On modules related to McCoy modules

Proposition 2.1

If an R-module M is Gaussian, then M is a semicommutative module.

Proof

Let m ∈ M and a ∈ R such that m a = 0 . Since M is Gaussian, m R a ⊆ m R a R = m a R = { 0 } . Therefore, M is semicommutative.□

Remark 2.2

1. Note that R = Z 4 is a local ring with J ( R ) 2 = { 0 } , where J ( R ) is the Jacobson radical of R . Thus, the commutativity of R guarantees that R is Gaussian according to [24, Proposition 1.8]. But R is not reduced. Hence, Gaussian modules may not be reduced.

2. As stated in [25, p. 2268], the polynomial ring over Z , Z [ x ] is not Gaussian since Z [ x ] is not Prüfer by [26, Theorem 7.7]. However, Z [ x ] is a reduced module over itself. Thus, reduced modules do not imply Gaussian in general.

3. Referring to [25, Example 11], symmetric modules need not be Gaussian modules.

4. There exists a duo module which is not Gaussian by (2).

Example 2.3

Let R = Z 2 and M = m 1 R ⊕ m 2 R be the free R -module with basis { m 1 , m 2 } . One can easily check that M is Gaussian. Consider the R -module endomorphism h of M induced by h ( m 1 ) = m 2 and h ( m 2 ) = m 1 . Then, h ( m 1 R ) = m 2 R ⊈ m 1 R . Thus, M is not duo.

Example 2.4

Let E = Z 2 ⟨ a , b ⟩ be the free algebra with identity and noncommuting indeterminates a , b over Z 2 . Set R = E / I where I is the ideal of E generated by the relation b a = 1 . We denote r = r + I for simplicity. Now consider the cyclic free R -module L = m R and the R -submodule N = m a R of L . Then, the factor module M = L / N is isomorphic to R / a R as an R -module. Note that each element r ∈ R can be written in the following form:

where α ∈ Z 2 and g 0 ( x ) , … , h ( x ) ∈ Z 2 [ x ] . If h ( x ) = ∑ j = 0 t β j x j with β t ≠ 0 , then

Therefore, every R -submodule of R / a R contains 1. Thus, M ≅ R / a R is a simple R -module, which forces that M is a duo module. Finally, we claim that M is not abelian. Let e = a b in R . Then, e 2 = e in R and m ¯ e b = ( m ¯ a ) b 2 = 0 ¯ in M . But m ¯ b e = m ¯ ( b a ) b = m ¯ b ≠ 0 ¯ in M . Hence, M is not abelian, which implies that M is not semicommutative by [22, Lemma 2.7].

Example 2.5

Let E = Z 2 ⟨ a 0 , a 1 , a 2 , b 0 , b 1 , c ⟩ be the free algebra with identity and noncommuting indeterminates as labeled above over Z 2 . Set R = E / I where I is the ideal of E generated by the following relations: for each 0 ≤ i ≤ 2 , 0 ≤ j , k ≤ 1 ,

We denote r = r + I for simplicity. Applying Bergman’s diamond lemma [27], we can write each r ∈ R in the following unique reduced form:

where α ∈ Z 2 and f 0 ( x ) , … , f 4 ( x ) , g ( x ) , h 0 ( p ) ( x ) , … , h 4 ( p ) ( x ) ∈ Z 2 [ x ] for each p ∈ N . Note that if r c ∈ c R for some r ∈ R , then r must be zero or identity.

Let L = y R be the cyclic free R -module. Set M = L / N , where N = { 0 , y c , y c 2 , … } = y c R is the R -submodule of L generated by the relation y c = 0 . Then, we can write each m ¯ ∈ M uniquely in the following form:

First, we claim that an R -module M is not McCoy. Consider the following nonzero polynomials m ( x ) = y ¯ a 0 + y ¯ a 1 x + y ¯ a 2 x 2 ∈ M [ x ] and r ( x ) = b 0 + b 1 x ∈ R [ x ] . Then, m ( x ) r ( x ) = 0 by the first row in the relations. But there is no nonzero element r ∈ R such that y ¯ a 1 r = 0 by the relations. Thus, M is not McCoy.

Finally, we claim that M is a duo R -module. Referring to [11, Lemma 1.1], we will find all R -endomorphism of M . Since M is still a cycle R -module, for any R -endomorphism, we need to examine the image of y ¯ . Consider the correspondence h : M → M defined by h ( y ¯ r ) = y ¯ a 0 r for any r ∈ R . Note that h ( y ¯ 0 ) = y ¯ a 0 0 = 0 ¯ ≠ y ¯ a 0 c = h ( y ¯ c ) despite 0 + N = y c + N . Therefore, h is not well-defined. From the fact that if r c ∈ c R for some r ∈ R , then r must be zero or identity in R , we can notice that if h ′ is an R -endomorphism of M induced by h ′ ( y ¯ ) = y ¯ r ′ for some r ′ ∈ R , then r ′ must be zero or identity in R by the well-definedness. This means that if h ′ is an R -endomorphism of M , then h ′ must be either the zero or the identity endomorphism. Thus, M must be duo by [11, Lemma 1.1].

Remark 2.6

Based on Example 2.5, one may suspect that every duo module is cyclic. However, the example in [28, Section 3] eliminates the possibility of the suspicion.

Example 2.7

Let E = Z 2 ⟨ a 0 , a 1 ⟩ be the free algebra with identity and indeterminates a 0 , a 1 over Z 2 . Set R = E / I where I is the ideal of E generated by the following relations:

We simply identify r = r + I in R . Applying the diamond lemma [27], we can write each element r ∈ R in the following reduced form:

where α , α 0 , α 1 ∈ Z 2 . Now, let L = m 0 R ⊕ m 1 R ⊕ m 2 R be the free R -module with basis { m 0 , m 1 , m 2 } . Set M = L / N where N is the submodule of L generated by the following relations:

We also denote m = m + N in M for simplicity. Then, we can write each element m ∈ M uniquely in the following form:

where β 0 , β 1 , β 2 , γ 0 , γ 1 ∈ Z 2 . Since R is commutative, M is symmetric. Next, we claim that M is not McCoy. Consider the nonzero polynomials m ( x ) = m 0 + m 1 x + m 2 x 2 ∈ M [ x ] and r ( x ) = a 0 + a 1 x ∈ R [ x ] . Then, m ( x ) r ( x ) = 0 by the relations. It suffices to show that ann R ( m 1 ) = { 0 } . Suppose that m 1 r = 0 for some r ∈ R and write r = α + α 0 a 0 + α 1 a 1 . Then, m 1 r = m 1 ( α + α 0 a 0 + α 1 a 1 ) = 0 implies that α = α 0 = α 1 = 0 , indicating that r = 0 as required.

Example 2.8

Let R = Z 2 and let M = m 1 R ⊕ m 2 R ⊕ ⋯ be the infinite free module over R . Note that the R -module M is reduced and Gaussian. However, by [30, Exercise 4.3.3], M is injective. This implies that M is not a Dedekind finite module by [31, Proposition 5.7].

Definition 3.1

Let M be a right module over a ring R and M [ x ] be the corresponding polynomial module over R [ x ] . M is referred to as ZPZC if ann R [ x ] ( m ( x ) ) ≠ { 0 } implies ann R ( m i ) ≠ { 0 } for each m i , where m ( x ) = ∑ i = 0 p m i x i ∈ M [ x ] .

Example 3.2

Let E = Z 2 ⟨ a 0 , a 1 , b 0 , b 1 , b 2 ⟩ be the free algebra over Z 2 with identity and commuting indeterminates a 0 , a 1 , b 0 , b 1 , b 2 . Set R = E / I where I is the ideal of E generated by the following relations: for each i , j ∈ { 0 , 1 } and k , ℓ ∈ { 0 , 1 , 2 } ,

We identify r = r + I for simple expression. Then, we can write each element r ∈ R uniquely in the following reduced form:

where α , α 0 , α 1 , β 0 , β 1 , β 2 ∈ Z 2 . Now let L = m 0 R ⊕ m 1 R ⊕ n 0 R ⊕ n 1 R ⊕ n 2 R be the free R -module with basis { m 0 , m 1 , n 0 , n 1 , n 2 } . Set M = L / N where N is the R -submodule of L generated by the following relations:

For every m ∈ M , we also identify m = m + N for simple expression. Then, we can write each element m ∈ M uniquely in the following form:

where γ 0 , γ 1 , γ 2 , δ 0 , … , δ 6 ∈ Z 2 . For any m ∈ M , we can easily check that m must be annihilated by one of the elements in { b 0 , b 1 , b 2 } . Thus, M is ZPZC. Finally, consider the nonzero polynomials m ( x ) = m 0 + m 1 x + n 0 x 2 + n 1 x 3 + n 2 x 3 ∈ M [ x ] and r ( x ) = a 0 + a 1 x ∈ R [ x ] . Then, m ( x ) r ( x ) = 0 . But since ann R ( m 1 ) ⋂ ann R ( n 1 ) = { 0 } , we can conclude that M is not McCoy.

Theorem 3.3

(cf. [7, Proposition 2.4]). If a ring R is reduced and right π -duo, then every right R-module M is McCoy.

Proof

Let m ( x ) = ∑ i = 0 p m i x i ∈ M [ x ] and r ( x ) = ∑ j = 0 q r j x j ∈ R [ x ] \ { 0 } with m ( x ) r ( x ) = 0 and r 0 ≠ 0 . Since R is right π -duo, r 1 r 0 n = r 0 r 1 ′ for some n ∈ N and r 1 ′ ∈ R . From m 0 r 1 + m 1 r 0 = 0 , we obtain that

Similarly, from r 2 r 0 n 2 + n = r 0 n + 1 r 2 ′ and r 1 r 0 n 2 + n = r 0 n + 1 r 1 ″ for some r 2 ′ , r 1 ″ ∈ R , we obtain

Continuing in this manner, we arrive at the conclusion that m ( x ) r 0 k = 0 for some k ∈ N .□

Remark 3.4

According to [32, Proposition 1.9(4)], right π -duo rings are abelian. From the fact that every reduced ring is abelian, the converse of Theorem 3.3 does not hold true in general by [5, Theorem 7.1]. Moreover, the hypothesis of Theorem 3.3 is not superfluous by the following examples.

1. In Example 2.7, the R -module M is not McCoy where the ring R is right π -duo but not reduced.

2. Let R = Z 2 ⟨ a 0 , a 1 ⟩ be the free algebra with identity and noncommuting indeterminates a 0 , a 1 over Z 2 . Clearly, R is a domain and so reduced. Note that R is not right π -duo. Let L = m 0 R ⊕ m 1 R ⊕ m 2 R be the free R -module with basis { m 0 , m 1 , m 2 } . Set M = L / N where N is the submodule of L generated by the following relations:

   m 0 a 0 = 0 , m 0 a 1 + m 1 a 0 = 0 , m 1 a 1 + m 2 a 0 = 0 , m 2 a 1 = 0 .

Corollary 3.5

Let J ( R ) be the Jacobson radical of a ring R and M be a right module over R. Then, M is McCoy if one of the following conditions holds:

1. R is right π -duo with J ( R ) = { 0 } .

2. R is weakly right duo with J ( R ) = { 0 } .

3. R is semiprime and right duo.

4. R is semiprime and right Gaussian.

5. R is commutative with J ( R ) = { 0 } .

6. R is commutative semiprime.

7. R is a commutative domain.

Example 3.6

Let E = Z 2 ⟨ a 0 , a 1 , s ⟩ be the free algebra with identity and commuting indeterminates a 0 , a 1 , s over Z 2 . Set R = E / I where I is the ideal of E generated by the relations:

We denote r = r + I for simplicity. Then, we can write each r ∈ R in the following reduced form:

where α , α 0 , α 1 , β ∈ Z 2 . One can easily show that if an ideal A of R is nonprincipal, then either A s ⊆ { 0 } or A s ⊆ s R ⊆ A . Therefore, R is a commutative S -principal ideal ring, where S = { 1 , s } . Now let L = m 0 R ⊕ m 1 R ⊕ m 2 R be the free R -module with basis { m 0 , m 1 , m 2 } . Set M = L / N where N is the R -submodule of L generated by the following relations:

Consider the following nonzero polynomials m ( x ) = m 0 ¯ + m 1 ¯ x + m 2 ¯ x 2 ∈ M [ x ] and r ( x ) = a 0 + a 1 x ∈ R [ x ] . Then, m ( x ) r ( x ) = 0 by the relations. However, ann R ( m 1 ¯ ) = { 0 } . Hence, M R is not ZPZC.

Proposition 3.7

(see Example 4.4(3)). If R is a right uniserial ring and M is a ZPZC right R-module, then M is McCoy.

Proof

Let m ( x ) = ∑ i = 0 p m i x i ∈ M [ x ] and r ( x ) = ∑ j = 0 q r j x j ∈ R [ x ] \ { 0 } with m ( x ) r ( x ) = 0 . Then, the ZPZC condition guarantees that each ann R ( m i ) is nonzero. Since R is right uniserial, we can find a nonzero element c k such that c k ∈ ann R ( m k ) ⊆ ann R ( m i ) for all i . Hence, m i c k = 0 for all i , which implies that M is McCoy.□

Question 3.8

Can we delete the condition that M is ZPZC in Proposition 3.7?