2. The McCoy Condition in Composite Hurwitz Rings
We start this section with a simple result. The proof is straightforward; thus, we omit it.
Lemma 1. If is an extension of commutative rings with identity, then the following assertions hold.
- (1)
If T is a reduced ring, then so is R.
- (2)
If T is a torsion-free -module, then so is R.
Let be an extension of commutative rings with identity. We next study zero-divisors in composite Hurwitz rings and .
Proposition 1. Let be an extension of commutative rings with identity. Suppose that T is a reduced ring which is a torsion-free -module. If and are elements of such that , then for all .
Proof. Suppose that . Then, . Suppose that for some . Then, the coefficient of in is . Multiplying both sides by , . Since T is a reduced ring, . By the induction, for all . This also implies that .
We next suppose that for some . Let . Then, ; thus, . Suppose that for some . Note that the coefficient of in h is . By multiplying both sides by , . Since T is a torsion free -module, . Since T is a reduced ring, . By the induction, for all . This shows that .
Thus, by the induction, for all . □
We give examples which show that two conditions “T is a reduced ring” and “T is a torsion-free -module” in Proposition 1 are not superfluous.
Example 1. Let be the ring of integers modulo 6. Then, is a reduced ring which is not a torsion-free -module. Note that in but in . Hence, the condition “T is a torsion-free -module" in Proposition 1 is essential.
Example 2. Let be a set of indeterminates over , I the ideal of generated by the set and . For an element , let denote the homomorphic image of h in R.
- (1)
Suppose that there exist an integer and an element such that . Then, . Since m is a unit in , ; thus, . Hence, R is a torsion-free -module.
- (2)
Note that ; thus, . Suppose to the contrary that . Then, for some ; thus, by an easy calculation, the constant term of is 0 and, for each , the coefficient of in is . Therefore, the coefficient of in is , which is absurd. Hence, . Thus, R is not a reduced ring.
- (3)
Let and be elements of . Then, but . This shows that the condition “T is a reduced ring" in Proposition 1 is essential.
Let
R be a commutative ring with identity and
the set of nonzero zero-divisors of
R. Recall that
satisfies the
McCoy condition if for any
, there exists a nonzero element
such that
([
2]
Section 4).
Proposition 2. If is an extension of commutative rings with identity, then the following assertions are equivalent.
- (1)
For each , there exists a nonzero element such that in .
- (2)
T is a torsion-free -module.
- (3)
satisfies the McCoy condition.
Proof. (1) ⇒ (2) Suppose that T is not a torsion-free -module. Then, there exist an integer and a nonzero element a of T such that ; thus, . Hence, . However, for any nonzero element t of T. This is a contradiction to the hypothesis. Thus, T is a torsion-free -module.
(2) ⇔ (3) The equivalence appears in [
2] (Theorem 4.1).
(3) ⇒ (1) Let . Then, . Since satisfies the McCoy condition, there exists a nonzero element such that . □
3. PF-Rings
Let be an extension of commutative rings with identity. In this section, we give necessary and sufficient conditions for the rings and to be PF-rings. To do this, we first study some relations among , , R and T in the view of PF-rings.
Lemma 2. Let be an extension of commutative rings with identity and let D be either or . If D is a PF-ring, then R and T are both PF-rings and torsion-free -modules.
Proof. Let and . Then, with . Since D is a PF-ring, there exists an element such that . Hence, and . Thus, T is a PF-ring. A similar argument also shows that R is a PF-ring.
Let and suppose that there exists an integer such that . Then, ; thus, . Since D is a PF-ring, we can find an element such that . Hence, and , which indicates that . Thus, T is a torsion-free -module. By Lemma 1(2), R is also a torsion-free -module. □
Let R be a commutative ring with identity. Then, it is obvious that, if R is a torsion-free -module, then . Hence, by Lemma 2, we obtain
Corollary 1. Let be an extension of commutative rings with identity. If , then neither nor is a PF-ring.
We give necessary and sufficient conditions for the composite Hurwitz series ring to be a PF-ring.
Theorem 1. If is an extension of commutative rings with identity, then the following statements are equivalent.
- (1)
is a PF-ring.
- (2)
T is a torsion-free -module and if and for all satisfy for all , then there exists an element such that and for all .
- (3)
For each such that , there exists an element such that and .
Proof. (1) ⇒ (2) Let
and
. Then, by the assumption,
and
. Since
is a PF-ring, there exists an element
such that
and
. By Lemma 2 and [
12] (Lemma 2.2),
T is both a reduced ring and a torsion-free
-module. Thus, by Proposition 1,
and
for all
.
(2) ⇒ (3) Let and be elements of such that . Note that, by the assumption, T is a PF-ring; thus, T is a reduced ring. Since T is a torsion-free -module, by Proposition 1, for all . Hence, we can find an element such that and for all . Thus, and .
(3) ⇒ (1) This implication comes from the definition of PF-rings. □
Let R be a commutative ring with identity and let D be either or . Recall that R is a Noetherian ring if every ideal of R is finitely generated. For an , the content ideal of f is the ideal of R generated by the set and is denoted by .
Corollary 2. Let be an extension of commutative rings with identity. If T is a Noetherian ring, then the following assertions are equivalent.
- (1)
is a PF-ring.
- (2)
T is a torsion-free -module and, for each and , there exists an element such that and .
Proof. (1) ⇒ (2) Suppose that is a PF-ring. Then, by Lemma 2, T is a torsion-free -module. Let and . Then, with ; thus, by Theorem 1, there exists an element such that and . Thus, and .
(2) ⇒ (1) Let and . Since T is a Noetherian ring, and for some . Note that, by the hypothesis, T is a PF-ring; thus, T is a reduced ring. Since T is a torsion-free -module, Proposition 1 indicates that for all and . Therefore, by the hypothesis, for each and , there exists an element such that and . For each , set . Then, for all and , and .
Let and, for each , let . Then, for all . By an iterative calculation, it can be shown that for each , and for all and . Hence, and for all and . Since and , and . Thus, by Theorem 1, is a PF-ring. □
By Theorem 1 and Corollary 2, we can regain
Corollary 3. ([
12] (Theorem 2.5 and Corollary 2.7)).
If R is a commutative ring with identity, then the following assertions hold.- (1)
is a PF-ring if and only if for each with , there exists an element such that and .
- (2)
If R is a Noetherian ring, then is a PF-ring if and only if R is a PF-ring which is a torsion-free -module.
Corollary 4. Let be an extension of commutative rings with identity. If is a PF-ring, then and are PF-rings.
Proof. Let f and g be elements of such that . Then, and are elements of such that . Since is a PF-ring, by Theorem 1, there exists an element such that and . Note that, by Theorem 1, T is a torsion-free -module; thus, and . Thus, by Corollary 3(1), is a PF-ring.
Let f and g be elements of such that . Then, f and g are elements of such that . Since is a PF-ring, by Theorem 1, there exists an element such that and . Thus, by Corollary 3(1), is a PF-ring. □
We next study when the composite Hurwitz polynomial ring is a PF-ring.
Theorem 2. If is an extension of commutative rings with identity, then the following statements are equivalent.
- (1)
is a PF-ring.
- (2)
T is a torsion-free -module and if and for all and are such that for all and , then there exists an element such that and for all and .
- (3)
For each such that , there exists an element such that and .
Proof. (1) ⇒ (2) Let
and
. Then, by the assumption,
and
. Since
is a PF-ring, we can find an element
such that
and
. By Lemma 2 and [
12] (Lemma 2.2),
T is both a torsion-free
-module and a reduced ring. Thus, by Proposition 1,
and
for all
and
.
(2) ⇒ (3) Let and be elements of such that . Note that, by the hypothesis, T is a PF-ring; thus, T is a reduced ring. Since T is a torsion-free -module, Proposition 1 implies that for all and . Hence, we can find an element such that and for all and . Thus, and .
(3) ⇒ (1) This implication follows directly from the definition of PF-rings. □
Corollary 5. If is an extension of commutative rings with identity, then the following assertions are equivalent.
- (1)
is a PF-ring.
- (2)
T is a torsion-free -module and, for each and , there exists an element such that and .
Proof. (1) ⇒ (2) Suppose that is a PF-ring. Then, by Lemma 2, T is a torsion-free -module. Let and . Then, with . Hence, by Theorem 2, there exists an element such that and . Thus, and .
(2) ⇒ (1) Let and be elements of such that . Note that, by the hypothesis, T is a PF-ring; thus, T is a reduced ring. Since T is a torsion-free -module, by Proposition 1, for all and . Therefore, by the hypothesis, for each and , we can find an element such that and . For each , set . Then, for all and , and .
Let and, for each , let . Then, a routine calculation shows that for each , and for all and . Therefore, and for all and . Hence, and . Note that . Thus, by Theorem 2, is a PF-ring. □
Corollary 6. If is an extension of commutative rings with identity, then the following assertions hold.
- (1)
If is a PF-ring, then is a PF-ring.
- (2)
If T is a Noetherian ring, then is a PF-ring if and only if is a PF-ring.
Proof. (1) This is an immediate consequence of Theorems 1 and 2.
(2) The equivalence follows directly from Corollaries 2 and 5. □
Corollary 7. (cf. [
12] Theorem 2.6).
If R is a commutative ring with identity, then the following conditions are equivalent.- (1)
is a PF-ring.
- (2)
R is both a PF-ring and a torsion-free -module.
- (3)
For each with , there exists an element such that and .
Proof. The equivalences come directly from Theorem 2 and Corollary 5. □
Corollary 8. Let be an extension of commutative rings with identity. If is a PF-ring, then and are PF-rings.
Proof. The result follows from Lemma 2 and Corollary 7. □
4. PP-Rings
Let be an extension of commutative rings with identity. In this section, we give equivalent conditions for the rings and to be PP-rings. Our first result in this section is a characterization of idempotent elements in and .
Let R be a commutative ring with identity and let be the set of idempotent elements of R.
Lemma 3. If is an extension of commutative rings with identity, then .
Proof. Clearly,
; thus, it remains to prove that
. Let
. Then,
. Note that
[
13] (Proposition 2.3). Thus,
. □
Let be an extension of commutative rings with identity. We next study PP-properties in terms of relations among , , R, and T.
Lemma 4. Let be an extension of commutative rings with identity and let D be either or . If D is a PP-ring, then R and T are both PP-rings and torsion-free -modules.
Proof. Note that any PP-ring is a PF-ring; thus, by Lemma 2, R and T are torsion-free -modules. Let . Since D is a PP-ring, by Lemma 3, there exists an element such that . Let . Then, ; thus, for some . Therefore, , which means that . The reverse containment is obvious. Hence, . Thus, T is a PP-ring. A similar argument shows that R is a PP-ring. □
Let R be a commutative ring with identity. If , then R is not a torsion-free -module. Hence, by Lemma 4, we obtain
Corollary 9. Let be an extension of commutative rings with identity. If , then neither nor is a PP-ring.
Lemma 5. Let R be a commutative ring with identity. If R is a reduced ring, then the following conditions hold.
- (1)
The relation defined on R by if and only if is a partial order.
- (2)
Let be a sequence of idempotent elements of R. If e is a least upper bound of , which is an idempotent element of R, and a is any upper bound of , which is an idempotent element of R, then .
- (3)
If a sequence of idempotent elements of R has a least upper bound in R which is an idempotent element, then it is unique.
- (4)
If each increasing sequence of idempotent elements of R has the least upper bound in R, which is an idempotent element, then each sequence of idempotent elements of R has the least upper bound in R, which is an idempotent element.
- (5)
Any finite sequence of idempotent elements of R has the least upper bound in R which is an idempotent element.
Proof. (1) This assertion was shown in [
12] (Lemma 3.8).
(2) By the assumption, and for all ; thus, for all . Therefore, is an upper bound of . Note that ; thus, and . Hence, by the minimality of e. Thus, .
(3) This is an immediate consequence of (2).
(4) This appears in [
14] (Lemma 2.5).
(5) The result can be shown by a similar argument as in the proof of [
14] (Lemma 2.5). □
Example 3. Let , and R the subring of T generated by and .
- (1)
Let be an increasing sequence in and, for each , let . For each , let Let . Then, such that e is the least upper bound of .
- (2)
For each and , let For each , let . Then, is an increasing sequence in . Suppose to the contrary that there exists the least upper bound of in . Then, for all ; thus, for all . Since , we can find an integer such that for all . Let m be the smallest even integer such that and, for each , let Let . Then, and for all . This contradicts the fact that a is the least upper bound of . Thus, does not have a least upper bound in .
Let R be a commutative ring with identity. Then, denotes the set of regular elements of R.
Lemma 6. If is an extension of commutative rings with identity, then the following assertions are equivalent.
- (1)
For each , there exists an element such that .
- (2)
For each , there exist and such that .
Proof. (1) ⇒ (2) This implication was shown in the remark in the proof of [
14] (Proposition 2.6).
(2) ⇒ (1) Let . Then, there exist and such that . Let . Then, ; thus, . Since t is regular in T, ; thus, . Note that ; thus, . Thus, . □
We are ready to give a necessary and sufficient conditions for the composite Hurwitz series ring to be a PP-ring.
Theorem 3. If is an extension of commutative rings with identity, then the following statements are equivalent.
- (1)
is a PP-ring.
- (2)
R is a PP-ring, T is a torsion-free -module, for each , there exists an element such that , and any sequence in admits the least upper bound in that belongs to .
- (3)
R is a PP-ring, T is a torsion-free -module, for each , there exists an element such that , and any increasing sequence in admits the least upper bound in that belongs to .
Proof. (1) ⇒ (2) Suppose that is a PP-ring. Then, by Lemma 4, R is a PP-ring and T is a torsion-free -module. Let . Then, by Lemma 3, there exists an element such that ; thus, . Therefore, . Hence, . Let . Then, ; thus, for some . Therefore, . Hence, . Thus, .
Let be a sequence of idempotent elements of R and let . Then, . Since is a PP-ring, by Lemma 3, there exists an element such that . Now, we show that is the least upper bound of . Since , for all ; thus, . Hence, for all . Let be such that for all . Then, ; thus, . Therefore, . Let be such that . Then, ; thus, . Hence, . Thus, is the least upper bound of .
(2) ⇒ (1) Let . Then, by the hypothesis and Lemma 6, for each , there exist and such that . In addition, by the assumption, there exist an element such that e is the least upper bound of ; thus, for all . Therefore, for all . Hence, , which implies that . For the reverse containment, let . Note that, by the hypothesis, T is a PP-ring; thus, T is a reduced ring. Since T is a torsion-free -module, by Proposition 1, for all . Note that, by the assumption and Lemma 6, for each , there exist and such that ; thus, for all , . Since and are regular elements of T for all , for all ; thus, for all . This shows that for all . Since e is the least upper bound of , Lemma 5(2) indicates that for all ; thus, for all . Therefore, for all , which shows that for all . Hence, , which implies that . Consequently, . Thus, is a PP-ring.
(2) ⇒ (3) This implication is clear.
(3) ⇒ (2) This implication was shown in Lemma 5(4). □
Lemma 7. Let R be a commutative ring with identity. If R is a reduced Noetherian ring and is an increasing sequence of , then there exists an integer such that is an upper bound of .
Proof. Let be an increasing sequence of and I the ideal of R generated by the set . Since R is a Noetherian ring, for some . Let k be an integer greater than n. Then, for some . Note that for all ; thus, . Hence, . Thus, is an upper bound of . □
Corollary 10. Let be an extension of commutative rings with identity. If R is a Noetherian ring, then is a PP-ring if and only if R is a PP-ring, T is a torsion-free -module and, for each , there exists an element such that .
Proof. The equivalence comes directly from Theorem 3 and Lemma 7. □
Corollary 11. (cf. [
12] Theorem 3.10).
If R is a commutative ring with identity, then the following assertions hold.- (1)
is a PP-ring if and only if R is both a PP-ring and a torsion-free -module and any (increasing) sequence in admits the least upper bound in that belongs to .
- (2)
If R is a Noetherian ring, then is a PP-ring if and only if R is both a PP-ring and a torsion-free -module.
Proof. (1) The equivalence is an immediate consequence of Theorem 3.
(2) The equivalence is an immediate consequence of (1) and Lemma 7. □
Corollary 12. If is an extension of commutative rings with identity, then the following assertions hold.
- (1)
is a PP-ring if and only if is a PP-ring, T is a torsion-free -module and, for each , there exists an element such that .
- (2)
If is a PP-ring, then is a PP-ring.
Proof. (1) This result follows directly from suitable combinations of Lemma 1(2), Theorem 3 and Corollary 11(1).
(2) Suppose that is a PP-ring. Then, by Lemma 4, T is both a PP-ring and a torsion-free -module. Let be a sequence in and let . Then, . Since is a PP-ring, Lemma 3 guarantees the existence of an element such that . We now show that is the least upper bound of . Since , for all ; thus, for all . Hence, for all . Let be such that for all . Then, ; thus, . Therefore, . Let be such that . Then, for some . Note that ; thus, . Hence, is the least upper bound of that belongs to . Thus, by Corollary 11(1), is a PP-ring. □
We next study the equivalent condition for the composite Hurwitz polynomial ring to be a PP-ring.
Theorem 4. If is an extension of commutative rings with identity, then the following statements are equivalent.
- (1)
is a PP-ring.
- (2)
R is a PP-ring, T is a torsion-free -module and, for each , there exists an element such that .
Proof. (1) ⇒ (2) Suppose that is a PP-ring. Then, by Lemma 4, R is a PP-ring and T is a torsion-free -module. Let . Then, by Lemma 3, there exists an element such that ; thus, . Hence, . Let . Then, ; thus, for some . Hence, , which shows that . Thus, .
(2) ⇒ (1) Let . Then, by the hypothesis and Lemma 6, for each , there exist and such that . Note that, by Lemma 5(5), has the least upper bound in R which is an idempotent element. Let be such that d is the least upper bound of . Then, for all . Hence, , which means that . For the reverse containment, let . Note that, by the hypothesis, T is a PP-ring; thus, T is a reduced ring. Since T is a torsion-free -module, Proposition 1 indicates that for all and . Note that, by the hypothesis and Lemma 6, for each , there exist and such that . Since and are regular elements of T and for all and , for all and ; thus, , or equivalently, for all and . Since d is the least upper bound of , Lemma 5(2) guarantees that for all ; thus, for all . Therefore, . Hence, . Consequently, . Thus, is a PP-ring. □
Corollary 13. If is an extension of commutative rings with identity, then the following assertions hold.
- (1)
If is a PP-ring, then is also a PP-ring.
- (2)
If R is a Noetherian ring, then is a PP-ring if and only if is a PP-ring.
Proof. (1) The equivalence follows directly from Theorems 3 and 4.
(2) This equivalence comes from Corollary 10 and Theorem 4. □
Corollary 14. ([
12] Theorem 3.7).
Let R be a commutative ring with identity. Then, is a PP-ring if and only if R is both a PP-ring and a torsion-free -module. Proof. The equivalence comes directly from Theorem 4. □
Corollary 15. If is an extension of commutative rings with identity, then the following assertions hold.
- (1)
is a PP-ring if and only if is a PP-ring, T is a torsion-free -module and, for each , there exists an element such that .
- (2)
If is a PP-ring, then is a PP-ring.
Proof. (1) This equivalence can be obtained by suitable combinations of Lemma 1(2), Theorem 4 and Corollary 14.
(2) This result follows directly from Lemma 4 and Corollary 14. □