup rings. We study the arithmetic properties
and determine sufficient conditions under which a semigroup semiring is atomic, has finite factorization, or has bounded
factorization. We also present a semigroup-semiring analog (although not a generalization) of the Gauss lemma on primitive
polynomials.
Напiвгруповi напiвкiльця визначаються по аналогiї з груповими кiльцями та напiвгруповими кiльцями. Вивчено
арифметичнi властивостi та отримано достатнi умови, за яких напiвгрупове напiвкiльце є атомним, має скiнченну
факторизацiю або має обмежену факторизацiю. Також наведено напiвгрупово-напiвкiльцевий аналог (хоча i не
узагальнення) гауссiвської леми про примiтивнi полiноми.
1. Introduction. The study of group rings is very popular and active, see, e.g., [7]. Similarly, the
study of semigroup rings has a rich history, see, e.g., [8]. Also, the study of semirings is very active
not only in mathematics but in computer science and control theory, see, e.g., [10]. In this context, it
seems natural to study semigroup semirings; however relatively little work (see, e.g., [4, 15]) has been
done in this area. We propose to study the arithmetic properties of semigroup semirings, specifically
factorization into irreducibles and primes. As tools we will use tools from semirings as well as from
semigroups.
This work is organized as follows. Section 1 collects basic facts and tools about semirings and
factorization theory. Section 2 introduces semigroup semirings and develops several tools to find
atoms. Section 3 introduces content sets and maximal common divisors, which generalize the notion
of greatest common divisors. These tools are sufficient to determine necessary conditions for a
semigroup semiring to be atomic, finite factorization, and bounded factorization. We also prove a
semiring analog of the Gauss lemma on primitive polynomials.
Definition 1.1. We call R = (R,+,×) a semiring if it satisfies the following:
(1) (R,+) is a commutative monoid with identity 0;
(2) (R,×) is a commutative monoid with identity 1. We abbreviate × via juxtaposition;
(3) for all a ∈ R, a0 = 0;
(4) for all a, b, c ∈ R, a(b+ c) = (ab) + (ac).
Following [10], we call R an information algebra if it also satisfies the following:
(5) for all a, b ∈ R, if a+ b = 0 then a = b = 0;
(6) for all a, b ∈ R, if ab = 0 then a = 0 or b = 0.
For convenience, set R∗ = R \ {0}. Properties (5), (6) are equivalent to R∗ being closed under
+,×.
Lemma 1.1. Let R be an information algebra, and let a1, a2, . . . , ak, b1, b2, . . . , bk ∈ R. Then∑k
i=1
aibi is nonzero if and only if there is some i ∈ [1, k] with ai and bi each nonzero.
c© V. PONOMARENKO, 2015
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2 213
214 V. PONOMARENKO
Proof. If the sum is zero, then each summand is zero by property (5), and hence for each i either
ai = 0 or bi = 0 by property (6). The other direction is trivial.
Let (M, ·) be a commutative monoid with identity e. We say that binary relation R on M respects
· if for all a, b, c ∈M, aRb implies that (a · c)R(b · c). We define the relation | on M via a | b if there
is some c ∈ M with a · c = b. We say that a is a unit if a | e and a nonunit otherwise. It is easy to
see that the product of units is a unit and the product of nonunits is a nonunit. We say that a and b
are associates if a | b and b | a.
We now recall terms from the arithmetical theory of semigroups. For an introduction, see [6].
We say that nonunit a is an atom if a = b · c implies that either b or c is a unit. We say that (M, ·)
is atomic if every nonunit may be factored into atoms in at least one way. We say that nonunit a is
prime if a | b · c implies that a | b or a | c. We say that (M, ·) is reduced if e is the only unit.
For a in atomic (M, ·),we say that a1·a2 . . . ak is a factorization of a if the product is a and each ai
is an atom. Two factorizations a1·a2 . . . ak and a′1·a′2 . . . a′k are equivalent if there is some permutation
π ∈ Sk, and ai, a′π(i) are associates for 1 ≤ i ≤ k. This is an equivalence relation on the set of all
factorizations of an element. For nonunit a, we call F (a) a factorization set of a if every equivalence
class of factorizations of a has exactly one representative in F (a).We define the length set L(a) = {k :
a1 · a2 . . . ak ∈ F (a)}. Note that |F (a)| and L(a) are independent of the choice of equivalence class
representatives for F (a). We say that a has unique factorization (UF) if |F (a)| = 1; we say that a has
finite factorization (FF) if |F (a)| <∞; we say that a has bounded factorization (BF) if |L(a)| <∞.
We say that (M, ·) is UF, FF, BF if each nonunit in M is. We define the elasticity ρ of a ∈ M via
ρ(a) =
supL(a)
minL(a)
. We define the elasticity ρ(M) = supa∈M ρ(a), and say M is fully elastic if every
x ∈ Q ∩ [1, ρ(M)) is the elasticity of some element in M.
We say that (M, ·) is yoked if for all a, b ∈ M there is some c ∈ M where either a · c = b or
a = b · c. We say that (M, ·) is cancellative if a · c = b · c implies that a = b. Following [3], we say
that (M, ·) is weakly cancellative if a · b = a implies that b = e. This property has also been called
plain, in [10].
Lemma 1.2. (M, ·) is weakly cancellative if and only if the following holds:
∀a, b, c ∈M, with unit a. If a · c = b · c, then a = b.
Proof. Let d ∈ M with d · a = e. Then c = d · a · c = d · b · c. Applying weak cancellativity,
d · b = e and hence b = a · d · b = a · e = a. The other direction follows by taking a = e.
Our primary focus is on information algebras R. We study the multiplicative monoid (R∗,×),
and say that R is atomic, cancellative, reduced, etc. if (R∗,×) is. We say that a ∈ R∗ is a unit, atom,
prime, etc. if it is in (R∗,×).
Example 1.1. The set of nonnegative integers N0 is a UF, cancellative, reduced, information
algebra under the usual +,×.
Example 1.2. For B = {0, 1}, we set 0 + 0 = 0 × 0 = 0 × 1 = 1 × 0 = 0, 0 + 1 = 1 + 0 =
= 1 + 1 = 1 × 1 = 1, and 1 > 0. This is an UF, cancellative, reduced, yoked information algebra
known as the Boolean semiring.
Example 1.3. Let (R,+,×) be the set of fractional ideals of a Dedekind domain D. + has
identity (0), and × has identity D. (R,+,×) is an atomic information algebra.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
ARITHMETIC OF SEMIGROUP SEMIRINGS 215
Example 1.4. Let R be a partially-ordered semiring (or ring). Set R+ = {r ∈ R : r ≥ 0}, the
positive cone of R. R+ is an information algebra.
Example 1.5. Let R be a semiring that is not a ring. Then {1+r : r ∈ R}∪{0} is an information
algebra.
Proposition 1.1. Let (M, ·) be a commutative monoid. Then the following hold:
1. The relation | is a preorder, with u | a for all units u and all a ∈M.
2. If a = u · b for some unit u, then a, b are associates. If (M, ·) is also weakly cancellative,
then the opposite implication also holds.
3. Let a ∈M. If a is an atom, then for all b | a, either b is a unit or an associate of a. If (M, ·)
is also weakly cancellative, then the opposite implication also holds.
4. Let a, b ∈M be associates. If a is an atom, then b is an atom.
5. If (M, ·) is weakly cancellative, then every prime is an atom.
6. If (M, ·) is UF, then every atom is prime.
7. If (M, ·) is reduced and weakly cancellative, then | is a partial order. If (M, ·) is yoked, then
| is a weak order. If (M, ·) is reduced, weakly cancellative, and yoked, then | is a total order.
8. The relation | respects · .
9. If there exists a partial order ≤ on M with least element e that respects ·, then (M, ·) is
reduced. Further, ≤ is a refinement of | . That is, if a | b then a ≤ b.
Proof. (1) Straightforward.
(2) Suppose that a = b · u for some unit u. Hence b | a. Let v be such that u · v = e. We have
a · v = b · u · v = b, and thus a | b so a, b are associates. Suppose now that (M, ·) is weakly
cancellative and a, b are associates. Then there are d, d′ with a = b · d, b = a · d′. Combining, we
have a = b · d = a · (d′ · d), and hence e = d′ · d. Thus d is a unit and a = b · d.
(3) Suppose that a is an atom. Let b | a for nonunit b that is a nonassociate of a. Then a = b · c
for some c ∈ M. Since a is an atom and b is a nonunit, then c must be a unit. By (2), a, b are
associates, contrary to hypothesis.
Now let a be a nonatom that satisfies the condition. Write a = b · c, a product of two nonunits.
By hypothesis, b must be a unit or associate of a, hence an associate of a. By (2), b = a · u for some
u ∈ M. Hence a = b · c = a · (u · c), and by weak cancellativity u · c = e. Hence c is a unit, in
violation of hypothesis.
(4) Suppose otherwise; we write b = u · v for some nonunits u, v. Since a, b are associates there
is some c ∈M with a = b · c = u · v · c. Hence a is a product of nonunits u, (v · c), a contradiction.
(5) Let a ∈ M be prime. Suppose a = b · c for nonunits b, c. Since a is prime, we assume
without loss that a | b. Hence a, b are associates and by (2) there is some unit u with b = a · u. Now
a = b · c = a · u · c, and by weak cancellativity u · c = e and hence c is a unit. This contradiction
shows that a is an atom.
(6) Let a ∈ M be an atom, and suppose a | b · c. There is some d ∈ M with a · d = b · c. We
take factorizations d = (d1 · d2 . . .), b = (b1 · b2 . . .), c = (c1 · c2 . . .). We have two factorizations
a · (d1 · d2 . . .) = (b1 · b2 . . .) · (c1 · c2 . . .); by UF these must be equivalent. Without loss we may
assume that a, b1 are associates; but then a | b. Hence a is prime.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
216 V. PONOMARENKO
(7) Suppose that a | b and b | a. Then by (2), a = u · b for some unit u, and u = e since (M, ·)
is reduced. Therefore a = b and | is antisymmetric. Suppose that (M, ·) is yoked. Then for all
a, b ∈M, either a | b or b | a and hence | is a total preorder (aka a weak order). A total partial order
is a total order.
(8) Straightforward.
(9) First, note that if u is a unit different from e we have u · v = e and e < u = u · e < u · v = e,
a contradiction. Hence (M, ·) is reduced. Suppose now that a | b. Then there is some c ∈ M with
b = a · c. We now have a = a · e ≤ a · c = b.
For information algebras R, we will use |+ to denote the preorder | on the reduced monoid (R,+).
Proposition 1.2. Suppose that (M, ·) has UF. Then (M, ·) is weakly cancellative if and only if
it is cancellative.
Proof. One direction is trivial. For the other, suppose that a, b are nonunits and a · c = b · c. If
c is a unit we multiply by c−1 and get a = b; otherwise we take factorizations a = (a1 · a2 . . .),
b = (b1 · b2 . . .), c = (c1 · c2 . . .). By UF the factorization (a1 · a2 . . .) · (c1 · c2 . . .) is equivalent
to (b1 · b2 . . .) · (c1 · c2 . . .). Hence we may assume without loss that ai, bi are associates for all i,
and thus a, b are associates. By Proposition 1.1(2), there is some unit u with a = u · b. But then
u · b · c = b · c, and by weak cancellativity u = e and hence a = e · b = b.
Definition 1.2. Given semirings R1, R2, we say φ : R1 → R2 is a morphism if it satisfies the
following, for all a, b ∈ R1 :
(1) φ(a+ b) = φ(a) + φ(b),
(2) φ(ab) = φ(a)φ(b),
(3) φ(0) = 0, φ(1) = 1.
For morphism φ and property P, we say that P carries forward if P (a) implies P (φ(a)). We say
that P carries backward if P (φ(a)) implies P (a). We say that P is preserved if it carries forward
and backward. We make similar statements for semiring properties.
Proposition 1.3. Let φ : R1 → R2 be a morphism. Then units and associates carry forward.
If units are preserved, then atoms carry backward. If φ is injective, then cancellation and weak
cancellation carries backward. If φ is surjective, then yoked carries forward.
Proof. (Units) Let a ∈ R1 be a unit. Choose b ∈ R1 with ab = 1. Then φ(a)φ(b) = φ(1) = 1,
so φ(a) is a unit.
(Associates) Let a, b ∈ R1 be associates. There must be c, d ∈ R1 with a = bc, b = ad. We have
φ(a) = φ(b)φ(c) and φ(b) = φ(a)φ(d).
(Atoms) Suppose that φ(a) is an atom. a cannot be a unit since units carry forward. Suppose
that a is not an atom. Then we write a = a1a2 for nonunits a1, a2 ∈ R1. Then φ(a) = φ(a1)φ(a2),
a product of nonunits since units carry backward. Hence φ(a) is a nonatom, contrary to hypothesis.
(Cancellation) Suppose that ac = bc. Then φ(a)φ(c) = φ(b)φ(c), and by cancellation on R2,
φ(a) = φ(b). Since φ is injective, a = b.
(Weak cancellation) Suppose that ab = a. Then φ(a)φ(b) = φ(a), and by weak cancellation on
R2, φ(b) = 1. Since φ is injective, b = 1.
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
ARITHMETIC OF SEMIGROUP SEMIRINGS 217
(Yoked) Let a, b ∈ R2. Since φ is surjective, there are a′, b′ ∈ R1 with φ(a′) = a, φ(b′) = b.
Since R1 is yoked, without loss we let c ∈ R1 with a′c = b′. Then φ(a′)φ(c) = φ(b′) and aφ(c) = b.
Hence R2 is yoked.
Proposition 1.4. LetR1, R2 be semirings. Let φ : R1 → R2 be a bijective morphism. Then φ−1 :
R2 → R1 is a (bijective) morphism. Also, φ preserves units, associates, atoms, primes, and UF.
Proof. Let a, b ∈ R2. φ(φ
−1(a) + φ−1(b)) = φ(φ−1(a)) + φ(φ−1(b)) = a + b. Hence
φ−1(a + b) = φ−1(φ(φ−1(a) + φ−1(b))) = φ−1(a) + φ−1(b). Similarly, φ(φ−1(a)φ−1(b)) =
= φ(φ−1(a))φ(φ−1(b)) = ab. Hence φ−1(ab) = φ−1(φ(φ−1(a)φ−1(b))) = φ−1(a)φ−1(b). Hence
φ−1 is a morphism.
φ preserves units, associates, and atoms by applying Proposition 1.3 to φ and φ−1.
(Primes) Suppose that a is a prime but φ(a) is a nonprime. Hence there are b, c, d ∈ R2 with
φ(a)d = bc, but φ(a) - b and φ(a) - c. Because φ is surjective, there are b′, c′, d′ ∈ R1 with φ(b′) = b,
φ(c′) = c, φ(d′) = d. φ(ad′) = φ(a)φ(d′) = φ(a)d = bc = φ(b′)φ(c′) = φ(b′c′), hence since φ is
injective ad′ = b′c′. Since a is prime, without loss a | b′, i.e., there is some h ∈ R1 with ah = b′.
Now φ(a)φ(h) = φ(b′) = b, so φ(a) | b, a contradiction. Hence φ carries primes forward; the other
direction follows from φ−1.
(UF) Suppose R2 is UF. Let a ∈ R1 have two inequivalent factorizations a1a2 . . . ak, a′1a
′
2 . . . a
′
j .
But then φ(a1)φ(a2) . . . φ(ak), φ(a′1)φ(a
′
2) . . . φ(a
′
j) are inequivalent factorizations of φ(a), a con-
tradiction. Hence φ carries UF backward; the other direction follows from φ−1.
Note that for any morphism φ, if a | b then φ(a) | φ(b). Recalling an important tool from
factorization theory, we say that φ is a divisor morphism if the converse also holds.
Definition 1.3. We say that morphism φ : R1 → R2 is a divisor morphism if it satisfies the
following: For all a, b ∈ R1, if φ(a) | φ(b) (in R2), then a | b (in R1).
Proposition 1.5. Let φ : R1 → R2 be a divisor morphism. Then units and associates are
preserved, while atoms and primes carry backward. If φ is surjective, then atoms and primes are
preserved.
Proof. It is straightforward to prove that units, associates, and primes carry forward.
Let a ∈ R1 and suppose that φ(a) is a nonatom. Write φ(a) = bc, and by surjectivity there
are b′, c′ ∈ R1 with φ(b′) = b, φ(c′) = c. Hence φ(a) = φ(b′c′) and φ(a) | φ(b′c′), φ(b′c′) | φ(a).
Therefore a, b′c′ are associates. b′, c′ cannot be units since units carry forward; hence a is a nonatom.
Let a ∈ R1 be prime. Suppose that φ(a) | bc; by surjectivity there are b′, c′ ∈ R1 with φ(b′) = b,
φ(c′) = c. Because φ(a) | bc = φ(b′c′), a | b′c′; since a is prime we may assume without loss that
a | b′. But then φ(a) | φ(b′) = b.
Proposition 1.6. Let R1, R2 be atomic semirings, and let φ : R1 → R2 be a divisor morphism.
For all a ∈ R1, supL(a) ≤ supL(φ(a)). Further, BF and FF carry backward.
Proof. Let a ∈ R1, k ∈ L(a). Choose a1a2 . . . ak ∈ F (a). φ(a) = φ(a1)φ(a2) . . . φ(ak).
L(φ(a)) ⊇ L(φ(a1)) + L(φ(a2)) + . . . + L(φ(ak)) ≥ 1 + 1 + . . . + 1 ≥ k. Hence supL(a) ≤
≤ supL(φ(a)), and if the right-hand side is finite (i.e., a is BF) then so is the left-hand side.
Let a ∈ R1, a1a2 . . . ak ∈ F (a). Since φ preserves units, φ(a) = φ(a1)φ(a2) . . . φ(ak), a product
of nonunits. Suppose now that F (φ(a)) is finite, but F (a) is infinite. Since each element of F (φ(a))
has only finitely many partitions as above, there must be at least two (in fact infinitely many) partitions
ISSN 1027-3190. Укр. мат. журн., 2015, т. 67, № 2
218 V. PONOMARENKO
a1a2 . . . ak, a
′
1a
′
2 . . . a
′
k ∈ F (a) both mapping to (b11b
2
1 . . .)(b
1
2b
2
2 . . .) . . . (b
1
kb
2
k . . .) ∈ F (φ(a)), where
(b1i b
2
i . . .) ∈ F (φ(ai))∩F (φ(a′i)). But then φ(ai) = φ(a′i), and hence ai, a′i are associates, and hence
a1a2 . . . ak, a
′
1a
′
2 . . . ak are equivalent yet both in F (a), a contradiction.
The following result may be found in [9]; we include its short proof for completeness.
Proposition 1.7. LetR be an information algebra. Then φ : R→ B given by φ(0) = 0, φ(r) = 1
(for all other r) is a morphism.
Proof. If a = b = 0, then φ(a+ b) = φ(ab) = 0 = φ(a)+φ(b) = φ(a)φ(b). If a, b are nonzero,
then a+ b, ab are both nonzero, so φ(a+ b) = φ(ab) = 1 = φ(a) + φ(b) = φ(a)φ(b). If one of a, b
is zero, but the other is nonzero, then a+ b is nonzero and ab = 0. So φ(a+ b) = 1 = φ(a) + φ(b)
while φ(ab) = 0 = φ(a)φ(b).
2. Semigroup semirings. Semigroup semirings may be seen as analogous to group rings and
semigroup rings. We propose to study their arithmetic properties, namely factorization into atoms.
Definition 2.1. Given information algebra R and reduced commutative monoid (S,+) with
identity e, we define the semigroup semiring R[X;S] =
{∑
s∈S
rsX
s : rs ∈ R
}
, where we insist
that all but finitely many coefficients rs = 0. Notationally, we writeA =
∑
s∈S
asX
s, where elements
of R[X;S] are denoted by capital letters and their coefficients by lower case versions. We will often
abbreviate
∑
s∈S
by just
∑
. We define +, × as follows:
(1) A+B =
∑
(as + bs)X
s, with identity 0 =
∑
0Xs,
(2) A×B =
∑
rsX
s, where rs =
∑
u+v=s
aubv. This has identity 1 = 1Xe.
Example 2.1. Let R be an arbitrary information algebra, and let S = N0. Then R[X;S] is the
semiring of polynomials with coefficients from R. These have been studied in [2, 14].
Example 2.2. Let S be an arbitrary reduced commutative monoid, and let R = B. Then R[X;S]
models the semiring of subsets of S, with operations union (+) and Minkowski addition (×). These
have been studied in [5, 13].
Example 2.3. Let S be an arbitrary reduced commutative monoid, and letR = N0. ThenR[X;S]
models the semiring of multisets from S, with operations multiset union (+) and multiset addition (×).
Proposition 2.1. Semigroup semiring R[X;S] is an information algebra.
Proof. (1) (R[X;S],+) is closed and commutative with identity 0 because (R,+) is.
(2) (R[X;S],×) is closed and commutative because R is a semiring. We have 1 = 1Xe +
+
∑
0Xs, so 1A =
∑
rsX
s where rs has only one nonzero summand, namely 1as. Hence
1A = A.
(3) A0 =
∑
rsX
s, where rs =
∑
u+v=s
au0 = 0. Hence A0 = 0.
(4) A(B + C) =
∑
rsX
s, where rs =
∑
u+v=s
au(bv + cv) =
∑
u+v=s
(aubv) + (aucv) =
=
∑
u+v=s
aubv+
∑
u+v=s
aucv, repeatedly using the semiring properties ofR. Hence A(B+C) =
= (AB) + (AC).
(5) Let A,B ∈ R[X;S]. If A+B = 0, then (a+ b)s = 0 for all s ∈ S; hence since R is positive
as = bs = 0 and thus A = B = 0.
(6) Let A,B ∈ R[X;S], and set C = AB. If C = 0, then
∑
u+v=s
aubv = 0 for all s ∈ S.
Suppose that A 6= 0 and B 6= 0. Then there are u, v ∈ S with au 6= 0 and bv 6= 0. By Lemma 1.1,
cu+v 6= 0 and hence C 6= 0, which contradicts hypothesis. Hence either A = 0 or B = 0.
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ARITHMETIC OF SEMIGROUP SEMIRINGS 219
Proposition 2.2. Let R[X;S] be a semigroup semiring. The natural embedding φ : R →
→ R[X;S] given by φ(a) = aXe is a divisor morphism. Also, the natural projection ψ : R[X;S]→
→ R given by ψ
(∑
asX
s
)
= ae is a morphism.
Proof. (1) Let a, b ∈ R. φ(ab) = (ab)Xe = abXe+e = (aXe)(bXe) = φ(a)φ(b). φ(a + b) =
= (a + b)Xe = aXe + bXe = φ(a) + φ(b). If φ(a) | φ(b) then there is some C ∈ R[X;S] with
φ(a)C = φ(b). Since S is reduced, cs = 0 for s 6= e, and hence C = ceX
e. We have aece = be, so
a | b.
(2) Let A,B ∈ R[X;S]. ψ(A+B) = (A+B)e = Ae+Be = ψ(A)+ψ(B). ψ(AB) = (AB)e =
=
∑
u+v=e
aubv = aebe = ψ(A)ψ(B), where u+ v = e implies u = v = e since S is reduced.
Proposition 2.3. Let R[X;S] be a semigroup semiring. Then the following hold:
(1) aXe is a unit in R[X;S] if and only if a is a unit in R;
(2) aXe is an atom in R[X;S] if and only if a is an atom in R;
(3) if R[X;S] is cancellative/weakly cancellative/yoked, then so is R.
Proof. Propositions 1.3, 1.5, and 2.2.
Definition 2.2. Let R[X;S] be a semigroup semiring, and let A ∈ R[X;S]. Set Supp(A) =
= {s ∈ S : as 6= 0}, the support of A. If e ∈ Supp(A), then we call A elementary. If | Supp(A)| = 1,
then we call A monomial. We call elementary monomials constants.
The product of two elementary elements is elementary, and the product of two monomials is a
monomial. The converses of these statements are given in Propositions 2.4 and 2.6.
Proposition 2.4. Let R[X;S] be a semigroup semiring. The following are equivalent:
1. For all A,B ∈ R[X;S], if AB is a monomial then A, B are each monomials.
2. S is cancellative.
Proof. If S is not cancellative, let a, b, c ∈ S satisfy a + b = a + c and b 6= c. Then
(1Xa)(1Xb + 1Xc) = 1Xa+b + 1Xa+c = 2Xa+b gives a monomial product, yet 1Xb + 1Xc
is not a monomial. Conversely, suppose that S is cancellative and B is not a monomial. Let
a ∈ Supp(A), b 6= c ∈ Supp(B). Both a+ b and a+ c are in Supp(AB), and a+ b 6= a+ c since S
is cancellative and b 6= c. Hence AB is not a monomial.
Proposition 2.5. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. Then
Supp(A+B) = Supp(A) ∪ Supp(B) and Supp(AB) = Supp(A) + Supp(B).
Proof. Set C = A+ B. Let s ∈ Supp(C). Then cs = as + bs, and since cs 6= 0, either as 6= 0(
and s ∈ Supp(A)
)
or bs 6= 0
(
and s ∈ Supp(B)
)
. Hence s ∈ Supp(A) ∪ Supp(B). Now let
s ∈ Supp(A) ∪ Supp(B). Without loss s ∈ Supp(A). We have cs = as + bs. Since as 6= 0, cs 6= 0
so s ∈ Supp(C).
Now, set D = AB. Let s ∈ Supp(D). Then ds =
∑
u+v=s
aubv. Since ds 6= 0, some summand
corresponding to u′+v′ = s is nonzero. Hence au′ , bv′ are nonzero, so u′ ∈ Supp(A), v′ ∈ Supp(B)
and s = u′ + v′ ∈ Supp(A) + Supp(B). Now let u + v ∈ Supp(A) + Supp(B). By Lemma 1.1,
since au, bv are nonzero, du+v 6= 0, and so u+ v ∈ Supp(D).
Lemma 2.1. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. AB is a constant
if and only if A,B are.
Proof. Set C = AB. If C is a constant, then Supp(C) = {e} = Supp(A) + Supp(B), by
Proposition 2.5. Since S is reduced, Supp(A) = Supp(B) = {e}, and hence A, B are constant. The
other direction is similar.
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220 V. PONOMARENKO
Proposition 2.6. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. If A + B is
elementary, then at least one of A, B is elementary. If AB is elementary, then both A, B are
elementary.
Proof. The first statement follows from Proposition 2.5. Let e ∈ Supp(AB) = Supp(A) +
+ Supp(B). By Proposition 2.5 again, there are a ∈ Supp(A), b ∈ Supp(B) with a+ b = e. Since
S is reduced, a = b = e.
Proposition 2.7. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. Suppose that
S is cancellative. Then |Supp(AB)| ≥ max
(
|Supp(A)|, |Supp(B)|
)
, with equality if one of A, B
is a monomial.
Proof. Without loss we assume |Supp(A)| ≥ | Supp(B)|. Let r ∈ Supp(B). We have
| Supp(AB)| = | Supp(A) + Supp(B)| ≥ |Supp(A) + {r}|. For a, b ∈ Supp(A), if a+ r = b+ r
then a = b since S is cancellative. Therefore | Supp(A) + {r}| = | Supp(A)|. If B is a monomial,
then the inequality is an equality.
Corollary 2.1. Let R[X;S] be a semigroup semiring with S cancellative. Then every divisor of
a monomial is again a monomial.
Theorem 2.1. Let R[X;S] be a semigroup semiring. The set of units of R[X;S] is U = {rXe :
r is a unit of R}. In particular, each unit is a constant.
Proof. By Proposition 2.3, each element of U is a unit in R[X;S]. Suppose now that 1 = AB.
By Proposition 2.5, Supp(A) + Supp(B) = {e}. Since S is reduced, Supp(A) = Supp(B) = {e}.
Hence A = aXe, B = bXe, and AB = (ab)Xe. Hence A,B ∈ U.
Proposition 2.8. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. If AB is
elementary, then (beA+ aeB) |+ (AB + (aebe)X
e) and beA |+ AB.
Proof. Let s ∈ Supp(A). By Proposition 2.6, B is elementary and hence e ∈ Supp(B). Thus
s = s+ e ∈ Supp(A) + Supp(B) = Supp(AB). Set C = AB. We have asbe |+ cs. Further, for all
s 6= e, asbe + bsae |+ cs, while for s = e, asbe = cs so asbe + aebs = cs + aebe.
The following result simplifies the search for divisors of an elementary element, and is the key
to finding elementary atoms (as will be seen in Corollary 3.1).
Theorem 2.2. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. If AB is elemen-
tary, then Supp(A) ∪ Supp(B) ⊆ Supp(AB).
Proof. Apply Proposition 2.8 to A, B.
Proposition 2.9. Let R1[X;S], R2[X;S] be semigroup semirings, and let φ : R1 → R2 be a
morphism. Define φ′ : R1[X;S] → R2[X;S] via φ′
(∑
asX
s
)
=
∑
φ(as)X
s. Then φ′ is a
morphism. Further, if φ preserves units, then so does φ′.
Proof. Let A,B ∈ R1[X;S]. Then
φ′(A+B) = φ′
(∑
(as + bs)X
s
)
=
∑
φ(as + bs)X
s =
=
∑
(φ(as) + φ(bs))X
s =
∑
φ(as)X
s +
∑
φ(bs)X
s = φ′(A) + φ′(B),
φ′(AB) = φ′
(∑
csX
s
)
=
∑
s∈S
φ(cs)X
s =
=
∑
s∈S
Xs
∑
u+v=s
φ(au)φ(bv) =
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ARITHMETIC OF SEMIGROUP SEMIRINGS 221
=
∑
φ(as)X
s
∑
φ(bs)X
s = φ′(A)φ′(B),
φ′(0) = φ′(0Xe) = φ(0)Xe = 0Xe = 0, and φ′(1) = φ′(1Xe) = φ(1)Xe = 1Xe = 1.
Suppose now that φ preserves units. By Proposition 1.3, φ′ carries units forward. Suppose now
that A ∈ R1[X;S] and φ′(A) is a unit. By Theorem 2.1, φ′(A) = rXe = φ(ae)X
e, and φ(ae) is a
unit in R2. Since φ preserves units, ae is a unit, and hence A = aeX
e is a unit.
Definition 2.3. Let S be a reduced commutative monoid with identity e. We call ν : S → N0 a
valuation if ν(s · t) = ν(s) + ν(t) for all s, t ∈ S.
Note that if ν is a valuation then ν(e) = ν(e · e) = ν(e) + ν(e), so ν(e) = 0. Valuations are
useful in our context because of the following.
Proposition 2.10. LetR[X;S] be a semigroup semiring, let ν be a valuation on S, and let ρ ∈ R.
Set ν : R[X;S]→ R via ν(A) =
∑
s∈Supp(A)
asρ
ν(s). Then ν : R[X;S]→ R is a morphism.
Proof. We have
ν(A+B) =
∑
s∈Supp(A+B)
(as + bs)ρ
ν(s) =
=
∑
s∈Supp(A)
asρ
ν(s) +
∑
s∈Supp(B)
bsρ
ν(s) = ν(A) + ν(B),
ν(1) = ν(1Xe) = 1ρν(e) = 1ρ0 = 1,
and ν(0) is an empty sum so ν(0) = 0,
ν(AB) =
∑
s∈Supp(A)+Supp(B)
∑
u+v=s
aubvρ
ν(s) =
=
∑
s∈Supp(A)+Supp(B)
∑
u+v=s
auρ
ν(u)bvρ
ν(v) =
∑
s∈Supp(A)
auρ
ν(u)
∑
s∈Supp(B)
bvρ
ν(v) = ν(A)ν(B).
Proposition 2.11. Let (S, ·) be a a reduced commutative monoid with identity e. Let σ ∈ S be
prime. For all s ∈ S, set ν(s) = sup{n : σn | s}. Suppose that the following two conditions hold:
(1) for all s ∈ S, ν(s) ∈ N0;
(2) for all s, t ∈ S, if σ · s = σ · t then s = t.
The ν is a valuation on S.
Proof. Let u · v = s; we write u = σν(u) · u′, v = σν(v) · v′ for some u′, v′ ∈ S. Certainly
ν(u′) = ν(v′) = 0, by maximality of ν.We have s = u·v = σν(u)+ν(v) ·u′ ·v′, so ν(s) ≥ ν(u)+ν(v),
and we write s = σν(u)+ν(v) · s′ = σν(u)+ν(v) · u′ · v′. By hypothesis (2), s′ = u′ · v′. Since σ is
prime, if σ | s′ then σ | u′ or σ | v′, contrary to ν(u′) = ν(v′) = 0. Hence ν(s′) = 0 and thus
ν(s) = ν(u) + ν(v).
Note that if S is BF, then the first condition of Proposition 2.11 holds. If S is cancellative, then
the second condition of Proposition 2.11 holds.
Example 2.4. Consider the multiplicative submonoid of the quadratic integer ring Z
[√
−5
]
. It
has units U = {1,−1}. Set S =
(
Z[
√
−5]/U, ·
)
, a reduced commutative semigroup with identity
e = [1]. Set σ =
[
6 +
√
−5
]
∈ S. It is prime since it has norm 36 + 5 = 41, a rational prime. S
is BF and cancellative, hence we define ν as in Proposition 2.11. Consider N[X;S], the semiring of
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222 V. PONOMARENKO
multisets from S. Set A = 1Xe + 2Xσ + 3Xσ·σ. Taking ρ = 1, we have ν(A) = 6, which is not
helpful. However, if we take ρ = 2 we have ν(A) = 17, a rational prime. ν preserves units, since if
ν(B) = 0 then bs = 0 for all s. Hence by Proposition 1.3, A is an atom.
Proposition 2.12. Let S be a reduced commutative semigroup with identity e. For valuations ν1,
ν2, we define operation ◦ via (ν1 ◦ν2)(s) = ν1(s)+ν2(s). Under this operation, the set of valuations
on S is a nonempty reduced commutative semigroup with identity ν0 : S → {0}.
Proof. We have (ν1 ◦ ν2)(s · t) = ν1(s · t) + ν2(s · t) = ν1(s) + ν1(t) + ν2(s) + ν2(t) =
= (ν1(s) + ν2(s)) + (ν1(t) + ν2(t)) = (ν1 ◦ ν2)(s) + (ν1 ◦ ν2)(t), so ν1 ◦ ν2 is a valuation. ◦
is commutative by symmetry. We have (ν0 ◦ ν)(s) = ν0(s) + ν(s) = 0 + ν(s) = ν(s), so ν0 is
neutral. Lastly, if ν1 ◦ ν2 = ν0, then ν1(s) + ν2(s) = 0 for all s, but since ν1(s), ν2(s) ∈ N0,
ν1(s) = ν2(s) = 0 and ν1 = ν2 = ν0.
3. Primitivity and maximal common divisors.
Definition 3.1. Let R[X;S] be a semigroup semiring, and A ∈ R[X;S] be nonzero. We call
the R-content of A the set Rc(A) = {r ∈ R : r | as for all s ∈ Supp(A)}. We say that A is
R-primitive if there is no nonunit in Rc(A). We call the S-content of A the set Sc(A) = {t ∈ S :
t | s for all s ∈ Supp(A)}. We say that A is S-primitive if Sc(A) = {e}. We say that A is primitive
if it is both R-primitive and S-primitive.
Lemma 3.1. Let R[X;S] be a semigroup semiring, and let A ∈ R[X;S] be nonzero. r ∈ Rc(A)
if and only if rXe | A. s ∈ Sc(A) if and only if 1Xs | A.
Proof. If r ∈ Rc(A), then for all s ∈ Supp(A) we define a′s via as = ra′s. We have A =
=
∑
s∈Supp(A)
ra′sX
s = (rXe)
∑
s∈Supp(A)
a′sX
s. If A = (rXe)B =
∑
s∈Supp(B)
rbsX
s, so
r | as for all s ∈ Supp(A) = Supp(B) and hence r ∈ Rc(A).
If s ∈ Sc(A), then for all t ∈ Supp(A) we define t′ via t = s+ t′. We have
A =
∑
t∈Supp(A)
atX
t =
∑
t′ : s+t′∈Supp(A)
atX
s+t′ = (1Xs)
∑
t′ : s+t′∈Supp(A)
as+t′X
t′ .
Now, if A = (1Xs)B =
∑
t∈Supp(B)
btX
s+t, then Supp(A) = s+ Supp(B), so s ∈ Sc(A).
Proposition 3.1. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. Then
Rc(A)Rc(B) ⊆ Rc(AB), Rc(A) ∩Rc(B) ⊆ Rc(A+B), and Sc(A) + Sc(B) ⊆ Sc(AB).
Proof. We apply Lemma 3.1 repeatedly. If da ∈ Rc(A), db ∈ Rc(B), then daXe | A, dbXe | B
and hence (daX
e)(dbX
e) = (dadb)X
e | AB so dadb ∈ Rc(AB). If d ∈ Rc(A) ∩ Rc(B), then
dXe | A, dXe | B so A = dXeA′, B = dXeB′ and A + B = dXe(A′ + B′), so dXe | (A + B)
and hence d ∈ Rc(A + B). If ta ∈ Sc(A), tb ∈ Sc(B), then 1Xta | A, 1Xtb | B and hence
(1Xta)(1Xtb) = 1Xta+tb | AB so ta + tb ∈ Sc(AB).
Corollary 3.1. Let R[X;S] be a semigroup semiring, and let A,B ∈ R[X;S]. If AB is R-
primitive (resp. S-primitive), then A and B are each R-primitive (resp. S-primitive).
Theorem 3.1. Let R[X;S] be a semigroup semiring, and let A ∈ R[X;S] be primitive. Let
B ∈ B[X;S] with Supp(A) = Supp(B). Suppose B is an atom. Then A is an atom.
Proof. Combining Propositions 1.7 and 2.9, we get morphism φ : R[X;S]→ B[X;S] that maps
A to B. Suppose that A = A1A2, a product of nonunits. Then φ(A) = B = φ(A1)φ(A2). Since B
is an atom, without loss φ(A1) is a unit. The only unit of B[X;S] is 1Xe, and hence A1 = rXe for
some nonzero r. By Corollary 3.1, A1 is primitive and thus r is a unit in R. But then A1 is a unit by
Proposition 2.3, which is a contradiction. Hence A is an atom.
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ARITHMETIC OF SEMIGROUP SEMIRINGS 223
Proposition 3.2. LetR[X;S] be a semigroup semiring. Every elementary element is S-primitive.
If S is also yoked, then the opposite implication also holds.
Proof. Let A ∈ R[X;S]. Suppose that A is elementary, and let t ∈ Sc(A). e ∈ Supp(A) since
A is elementary, so t | e and t = e since S is reduced. Hence Sc(A) contains no nonunits, and A
is S-primitive. Suppose now that S is yoked and A is S-primitive. By Proposition 1.1, | is a total
preorder on S, hence on Supp(A). We may therefore choose some t ∈ Supp(A) (not necessarily
uniquely) that is a common divisor of each element of Supp(A). Since A is S-primitive, t is a unit.
Since S is reduced, t = e. Therefore e ∈ Supp(A) and A is elementary.
The following result is a major tool for construction of elementary atoms in semigroup semirings.
We have no similarly simple construction for primitive non elementary atoms, lacking an analog to
Theorem 2.2.
Theorem 3.2. Let R[X;S] be a semigroup semiring, let A ∈ R[X;S] be elementary, and set
T = Supp(A) \ {e}. Let s ∈ T be “maximal ” in the sense that for all t ∈ T, s + t /∈ T. Suppose
that for all u, v ∈ T, u+ v 6= s. Then A is an atom.
Proof. Suppose otherwise, that A = BC, a product of nonunits. We first prove that B is not a
monomial. Suppose otherwise, thatB = bXt.We have b ∈ Rc(A), t ∈ Sc(A). SinceA is elementary,
by Proposition 3.2, A is primitive and so b, t are units, and by Proposition 2.3, B is a unit, contrary to
hypothesis. Hence B (and, similarly, C) is not a monomial. Since s ∈ Supp(A), by Proposition 2.5
there are u ∈ Supp(B), v ∈ Supp(C) with u + v = s. If u = e, then s ∈ Supp(C). Choose any
u′ ∈ Supp(B) \ {e}, which is nonempty since B is not a monomial. Then u′ + v = u′ + s ∈ T,
which contradicts the “maximality” of s. Hence u 6= e, and similarly v 6= e. Now by Theorem 2.2,
u, v ∈ T and u+ v = s. This contradicts hypothesis, and hence A is an atom.
Example 3.1. Consider R[X;N0], which are polynomials and admit a degree. Let A ∈ R[X;N0]
be elementary with deg(A) = n. Suppose that either
[
1,
n
2
]
or
[n
2
, n− 1
]
has no intersection with
Supp(A). Taking s = n in Theorem 3.2, we conclude that A is an atom.
The following is derived from an example in [14].
Proposition 3.3. Let R[X;S] be a semigroup semiring. Suppose that 1, 2(= 1 + 1), 3 are
pairwise distinct in R. Suppose further that there is some s ∈ S with s, 2s(= s + s), . . . , 12s all
distinct. Then the elasticity of R[X;S] is at least 3/2. In particular, R[X;S] is not UF.
Proof. Note that Xe+Xs+X2s+X3s+2X4s+2X5s+2X6s+3X7s+X8s+X9s+X10s =
= AB = CDE, for A = Xe +X3s +X5s +X6s, B = Xe +Xs +X2s +X4s, C = Xe +Xs,
D = Xe+X2s, E = Xe+2X4s+X7s. C,D,E are all atoms by Theorem 3.2. IfA were not an atom,
we write A = A1A2, and applying Proposition 2.2 we get Supp(A1) ∪ Supp(A2) ⊆ Supp(A) =
= {e, 3s, 5s, 6s}. Within this set 5s+e is the only factorization of 5s. Hence without loss we assume
that 5s ∈ Supp(A1), e ∈ Supp(A2). But now Supp(A2) = {e} else Supp(A) 6= {e, 3s, 5s, 6s}.
Hence A2 = uXe and A1 = aXe + bX3s + cX5s + dX6s. Since ua = 1, u is a unit in R, and
hence A2 is a unit in R[X;S]. By a similar argument, if B is not an atom, then B = (aXe + bXs +
+ cX2s)(a′Xe + b′Xs + c′X2s). So aa′ = 1, ab′ + ba′ = 1, ac′ + bb′ + ca′ = 1, bc′ + cb′ = 0,
cc′ = 1. So a, a′, c, c′ are units. But bc′ = 0 and cb′ = 0, so b = b′ = 0 and hence ab′ + ba′ = 0, a
contradiction.
Proposition 3.4. Let R[X;S] be a semigroup semiring. Suppose there is some r ∈ R∗ with
r + r = rr = r. Suppose there is some s ∈ S with s, 2s, 3s, . . . all distinct. Then R[X;S] has
infinite elasticity. Further, if there is some prime P ∈ R[X;S], then R[X;S] is fully elastic.
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224 V. PONOMARENKO
Proof. Let n ∈ N. Set A = 1Xe + rXs, B = 1Xe + rXs + rX2s + . . .+ rXns + rX(2n+1)s,
C = 1Xe+ rX(n+1)s+ rX(n+2)s+ . . .+ rX(2n+1)s. By Theorem 3.2, A,B,C are each atoms. We
have BC = 1Xe+rXs+rX2s+ . . .+rX(4n+2)s = A4n+2, hence ρ(R[X;S]) ≥ 4n+ 2
2
= 2n+1,
so ρ(R[X;S]) =∞. Now, let
p
q
≥ 1. If p = q we have ρ(P ) =
p
q
, otherwise we take n = p− q and
consider BCP 4q−2 = A4n+2P 4q−2. We have ρ(BCP 4q−2) =
4(p− q) + 2 + 4q − 2
2 + 4q − 2
=
p
q
.
Note that an r such as in Proposition 3.4 may be appended to any semiring R, defining x+ r =
= xr = r for all nonzero x ∈ R. Further, in [2] it is shown that N0[X;N0] has infinite elasticity.
Hence even for these very well-behaved R,S (cancellative, reduced, UF, greatest common divisors,
totally ordered), the resulting semigroup semiring has infinite elasticity.
At the other extreme, if |S| = 1 then we call R[X;S] trivial. Then the natural embedding φ :
R[X;S] → R is bijective, so by Proposition 1.4, R[X;S] inherits the properties of R, such as
elasticity and UF.
The evidence collected above and in Propositions 3.3 and 3.4 leads us to the following conjecture.
It stands in contrast to the (semi)group ring case, where UF is achievable (e.g., D[X] for any unique
factorization domain D).
Conjecture 3.1. All nontrivial atomic semigroup semirings have infinite elasticity. In particular,
no semigroup semiring is UF.
Proposition 3.5. Let R[X;S] be a semigroup semiring. 1Xs is an atom in R[X;S] if and only
if s is an atom in S. 1Xs is prime in R[X;S] if and only if s is prime in S.
Proof. Suppose s = r+ t, for nonzero r, t ∈ S. Then 1Xs = (1Xr)(1Xt), a product of nonunits
by Theorem 2.1. Suppose now that 1Xs = AB, for nonunits A,B. If Supp(A) = {e}, then ae
is a nonunit in R by Theorem 2.1. However, ae ∈ Rc(AB) by Proposition 3.1, hence ae | 1, a
contradiction. Hence there is some t ∈ Supp(A) with t 6= e. Similarly, there is some r ∈ Supp(B)
with r 6= e. By Proposition 2.5, r + t ∈ Supp(AB) = {s}, and hence r + t = s. Therefore s is not
an atom in S.
Suppose now that 1Xs is prime in R[X;S], and there are a, b, c ∈ S with a + s = b + c. Then
(1Xa)(1Xs) = (1Xb)(1Xc), and since 1Xs is prime, then without loss there is some A ∈ R[X;S]
with A(1Xs) = 1Xb. Then for all t ∈ Supp(A), t + s = b, so in particular s | b. Hence s is prime
in S. Suppose next that s is prime in S, and that there are A,B,C ∈ R[X;S] with A(1Xs) = BC.
Suppose there are u ∈ Supp(B), v ∈ Supp(C) with s - u and s - v. But then u+ v ∈ Supp(BC) =
= Supp(A) + s, so s | u + v yet s - u, s - v, which contradicts the primality of S. Hence either
s ∈ Sc(B) or s ∈ Sc(C) and hence 1Xs | B or 1Xs | C, and so 1Xs is prime.
Proposition 3.6. Let R[X;S] be a semigroup semiring. Let A = aXs be an arbitrary monomial.
A is an atom if exactly one of the following holds:
(1) a is an atom in R, and s = e
or
(2) a is a unit in R, and s is an atom in S.
Proof. One direction is given by Propositions 2.3 and 3.5. If neither a nor s is a unit, then we
write A = (aXe)(1Xs), a product of nonunits. Hence if A is an atom either a or s is a unit; then by
Propositions 2.3 and 3.5 again, one of the above must hold.
Proposition 3.7. Let R[X;S] be a semigroup semiring. Let r, r′ ∈ R, s, s′ ∈ S. Monomials
rXs, r′Xs′ are associates in R[X;S] if and only if
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ARITHMETIC OF SEMIGROUP SEMIRINGS 225
(1) r, r′ are associates in R, and
(2) s, s′ are associates in S.
Proof. Suppose first (1) + (2). Choose r′′ ∈ R, s′′ ∈ S with r = r′r′′, s = s′s′′. We have
rXs = (r′Xs′)(r′′Xs′′); hence rXs | r′Xs′ . Repeating in the other direction proves (1) and (2).
Suppose now rXs, r′Xs′ are associates. Choose A ∈ R[X;S] with rXsA = r′Xs′ . Then
(rXe)(1Xs)A = r′Xs′ , so by Lemma 3.1, (1) r ∈ Rc(r′Xs′) so r | r′, and (2) s ∈ Sc(r′Xs′) so
s | s′. Repeating in the other direction proves (1) and (2).
Proposition 3.8. Let R[X;S] be a semigroup semiring, with R, S atomic. If all monomials of
R[X;S] have UF/FF/BF, then R, S are also UF/FF/BF. If S is also cancellative, then the opposite
implication also holds.
Proof. Consider monomial rXs. By Proposition 3.6, we may reorder any factorization of
rXs into monomials canonically, as (r1X
e)(r2X
e) . . . (rkX
e) (1Xs1)(1Xs2) . . . (1Xsj ), where
r1r2 . . . rk is a factorization of r and s1 + s2 + . . . + sj is a factorization of s. By Proposition 3.7,
two such factorizations are equivalent in R[X;S] if and only if the corresponding factorizations are
each equivalent in R, S. Hence the set of factorizations of rXs into monomials is UF/FF/BF, if and
only if the same conditions hold on both factorizations of r, s. This gives one direction, since if
all factorizations of rXs are UF/FF/BF then surely all factorizations into monomials are. If S is
cancellative, then by Proposition 2.4, all factorizations of rXs are into monomials, which gives the
other direction.
Proposition 3.9. Let R[X;S] be a semigroup semiring, and let A ∈ R[X;S]. If A is an atom
then either A is a monomial or primitive.
Proof. Suppose otherwise. If A is not R-primitive, there is some nonunit r ∈ Rc(A), and by
Lemma 3.1, we write A = (rXe)B. If instead A is not S-primitive, there is some nonunit t ∈ Sc(A)
and by Lemma 3.1, we write A = (1Xt)B. In both cases, B is a nonunit by Theorem 2.1 since
|Supp(B)| = | Supp(A)|, which is greater than 1. Hence we have factored A into two nonunits,
which contradicts the hypothesis.
Hence, every atomic factorization of A ∈ R[X;S] is the product of monomial atoms (whose
product may or may not be a monomial) and the product of primitive atoms (whose product may or
may not be primitive). The former can be resolved by Corollary 2.1, and the latter by Theorem 3.11.
However another approach to factoring A into atoms is to first factor A into a monomial (each
factor of which must be monomial) times a primitive element (each factor of which must be primitive).
Then, we factor each part into atoms. We take this approach, with a sequence of results leading up
to Theorem 3.6, which gives sufficient conditions for R[X;S] to be atomic.
Definition 3.2. Given monoid (M, ·), suppose that for each finite set S ⊆ M, there is some d
that satisfies the following:
(1) for all s ∈ S, d | s;
(2) if there is some d′ with d | d′ and d′ | s ( for all s ∈ S), then d, d′ are associates.
In this case we say that (M, ·) has maximal common divisors.
In the ring theoretic context, maximal common divisors have been considered in [1, 11]. Every
FF monoid has maximal common divisors. A monoid may have maximal common divisors but no
greatest common divisors, as shown by the following.
Example 3.2. Consider the numerical monoid (M,+) for M = {0, 3, 4, 5, 6, . . .}. For S =
= {9, 10}, the set of common divisors is {0, 3, 4, 5, 6}, and the set of maximal common divisors is
{4, 5, 6}. There is no greatest common divisor of S.
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226 V. PONOMARENKO
Theorem 3.3. LetR[X;S] be a semigroup semiring, whereR and S each have maximal common
divisors and are weakly cancellative. Let A ∈ R[X;S]. Then we may write A = BC, where B is
monomial and C is primitive.
Proof. Let b be a maximal common divisor of {as : s ∈ Supp(A)}, and let t be a maximal
common divisor of Supp(A). Set B = bXt, and C =
∑
s∈Supp(A)
(as/b)X
(s−t). Evidently A = BC
and B is a monomial; it remains to prove that C is primitive. Suppose d is a common divisor of {cs :
s ∈ Supp(C)}. But then bd is a common divisor of {as : s ∈ Supp(A)}. Since b is maximal, in fact
b = bde for some unit e; hence d is a unit by weak cancellativity of R. Hence C is R-primitive.
Suppose now that r is a common divisor of Supp(C). But then r+t is a common divisor of Supp(A).
Since t is maximal, in fact t = t+ r + e for some unit e; hence r is a unit by weak cancellativity of
S. Hence C is S-primitive, and thus primitive.
Theorem 3.4. Let R[X;S] be a semigroup semiring, where R and S are each atomic. Let
A ∈ R[X;S] be a monomial. Then we may write A as a product of monomial atoms. Further, if
both R, S are UF/FF/BF, then so is A.
Proof. Let A = aXs. We factor into atoms a = a1a2 . . . am and also s = s1 + s2 + . . .
. . . + sn. By Proposition 3.6, we have a factorization of A into k = m + n atoms given by
(a1X
e)(a2X
e) . . . (amX
e)(1Xs1)(1Xs2) . . . (1Xsn). On the other hand, if A = A1A2 . . . Ak, then
by Proposition 3.6, for some m+ n = k this corresponds to a factorization of a into m atoms, and a
factorization of s into n atoms. Hence |F (A)| ≤ |F (a)| |F (s)| and supL(A) ≤ supL(a)+supL(s).
Note that if S is also cancellative, Corollary 2.1 provides a partial converse to Theorem 3.4: all
factorizations of a monomial are into other monomials.
The following surprising theorem gives a factorization of any primitive element into a bounded
number of atoms, even if R, S are not BF or atomic themselves.
Theorem 3.5. Let R[X;S] be a semigroup semiring, with S cancellative. Let A ∈ R[X;S] be a
primitive nonunit. Then we may write A as a product of k primitive atoms, for some k ≤ 2|Supp(A)|.
Proof. The proof is by induction on |Supp(A)|. If | Supp(A)| = 1, then A is a primitive
monomial, hence a unit by Proposition 2.3, which is contrary to hypothesis. Now suppose that
|Supp(A)| > 1. If A is an atom, there is nothing to prove. If not, write A = BC for nonunits
B, C. By Corollary 3.1, both B and C are primitive. If either were a monomial, then again
by Proposition 2.3 it would be a unit, contrary to supposition. By Proposition 2.7, |Supp(A)| ≥
≥ |Supp(B)| and |Supp(A)| ≥ |Supp(C)|. We now prove that these inequalities are strict. Suppose
that instead | Supp(B) + Supp(C)| = |Supp(A)| = |Supp(B)|. Let c ∈ Supp(C) \ {e}; we must
have Supp(B) + {c} = Supp(B) + Supp(C). For all a ∈ Supp(A), we write a = u + v for
some u ∈ Supp(B), v ∈ Supp(C). Since u + v ∈ Supp(A) = Supp(B) + {c}, there is some
u′ ∈ Supp(B) with a = u′ + c. Therefore c ∈ Sc(A), contradicting the primitivity of A. Hence
|Supp(A)| > | Supp(B)| and similarly |Supp(A)| > | Supp(C)|. The inductive hypothesis now
gives us two factorizations into primitive atoms B = B1B2 . . . Bm, C = C1C2 . . . Cn. Combining
and setting k = m + n, we have A = B1B2 . . . BmC1C2 . . . Cn. By the inductive hypothesis
m ≤ 2| Supp(B)| ≤ 2|Supp(A)|−1 and also n ≤ 2| Supp(A)|−1 so k = m+ n ≤ 2| Supp(A)|.
Theorem 3.6 provides sufficient conditions for semigroup semiring R[X;S] to be atomic, and also
BF. UF was addressed in Conjecture 3.1, while FF requires some machinery and will be addressed in
Theorem 3.7.
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ARITHMETIC OF SEMIGROUP SEMIRINGS 227
Theorem 3.6. Suppose that R, S are atomic and have maximal common divisors. Suppose that
R is weakly cancellative, and S is cancellative. Then semigroup semiring R[X;S] is atomic. Further,
if R, S are each BF, then so is R[X;S].
Proof. We first factor A ∈ R[X;S] into monomial and primitive parts with Theorem 3.3, then
factor each part into atoms with Theorem 3.4 and Theorem 3.5, respectively.
Definition 3.3. Let (M, ·) be a commutative monoid. Suppose that M admits a total order ≥
with least element e that respects ·. Suppose further that M does not contain any infinite descending
chain s1 > s2 > . . . . We then call M well-ordered. For semiring R, if (R,+) is well-ordered, and
also ≥ respects ×, then we call R well-ordered.
Well-ordered semigroups are characterized in [12]. A well-ordering on R yields a natural partial
ordering on R[X;S] via A ≥ B if as ≥ bs for all s ∈ S.
Let R be a well-ordered semiring, and let r ∈ R. Then there are finitely many r′ ∈ R with r′ |+ r.
Theorem 3.7. Let R[X;S] be a BF semigroup semiring. Suppose that R is well-ordered, (R,+)
is FF, and S is cancellative, yoked, and FF. Then R[X;S] is FF.
Proof. Let A ∈ R[X;S]. Suppose we have factorization set F (A) with |F (A)| = ∞. By
Proposition 3.9, either infinitely many nonassociate monomial atoms divide A, or infinitely many
nonassociate primitive atoms divide A. In the first case, by Proposition 3.6, either R or S lacks FF.
We therefore suppose the second case. By Proposition 3.2, A is elementary, and by Proposition 2.8
if B | A then there is some constant c ∈ R with cB |+ A. Since (R,+) is FF, there are only finitely
many r ∈ R with r |+ as for each s ∈ Supp(A). Since there are infinitely many nonassociate B
dividing A, we may construct c1, c2, . . . ∈ R and nonassociate B1, B2, . . . such that Bi | A and
c1B1 = c2B2 = . . . . If some ci = cj , then Bi = Bj , contrary to hypothesis. Because ≥ is a
well-ordering, c1 > c2 > . . . is impossible, hence we assume without loss that c1 < c2 < . . . . Let
s ∈ Supp(B1), and set bi = (Bi)s. We have c1b1 = c2b2 = . . . . If b1 ≤ b2 then c1b1 < c2b1 ≤ c2b2,
a contradiction. Hence b1 > b2 > . . . , which contradicts the well-ordering of R.
The gap between Theorems 3.6, 3.7, and the following partial converse remains open. Note also
Conjecture 3.1.
Theorem 3.8. Suppose that R[X;S] is an atomic semigroup semiring. Then R,S are atomic.
Further, if R[X;S] is UF/FF/BF, then so are R, S.
Proof. Propositions 2.3 and 3.8.
We turn now toward a semiring analog of the Gauss lemma on primitive polynomials. Note
that this does not directly generalize Gauss lemma since semigroup semirings as we define them are
information algebras (hence cannot be rings). In Theorem 3.11 we present the desired result, but first
we need some machinery.
Proposition 3.10. Let (M, ·) be a commutative monoid with UF. Then (M, ·) has greatest com-
mon divisors. Further, if X,Y ⊆ M are nonempty and finite, then gcd(X · Y ) = gcd(X) · gcd(Y ),
up to associates.
Proof. Let S ⊆ M be finite. Choose a factorization of
∏
S, and let Q be the finite set of
primes from this factorization. Choose P ⊆ Q so that each q ∈ Q has some associate p ∈ P, but the
elements of P are pairwise nonassociate. By UF, any prime from any factorization of an element of
S is associate to some element of Q, hence some element of P. Hence for each s ∈ S we have a
unique associated prime factorization
∏
p∈P
pν(p), where ν(p) ∈ N0.
We claim that d | s if and only if d is associated to some
∏
p∈P
pγ(p), where 0 ≤ γ(p) ≤ ν(p).
If d is associated to an element of this form, then s is associated to d
∏
p∈P
pν(p)−γ(p), so d | s.
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228 V. PONOMARENKO
Suppose that d | s. If some prime divides d that is not associate to any prime in P, that prime divides∏
S, which contradicts the UF property. Hence d is associated to
∏
p∈P
pγ(p), where γ(p) ∈ N0.
Suppose that γ(p) > ν(p) for some p. Then pγ(p) | s, so we write s = pγ(p)a, for some a ∈M. But
this contradicts UF, since any factorization of pγ(p)a must have at least γ(p) primes associate to p,
and so p is associate to some other element of P. Now, set µ(p) = mins∈S{ν(p)}, d =
∏
p∈P
pµ(p).
By the above, d is a common divisor for S, and if d′ is any other divisor of s, then d′ | d.
Next, for S = X ∪ Y, we set µX(p) = mins∈X{ν(p)}, µY (p) = mins∈Y {ν(p)}, µX·Y (p) =
= mins∈X·Y {ν(p)}. Choose x ∈ X with x = pµX(p)x′ and ν(x′) = 0, and y ∈ Y with y = pµY (p)y′
and ν(y′) = 0. We have pµX(p)+µY (p) | xy, so µX·Y (p) ≥ µX(p) + µY (p). Suppose that ps | xy
with s > µX(p) + µY (p). We write ps | pµX(p)+µY (p)x′y′. But now, by UF, p | x′y′ and so p | x′
or p | y′, which contradicts ν(x′) = ν(y′) = 0. Hence µX·Y (p) = µX(p) + µY (p) and hence∏
p∈P
pµX·Y (p) =
∏
p∈P
pµX(p) ·
∏
p∈P
pµY (p). Each element of gcd(X · Y ) is associated to this
product, as is each element of gcd(X) · gcd(Y ).
The following is an Sc-analog of the Gauss lemma on primitive polynomials.
Theorem 3.9. Let R[X;S] be a semigroup semiring. Suppose that S has UF. Then Sc(AB) =
= Sc(A) + Sc(B). In particular, if A, B are both S-primitive, then AB is S-primitive.
Proof. gcd(Supp(AB)) = gcd(Supp(A) + Supp(B)) by Proposition 2.5. By Proposition 3.10,
this is associate to gcd(Supp(A)) + gcd(Supp(B)). If c ∈ Sc(AB), then c divides each element
of Supp(AB), hence c | gcd(Supp(AB)) = gcd(Supp(A)) + gcd(Supp(B)). Hence there is some
a ∈ gcd(Supp(A)), b ∈ gcd(Supp(B)) with c | a+ b and hence c ∈ Sc(A) + Sc(B). On the other
hand, if a ∈ Sc(A), b ∈ Sc(B), then a+b divides each element of Supp(AB), hence a+b ∈ Sc(AB).
Definition 3.4. Let R be a semiring. Following [10], we call R a PLIS-semiring if the following
property holds: For all a, b ∈ R, if d is a common divisor of a, a+ b, then d is a divisor of b.
Note that every ring is a PLIS-semiring. The following lemma extends Proposition 3.1, for
PLIS-semirings.
Lemma 3.2. Let R[X;S] be a semigroup semiring, where R is a PLIS-semiring. Let A,B ∈
∈ R[X;S]. Then Rc(A) ∩Rc(A+B) ⊆ Rc(B).
Proof. Let d ∈ Rc(A) ∩ Rc(A + B). Hence, for each s ∈ S, we have d | as, d | (as + bs).
Because R is a PLIS-semiring, d | bs; hence d ∈ Rc(B).
Definition 3.5. Let R[X;S] be a semigroup semiring. Suppose that S is well-ordered. We define
the degree function, deg : R[X;S]→ S via deg(A) = max{s : s ∈ Supp(A)}.
We now present an Rc-analog of the Gauss lemma on primitive polynomials.
Theorem 3.10. Let R[X;S] be a semigroup semiring. Suppose that R is a UF PLIS-semiring,
and S is well-ordered. Then Rc(AB) = Rc(A)Rc(B). In particular, if A, B are both R-primitive,
then AB is R-primitive.
Proof. If A, B are not R-primitive, we choose a to be a greatest common divisor of Rc(A), b to
be a greatest common divisor of Rc(B). By Lemma 3.1 we consider A/(aXe), B/(bXe). We claim
that these are each R-primitive; suppose d ∈ Rc(A/(aXe)). But then da ∈ Rc(A), and hence da | a.
By unique factorization, d is a unit. Applying the theorem to R-primitive A/(aXe), B/(bXe), we
conclude that AB/(abXe) is R-primitive, and hence Rc(AB) = Rc(A)Rc(B).
We assume henceforth that A,B are R-primitive. Set C = AB. The proof proceeds by recursion
on deg(C). If deg(C) = e, then C is a constant. By Proposition 2.6, A,B are also constants and
hence Rc(C) = Rc(A)Rc(B). If either A or B is a monomial, then again Rc(C) = Rc(A)Rc(B).
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ARITHMETIC OF SEMIGROUP SEMIRINGS 229
Suppose deg(C) > e, and some nonunit r ∈ Rc(C). Since R has UF, we choose prime p
dividing r. Let u ∈ Supp(A), v ∈ Supp(B). We have deg(A) + deg(B) ≥ deg(A) + v ≥ u + v,
with equality only if deg(A) = u, deg(B) = v. Hence cdeg(C) = adeg(A)bdeg(B). Since p | cdeg(C)
and p is prime, without loss we may assume that p | adeg(A). Set A′ =
∑
s∈Supp(A)\{deg(A)}
asX
s,
A′′ = adeg(A)X
deg(A). These are each nontrivial since A is not a monomial. We have p ∈ Rc(A′′) =
= Rc(A′′){1} ⊆ Rc(A′′B). But also p ∈ Rc(C) = Rc(A′B + A′′B), so by Lemma 3.2, p ∈
∈ Rc(A′B).
But now we have A′, B,A′B with deg(C) > deg(A′B). We continue recursively; however since
S is well-ordered there cannot be an infinite descending chain. Hence we may assume by recursion
that Rc(A′B) = Rc(A′)Rc(B) = Rc(A′). We have p ∈ Rc(A′)∩Rc(A′′) ⊆ Rc(A′+A′′) = Rc(A),
which contradicts the R-primitivity of A.
Theorem 3.11. Let R[X;S] be a semigroup semiring. Suppose that R is a UF PLIS-semiring,
and S is UF and well-ordered. Then Rc(AB) = Rc(A)Rc(B) and Sc(AB) = Sc(A) + Sc(B). In
particular, if A, B are both primitive, then AB is primitive.
Proof. Combine Theorems 3.9 and 3.10.
Acknowledgement. The author would like to acknowledge the hospitality of the Institut für
Mathematik und Wissenschaftliches Rechnen at Karl – Franzens – Universität Graz, where he was
visiting while this work was completed.
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