Abstract
In this paper we introduce and study right \(\mathcal {Z}\)-Armendariz rings. A ring R is said to be right \(\mathcal {Z}\)-Armendariz if f(x)g(x) = 0 implies that ab is a right singular element of R, where f(x) and g(x) belong to R[x] and a, b are arbitrary coefficients of f(x), g(x). Then we construct some examples of right \(\mathcal {Z}\)-Armendariz rings by a given one. Finally, we extend this notion for modules.
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1 Introduction
In this paper, all rings are associative with identity 1 ≠ 0 and all modules are unital. Let R be a ring. The set of nilpotent elements of R is denoted by Nil(R). A right ideal I of R is essential, if \(I \cap I^{\prime } \ne 0\) for any nonzero right ideal \(I^{\prime }\) of R. An element x ∈ R is called right singular, if annr(x) = {a ∈ R | xa = 0} is an essential right ideal of R. The set of all right singular elements of R is a two-sided ideal and is denoted by \(\mathcal {Z}(R_{R})\).
In [8], Rege and Chhawachharia introduced the notion of Armendariz rings. A ring R is Armendariz, if whenever \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\) and \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j}\) are in R[x], the equation f(x)g(x) = 0 implies that aibj = 0 for every \(i=0,1,\dots ,m\) and \(j=0,1,\dots ,n\). In [3, Lemma 1] the authors proved that every reduced ring is Armendariz and in [6, Lemma 7] it is proved that every Armendariz ring is Abelian. Motivated by this definition, we call a ring R right \(\mathcal {Z}\)-Armendariz, if the above equation implies that \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\). It turns out that this notion is not left-right symmetric. We prove that the property of being right \(\mathcal {Z}\)-Armendariz is closed under direct products and finite subdirect products but it is not a Morita invariant property. By an example we show that this property is not preserved under homomorphic images. Also we will prove that a ring R is right \(\mathcal {Z}\)-Armendariz if and only if the polynomial ring R[x] is so. However, if R is right \(\mathcal {Z}\)-Armendariz, then R[[x]], the ring of formal power series over R, is not necessarily right \(\mathcal {Z}\)-Armendariz.
A right R-module MR is called Armendariz ([1, Proposition 12]), if f(x)g(x) = 0 implies that mirj = 0 for any \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\), where \(f(x)= {\sum }_{i=0}^{m} m_{i}x^{i} \in M[x]\) (the corresponding polynomial module over R[x]) and \(g(x)={\sum }_{j=0}^{n} r_{j}x^{j} \in R[x]\). Generalizing this notion, an R-module MR is called \(\mathcal {Z}\)-Armendariz, if the above equation implies that \(m_{i}r_{j} \in \mathcal {Z}(M_{R})\) for every \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). We show that an R-module M is \(\mathcal {Z}\)-Armendariz if and only if every (finitely generated) submodule of it is \(\mathcal {Z}\)-Armendariz, and we prove that every right module over a right duo-ring is \(\mathcal {Z}\)-Armendariz. It is proved that the class of \(\mathcal {Z}\)-Armendariz modules is closed under direct sums but it is not closed under infinite direct products. Also it turns out that when R is a right \(\mathcal {Z}\)-Armendariz ring, flat R-modules and also semisimple R-modules are \(\mathcal {Z}\)-Armendariz.
2 \(\mathcal {Z}\)-Armendariz Rings
In this section, we focus on right \(\mathcal {Z}\)-Armendariz rings and prove some related results. Then we construct some examples of right \(\mathcal {Z}\)-Armendariz rings.
Definition 1
A ring R is called right \(\mathcal {Z}\)-Armendariz, if for every \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\) and \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j}\) in R[x], the equation f(x)g(x) = 0 implies that \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\) for every \(i=0,1,\dots ,m\) and \(j=0,1,\dots ,n\).
We define left \(\mathcal {Z}\)-Armendariz rings similarly. If a ring R is both left and right \(\mathcal {Z}\)-Armendariz, then we say that R is a \(\mathcal {Z}\)-Armendariz ring.
Obviously every Armendariz ring is \(\mathcal {Z}\)-Armendariz. On the other hand, if R is a right \(\mathcal {Z}\)-Armendariz ring which is right nonsingular, then clearly it is Armendariz. In the following example we show that every commutative ring is \(\mathcal {Z}\)-Armendariz and in Example 4, we generalize this result.
Example 1
Every commutative ring R is \(\mathcal {Z}\)-Armendariz.
Let \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R [x]\) and f(x)g(x) = 0, which implies that
So \(a_{0}b_{0} = 0 \in \mathcal {Z}(R)\). Multiplying the equation a0b1 + a1b0 = 0 by a1b0, we have (a1b0)2 = 0. Since all nilpotent elements of a commutative ring are singular, the nilpotent elements a1b0 and a0b1 belong to \(\mathcal {Z}(R)\). Now multiplying the equation a0b2 + a1b1 + a2b0 = 0 by a2b0, we have (a2b0)2 = −a2b0a1b1 ∈Nil(R) so that \(a_{2}b_{0} \in \text {Nil}(R) \subseteq \mathcal {Z}(R)\). By continuing this processes we obtain that \(a_{i}b_{j} \in \text {Nil}(R) \subseteq \mathcal {Z}(R)\), for every \(i= 0,1,\dots , m\) and \(j=0,1, \dots ,n\). So R is \(\mathcal {Z}\)-Armendariz.
The following example shows that a (commutative) \(\mathcal {Z}\)-Armendariz ring need not be Armendariz.
Example 2
Let \(R= \mathbb {Z}_{8} (+) \mathbb {Z}_{8}\) with componentwise addition and multiplication \((a,b)(a^{\prime },b^{\prime }) = (aa^{\prime }, ab^{\prime } + ba^{\prime })\). By [8, Example 3.2], R is not Armendariz and by Example 1, it is \(\mathcal {Z}\)-Armendariz.
Example 3
For any ring R and n ≥ 2, Mn(R), the ring of all n × n matrices and also the ring of all n × n upper (lower) triangular matrices over R are not right \(\mathcal {Z}\)-Armendariz.
Let S = Mn(R) and Eij ∈ S be the matrix unit with 1 in the (i, j)th entry and 0 elsewhere. Let f(x) = E12 + E11x and g(x) = E12 − E22x ∈ S[x]. We have f(x)g(x) = 0, but \(E_{11}E_{12} = E_{12} \notin \mathcal {Z}(S_{S})\), since annr(E12) ∩ E22S = 0. A similar proof can be used for the ring of n × n upper (lower) triangular matrices over R.
Proposition 1
Let {Ri}i∈Ibe a family of rings and\(R= {\prod }_{i \in I} R_{i} \). Then R is right\(\mathcal {Z}\)-Armendariz if and only if each Ri is so.
Proof
The proof follows from the fact that \(\mathcal {Z}(R_{R})= {\prod }_{i \in I} \mathcal {Z}({R_{i}}_{R_{i}})\). □
To show that the class of right \(\mathcal {Z}\)-Armendariz rings is closed under finite subdirect products, we need the following lemma.
Lemma 1
Let\(I_{1},\dots , I_{t} \)be ideals of a ring R such that\(\cap _{k=1}^{t} I_{k}=0 \). If x + Ikis a right singular element of the ring\(\frac {R}{I_{k}}\)for each\(k=1, \dots , t\), then\(x \in \mathcal {Z}(R_{R})\).
Proof
Let 0≠y ∈ R. Since \(\cap _{k=1}^{t} I_{k}=0\), we can assume that y∉I1. So there exists r1 ∈ R such that yr1∉I1 and xyr1 ∈ I1. If yr1 ∈ Ik for \( i=2, \dots , t\), then \(xyr_{1} \in \cap _{k=1}^{t} I_{k}=0\). If yr1∉I2, then there exists r2 ∈ R such that yr1r2∉I2 and xyr1r2 ∈ I1 ∩ I2. By continuing this process, we can find r ∈ R with yr≠ 0 and xyr = 0. Thus, \(x \in \mathcal {Z}(R_{R})\). □
Theorem 1
A finite subdirect product of right\(\mathcal {Z}\)-Armendariz rings is right\(\mathcal {Z}\)-Armendariz.
Proof
Suppose that I1,⋯ , It are ideals of a ring R such that \(\cap _{k=1}^{t} I_{k}=0\) and for each \(k=1, \dots , t\), the ring \(\frac {R}{I_{k}}\) is right \(\mathcal {Z}\)-Armendariz. Let \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R[x]\) and f(x)g(x) = 0. Then aibj + Ik is a right singular element of the ring \(\frac {R}{I_{k}}\), for all \(k=1, \dots , t\). So by Lemma 1, \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\) for \( i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). Therefore, R is right \(\mathcal {Z}\)-Armendariz. □
Suppose that \(I_{1}, \dots , I_{n}\) are ideals of a ring R such that \(\frac {R}{I_{1}}, \dots , \frac {R}{I_{n}}\) are right \(\mathcal {Z}\)-Armendariz rings. Then \(\frac {R}{\cap _{k=1}^{n} I_{k}}\), as a subdirect product of \(\frac {R}{I_{1}}, \dots , \frac {R}{I_{n}}\) is right \(\mathcal {Z}\)-Armendariz.
Remark 1
In general, a subdirect product of right \(\mathcal {Z}\)-Armendariz rings is not necessarily right \(\mathcal {Z}\)-Armendariz. For example, let \(R=\left [\begin {array}{cc} \mathbb {Z}& \mathbb {Z}\\ 0 & \mathbb {Z} \end {array}\right ]\). By Example 3, R is not right \(\mathcal {Z}\)-Armendariz. For any n ≥ 1, suppose that \(I_{n}= \left [\begin {array}{cc} 0&n\mathbb {Z}\\ 0&0 \end {array}\right ]\). Then \(\cap _{n=1}^{\infty } I_{n}=0\), which implies that R is a subdirect product of \(\left \{\frac {R}{I_{n}}\right \}_{n=1}^{\infty }\). If \(R_{n} := \frac {R}{I_{n}}= \left [\begin {array}{cc} \mathbb {Z}& \mathbb {Z}_{n}\\ 0&\mathbb {Z} \end {array}\right ]\), then \(\mathcal {Z}({R_{n}}_{R_{n}}) = \left [\begin {array}{cc} 0& \mathbb {Z}_{n}\\ 0&0 \end {array}\right ]\). So \(\frac {R_{n}}{\mathcal {Z}({R_{n}}_{R_{n}})}\) is reduced and by [3, Lemma 1] it is Armendariz. As we shall see in Proposition 4, each \(\frac {R}{I_{n}}\) is right \(\mathcal {Z}\)-Armendariz for any n ≥ 1.
In the sequel, we use the following observation. Let R be a ring and S = R[X], where X is a set of commuting indeterminates over R. Then \(\mathcal {Z}(S_{S})= \mathcal {Z}(R_{R})[X]\), see [7, Exercise 7.35].
Proposition 2
Let R be a ring. Then R is a right\(\mathcal {Z}\)-Armendariz ring if and only if R[x] is so.
Proof
For the “only if part” let R be a right \(\mathcal {Z}\)-Armendariz ring and f(t) = f0 + f1t + ⋯ + fntn, g(t) = g0 + g1t + ⋯ + gmtm ∈ R[x][t] and f(t)g(t) = 0, where fi, gj ∈ R[x] for each \(i=0, 1, \dots , n\) and \(j=0, 1, \dots ,m\). We show that \(f_{i}g_{j} \in \mathcal {Z}(R[x]_{R[x]})\). Let k = degf0 + ⋯ + degfn + degg0 + ⋯ + deggm. Then f(xk) = f0 + f1xk + ⋯ + fnxkn, g(xk) = g0 + g1xk + ⋯ + gmxkm ∈ R[x] and f(xk)g(xk) = 0. So the product of each coefficient of fi with every coefficient of gj belongs to \(\mathcal {Z}(R_{R})\). Thus, \(f_{i}g_{j} \in \mathcal {Z}(R[x]_{R[x]})\).
For the “if part” suppose that the polynomial ring R[x] is right \(\mathcal {Z}\)-Armendariz and \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R[x]\) such that f(x)g(x) = 0. Consider \(F(t)= {\sum }_{i=0}^{m} f_{i} t^{i}\) and \(G(t)= {\sum }_{j=0}^{n} g_{j}t^{j} \in R[x][t]\), where fi = aixi and gj = bjxj. We have F(t)G(t) = 0, so that \(f_{i}g_{j} \in \mathcal {Z}(R[x]_{R[x]})\) which implies that \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\), for \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). Thus, R is right \(\mathcal {Z}\)-Armendariz. □
Corollary 1
A ring R is right\(\mathcal {Z}\)-Armendariz if and only if the polynomial ring S = R[{xα}α∈A] is right\(\mathcal {Z}\)-Armendariz.
Proof
Let R be a right \(\mathcal {Z}\)-Armendariz ring and f, g ∈ R[{xα}α∈A][t] with fg = 0. Then \(f, g \in T[t]=R[x_{\alpha _{1}}, \dots , x_{\alpha _{n}}][t]\) for some finite subset \(\{\alpha _{1}, \dots , \alpha _{n}\} \subseteq A\). By Proposition 2, the ring \(R[x_{\alpha _{1}}, \dots , x_{\alpha _{n}}]\) is right \(\mathcal {Z}\)-Armendariz, so that \(ab \in \mathcal {Z}(T_{T}) \subseteq \mathcal {Z}(S_{S})\) for each coefficient a of f and b of g. Therefore, S is right \(\mathcal {Z}\)-Armendariz. The converse is trivial. □
Remark 2
If R is a right \(\mathcal {Z}\)-Armendariz ring, then S = R[[x]], the formal power series ring over R, is not necessarily right \(\mathcal {Z}\)-Armendariz. For example, let K be a field and \(R= \frac {K \langle a,b \rangle }{\langle b^{2} \rangle }\). In [2, Example 1], it is shown that R is an Armendariz ring but R[[x]] is not. We show that S is not right \(\mathcal {Z}\)-Armendariz. Let u = (1 − ax) ∈ S. Clearly u is a unit in S with u− 1 = (1 + ax + a2x2 + a3x3 + ⋯ ) ∈ S and f = ubu− 1 is such that f2 = 0. In the polynomial ring S[y], (b + bfy)(b − fby) = 0 but \(bfb \notin \mathcal {Z}(S_{S})\), since annr(bfb) ∩ aS = 0. Hence, S = R[[x]] is not right \(\mathcal {Z}\)-Armendariz. Also S is an example of an Abelian ring which is not right \(\mathcal {Z}\)-Armendariz.
Proposition 3
Let R be a ring and G be a group. If the group ring RG or R[[x]] is right\(\mathcal {Z}\)-Armendariz, then so is R.
Proof
Let S be one of the rings RG or R[[x]]. We can show that \(\mathcal {Z}(S_{S}) \cap R \subseteq \mathcal {Z}(R_{R})\). Now the rest of the proof follows easily. □
Proposition 4
Let Ibe an ideal of a ring R such that the factor ring\(\bar {R} = \frac {R}{I}\)is Armendariz. Then for\(f_{1},f_{2}, \dots ,f_{n} \in R[x]\)the equation\(f_{1}f_{2} {\dots } f_{n} \in I[x]\)implies that\(a_{1}a_{2} {\dots } a_{n} \in I\), where aiis an arbitrary coefficient of fifor\(i=1, 2, \dots ,n\). In particular, if\(I \subseteq \mathcal {Z}(R_{R})\), then R is right\(\mathcal {Z}\)-Armendariz.
Proof
Suppose that \(f_{1}, f_{2}, \dots , f_{n} \in R[x]\) such that \(f_{1}f_{2} {\dots } f_{n} \in I[x]\). Then in \(\bar {R}[x]\), we have \(\bar {f_{1}} \bar {f_{2}} {\dots } \bar {f_{n}} = 0\). By [1, Proposition 1], \(a_{1}a_{2} {\dots } a_{n} \in I\) where ai is an arbitrary coefficient of fi for \(i=1, 2, \dots ,n\). □
Corollary 2
Let R be a ring. If Nil(R) is an ideal of R contained in\(\mathcal {Z}(R_{R})\), then R is right\(\mathcal {Z}\)-Armendariz.
Proof
The factor ring \(\frac {R}{\text {Nil}(R)}\) is reduced and by [3, Lemma 1], it is Armendariz. So by Proposition 4, R is right \(\mathcal {Z}\)-Armendariz. □
Recall that a ring R is right duo, if all right ideals are two-sided, also a ring R is called reversible, if ab = 0 implies that ba = 0 for all a, b ∈ R.
Example 4
Right duo rings and reversible rings are examples of right \(\mathcal {Z}\)-Armendariz rings. By an easy calculation, we can show that \(\frac {R}{\mathcal {Z}(R_{R})}\) is reduced, whenever R is a right duo or a reversible ring. So by [3, Lemma 1], it is Armendariz. Now, applying Proposition 4, we get that R is right \(\mathcal {Z}\)-Armendariz.
The next example shows that for a ring R, being \(\mathcal {Z}\)-Armendariz is not left-right symmetric and also it is not preserved under homomorphic images.
Example 5
Let \(R=\left [\begin {array}{cc} \mathbb {Z}_{2} &\mathbb {Z}_{2}\\0&\mathbb {Z}_{4} \end {array}\right ]\). Since \(\mathcal {Z}(R_{R}) = \left [\begin {array}{cc} 0&\mathbb {Z}_{2}\\0&2 \mathbb {Z}_{4} \end {array}\right ] = \text {Nil}(R)\), Corollary 2 implies that R is right \(\mathcal {Z}\)-Armendariz. However, it is not left \(\mathcal {Z}\)-Armendariz, because for f(x) = E12 + E11x and g(x) = E12 − E22x ∈ R[x], where Eij’s are those introduced in Example 3, we have f(x)g(x) = 0, but \(E_{11}E_{12} =E_{12} \notin \mathcal {Z}(_{R}R)\), since annl(E12) ∩ RE11 = 0. Note that R is an example of a noncommutative right \(\mathcal {Z}\)-Armendariz ring which is not Armendariz. Moreover, let \(I= \left [\begin {array}{cc} 0&0\\0&2 \mathbb {Z}_{4} \end {array}\right ]\). Then \(\frac {R}{I}\) is isomorphic to \(\left [\begin {array}{cc} \mathbb {Z}_{2}&\mathbb {Z}_{2}\\ 0&\mathbb {Z}_{2} \end {array}\right ]\) which is not right \(\mathcal {Z}\)-Armendariz by Example 3. Therefore, a homomorphic image of a right \(\mathcal {Z}\)-Armendariz ring need not be right \(\mathcal {Z}\)-Armendariz.
Every Armendariz ring is Abelian [6, Lemma 7]. But a \(\mathcal {Z}\)-Armendariz ring is not necessarily Abelian. For example, let \(R= \left [\begin {array}{cc} \mathbb {Z}_{4} & 2\mathbb {Z}_{4}\\ 0 & \mathbb {Z}_{4} \end {array}\right ]\). We have \(\mathcal {Z}(R_{R})= \mathcal {Z}(_{R}R)= \left [\begin {array}{cc} 2 \mathbb {Z}_{4} & 2 \mathbb {Z}_{4}\\ 0 & 2 \mathbb {Z}_{4} \end {array}\right ]\). So \(\frac {R}{\mathcal {Z}(R_{R}) } = \frac {R}{\mathcal {Z}(_{R}R)}\) is reduced and so it is Armendariz. Therefore, according to Proposition 4, R is \(\mathcal {Z}\)-Armendariz. However, R is not Abelian.
Now, we need the following lemma whose proof is the same as the proof of [5, Lemma 7].
Lemma 2
If a, b, care elements in a right\(\mathcal {Z}\)-Armendariz ring R such that ab = 0 and acnb = 0 for some\(n \in \mathbb {N}\), then\(acb \in \mathcal {Z}(R_{R})\).
Proof
We have f(x)g(x) = 0, where f(x) = a(1 − cx) and g(x) = (1 + cx + ⋯ + cn− 1xn− 1)b. Thus, \(acb \in \mathcal {Z}(R_{R})\). □
Proposition 5
If R is a right\(\mathcal {Z}\)-Armendariz ring and idempotents lift modulo\(\mathcal {Z}(R_{R})\), then the ring\(\bar {R}= \frac {R}{\mathcal {Z}(R_{R})}\)is Abelian.
Proof
Let \(\bar {e} \in \text {Id}(\bar {R})\). By the hypothesis, we can assume that e ∈Id(R). Thus, it is sufficient to show that for any r ∈ R, \(er-re \in \mathcal {Z}(R_{R})\). Let a = e, b = (1 − e) and c = er(1 − e). Clearly ab = 0 and c2 = 0. By Lemma 2, \(er-ere = acb \in \mathcal {Z}(R_{R})\). Similarly, we have \(re-ere \in \mathcal {Z}(R_{R})\). So \( er-re \in \mathcal {Z}(R_{R})\). □
Proposition 6
Every right\(\mathcal {Z}\)-Armendariz ring is Dedekind-finite.
Proof
Suppose that R is a right \(\mathcal {Z}\)-Armendariz ring and uv = 1 for some u, v ∈ R. The element c = v(1 − vu) is nilpotent of nilpotency index two. If we put a = vu and b = 1 − vu, then by Lemma 2, \(v(1-vu)=acb \in \mathcal {Z}(R_{R})\). Thus, \(uv(1-vu)= (1-vu) \in \text {Id}(R) \cap \mathcal {Z}(R_{R})=~0\). □
Remark 3
Let R be a ring and Γ be an infinite set. Then the ring of column (respectively, row) finite Γ ×Γ matrices over R is not Dedekind-finite and so is neither left nor right \(\mathcal {Z}\)-Armendariz.
Note that the converse of Proposition 6 is not true in general. For example, the matrix ring Mn(F), where F is any field and n ≥ 2 is Dedekind-finite but is neither left nor right \(\mathcal {Z}\)-Armendariz (Example 3). Therefore, the class of (right) \(\mathcal {Z}\)-Armendariz rings lies strictly between the classes of Armendariz and Dedekind-finite rings.
Recall that a ring R is subdirectly irreducible, if every representation of R as a subdirect product of other rings is trivial, equivalently the intersection of all nonzero ideals of R is nonzero.
Example 6
A subdirectly irreducible ring is not necessarily right \(\mathcal {Z}\)-Armendariz. Let R be the ring of \( \mathbb {N} \times \mathbb {N}\) column finite matrices over a field F. Then R has exactly one nonzero proper ideal and so it is subdirectly irreducible. However, R is not right \(\mathcal {Z}\)-Armendariz.
For the rest of this section we construct some right \(\mathcal {Z}\)-Armendariz rings by a given one.
Proposition 7
Let R be a ring and e ∈ Id(R) such thateR(1 − e) = 0. If R is a right\(\mathcal {Z}\)-Armendariz ring, then so is S = eRe.
Proof
First, we show that \(\mathcal {Z}(R_{R}) \cap S \subseteq \mathcal {Z}(S_{S})\). Let \(a \in \mathcal {Z}(R_{R}) \cap S\) and 0≠s ∈ S. There exists r ∈ R such that sr≠ 0 and asr = 0. Thus, as(ere) = 0 and s(ere)≠ 0, since eR(1 − e) = 0. This implies that \(a \in \mathcal {Z}(S_{S})\). Now, suppose that \(f(x)={\sum }_{i=0}^{m} a_{i} x^{i}\) and \(g(x)= {\sum }_{j=0}^{n} b_{j} x^{j} \in S[x]\) such that f(x)g(x) = 0. So \(a_{i}b_{j} \in \mathcal {Z}(R_{R}) \cap S \subseteq \mathcal {Z}(S_{S})\) for every \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). □
Proposition 8
Let R be a ring and M be an ideal of R containing an element r such that annl(r) = 0. Then the ring\(S= \left \{\left [\begin {array}{cc} a&m\\0&a \end {array}\right ] ~|~ a \in R \ \text {and}\ m \in M \right \}\)is right\(\mathcal {Z}\)-Armendariz if and only if R is a right\(\mathcal {Z}\)-Armendariz ring.
Proof
By some calculations we can show that
Suppose that R is a right \(\mathcal {Z}\)-Armendariz ring and F(x)G(x) = 0, where \(F(x)= {\sum }_{i=0}^{m} \left [\begin {array}{cc} a_{i}&m_{i}\\0&a_{i} \end {array}\right ] x^{i}\) and \(G(x)= {\sum }_{j=0}^{n} \left [\begin {array}{cc} b_{j}&m^{\prime }_{j}\\0&b_{j} \end {array}\right ] x^{j}\). So we have f(x)g(x) = 0, where \(f(x)= {\sum }_{i=0}^{m} a_{i} x^{i}\) and \(g(x)={\sum }_{j=0}^{n} b_{j} x^{j}\). Since R is right \(\mathcal {Z}\)-Armendariz, \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\), which implies that \(\left [\begin {array}{cc} a_{i}&m_{i}\\0&a_{i} \end {array}\right ] \left [\begin {array}{cc} b_{j}&m^{\prime }_{j}\\0&b_{j} \end {array}\right ] \in \mathcal {Z}(S_{S})\) for every \(i=0,1,\dots ,m\) and \(j=0,1,\dots ,n\). Therefore, S is a right \(\mathcal {Z}\)-Armendariz ring. Now, suppose that S is a right \(\mathcal {Z}\)-Armendariz ring and f(x)g(x) = 0, where \(f(x)={\sum }_{i=0}^{m} a_{i} x^{i}\) and \(g(x)={\sum }_{j=0}^{n} b_{j} x^{j} \in R[x]\). So F(x)G(x) = 0 where \(F(x)= {\sum }_{i=0}^{m} \left [\begin {array}{cc} a_{i}&0\\0&a_{i} \end {array}\right ] x^{i}\) and \(G(x)= {\sum }_{j=0}^{n} \left [\begin {array}{cc} b_{j}&0\\0&b_{j} \end {array}\right ] x^{j} \in S[x]\). So \(\left [\begin {array}{cc} a_{i}&0\\0&a_{i} \end {array}\right ] \left [\begin {array}{cc} b_{j}&0\\0&b_{j} \end {array}\right ] \in \mathcal {Z}(S_{S})\) for every \(i=0,1,\dots ,m\) and \(j=0,1,\dots ,n\), which implies that \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\). Thus, R is right \(\mathcal {Z}\)-Armendariz. □
Corollary 3
Let R be a ring. Then R is right\(\mathcal {Z}\)-Armendariz if and only if the ring\(\frac {R[x]}{\langle x^{2} \rangle }\)is so.
Proof
The ring \(\frac {R[x]}{\langle x^{2} \rangle }\) is isomorphic to the ring \(S= \left \{\left [\begin {array}{cc} a&b\\0&a \end {array}\right ] ~|~ a,b \in R \right \}\). Now, apply Proposition 8. □
Proposition 9
For a ring R, the following are equivalent
-
(1)
R is right\(\mathcal {Z}\)-Armendariz;
-
(2)
\(\frac {R[x]}{\langle x \rangle ^{n}}\)is right\(\mathcal {Z}\)-Armendariz for every\(n \in \mathbb {N}\);
-
(3)
\(\frac {R[x]}{\langle x \rangle ^{n}}\)is right\(\mathcal {Z}\)-Armendariz for some\(n \in \mathbb {N}\).
Proof
The proof follows from the fact that the ring \(\frac {R[x]}{\langle x \rangle ^{n}}\) is isomorphic to the ring
and \(\mathcal {Z}(S_{S})= \left \{\left [\begin {array}{cccc} a_{1} & a_{2}&\dots &a_{n}\\ 0& a_{1} &\dots &a_{n-1}\\ \vdots & & & \vdots \\ 0 & 0&{\dots } &a_{1} \end {array}\right ] \in S ~|~ a_{1} \in \mathcal {Z}(R_{R}) \right \}\). □
Proposition 10
Let R be a ring and M be an ideal of R such that\(M \subseteq \mathcal {Z}(R_{R})\). Then R is right\(\mathcal {Z}\)-Armendariz if and only if the ring\(S= \left [\begin {array}{cc} R & M \\ 0 & R \end {array}\right ]\)is so.
Proof
It is not difficult to show that \(\mathcal {Z}(S_{S}) = \left \{ \left [\begin {array}{cc} a&m\\0&b \end {array}\right ] \in S ~|~ a,b \in \mathcal {Z}(R_{R}) \right \}\). The rest of the proof is similar to the proof of Proposition 8. □
Proposition 11
Let R and S be rings, RMS ( an (R, S)-(and\( T= \left [\begin {array}{cc} R&M\\0&S \end {array}\right ]\). If R is Armendariz, S is right\(\mathcal {Z}\)-Armendariz and \(\mathcal {Z}(M_{S})=M\), then T is right \(\mathcal {Z}\)-Armendariz.
Proof
First note that \(\left [\begin {array}{cc} 0&M\\0& \mathcal {Z}(S_{S}) \end {array}\right ] \subseteq \mathcal {Z}(T_{T})\). Now suppose that \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in T[x]\) and f(x)g(x) = 0, where \(a_{i} = \left [\begin {array}{cc} r_{i}& m_{i}\\ 0& s_{i} \end {array}\right ]\) and \(b_{j}= \left [\begin {array}{cc} r^{\prime }_{j} &m^{\prime }_{j}\\ 0 & s^{\prime }_{j} \end {array}\right ]\). Thus, \(({\sum }_{i=0}^{m} r_{i}x^{i})({\sum }_{j=0}^{n} r^{\prime }_{j}x^{j})=0\) in R[x] and \(({\sum }_{i=0}^{m} s_{i}x^{i})({\sum }_{j=0}^{n} s^{\prime }_{j}x^{j})=0\) in S[x]. Since R is Armendariz and S is right \(\mathcal {Z}\)-Armendariz, for any \(i=0,1, \dots , m\) and \(j=0,1, \dots , n\) we have \(r_{i}r^{\prime }_{j}=0\) and \(s_{i}s^{\prime }_{j} \in \mathcal {Z}(S_{S})\) and hence \(a_{i}b_{j} \in \mathcal {Z}(T_{T})\). □
Remark 4
Let R and S be rings, RMS be an (R, S)-bimodule and \(T= \left [\begin {array}{cc} R&M\\0&S \end {array}\right ]\). If MS is not a singular S-module, then T is not a right \(\mathcal {Z}\)-Armendariz ring. For if \(m \in M-\mathcal {Z}(M_{S})\), then
But \(\left [\begin {array}{cc} 1&0\\0&0 \end {array}\right ] \left [\begin {array}{cc} 0&m\\0&0 \end {array}\right ]= \left [\begin {array}{cc} 0&m\\0&0 \end {array}\right ] \notin \mathcal {Z}(T_{T})\).
Proposition 12
Let R be a ring and S be a multiplicatively closed set of central regular elements of R. Then R is right\(\mathcal {Z}\)-Armendariz if and only if the ring T = RS− 1is so.
Proof
It is easy to see that \(\frac {a}{s} \in \mathcal {Z}(T_{T})\) if and only if \(a \in \mathcal {Z}(R_{R})\). Now the rest of the proof is straightforward. □
Corollary 4
A ring R is right\(\mathcal {Z}\)-Armendariz if and only if the ring
is right\(\mathcal {Z}\)-Armendariz.
Proof
Consider the multiplicatively closed set
in \(R[x_{1}, x_{2}, \dots , x_{n}]\). Now, apply Proposition 12 and Corollary 1. □
Corollary 5
Let R be a ring. Then R[[x]] is a right\(\mathcal {Z}\)-Armendariz ring if and only if R((x)), the Laurent series ring over R, is so.
Proof
Use Proposition 12 when \(S=\{1, x, x^{2}, \dots \} \subseteq R[[x]]\). □
Proposition 13
Let R be a ring and consider the ring
with component-wise addition and multiplication. Then R is right\(\mathcal {Z}\)-Armendariz if and only if S is so.
Proof
Let R be right \(\mathcal {Z}\)-Armendariz. If
and F(x)G(x) = 0, then f(x)g(x) = 0, where \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\) and \(g(x)= {\sum }_{j=0}^{n} a^{\prime }_{j} x^{j} \in R[x]\). So \(a_{i}a^{\prime }_{j} \in \mathcal {Z}(R_{R})\) for \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). Similarly, we can show that \(b_{i}b^{\prime }_{j} \in \mathcal {Z}(R_{R})\). Thus, \( (a_{i},b_{i})(a^{\prime }_{j},b^{\prime }_{j}) \in \mathcal {Z}(R_{R}) \times \mathcal {Z}(R_{R}) \subseteq \mathcal {Z}(S_{S})\). So S is right \(\mathcal {Z}\)-Armendariz.
Now, suppose that S is right \(\mathcal {Z}\)-Armendariz. If \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R[x]\) and f(x)g(x) = 0, then \(({\sum }_{i=0}^{m} (a_{i},a_{i})x^{i})({\sum }_{j=0}^{n} (b_{j},b_{j}) x^{j})=0\) in S[x]. So for any i, j we have \((a_{i},a_{i})(b_{j},b_{j}) \in \mathcal {Z}(S_{S})\). Thus, \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\). Therefore, R is right \(\mathcal {Z}\)-Armendariz. □
3 \(\mathcal {Z}\)-Armendariz Modules
Recall that a right R-module M is Armendariz if f(x)g(x) = 0 implies that mr = 0, where f(x) ∈ M[x], g(x) ∈ R[x], m is an arbitrary coefficient of f(x) and r is an arbitrary coefficient of g(x) [1]. In [4], it is shown that the class of Armendariz modules is closed under direct products and submodules, and also every flat module over an Armendariz ring is Armendariz. In general a homomorphic image of an Armendariz module need not be Armendariz [4, Example 2.12]. However, as we shall see below, for an Armendariz module MR, the factor module \(\frac {M}{\mathcal {Z}(M_{R})}\) is Armendariz too. But first we need a lemma.
Lemma 3
Let MRbe a right R-module. Then\(\mathcal {Z}(M[x]_{R[x]}) = \mathcal {Z}(M_{R})[x]\).
Proof
The proof is similar to [7, Exercise 7.35], for the right singular ideal of a polynomial ring. □
Proposition 14
If MRis an Armendariz R-module, then so is\(\bar {M} = \frac {M}{\mathcal {Z}(M_{R})}\).
Proof
Assume that \(f(x) = {\sum }_{i=0}^{m} m_{i} x^{i} \in M[x]\) and \(g(x) = {\sum }_{j=0}^{n} r_{j}x^{j} \in R[x]\) such that \(f(x)g(x) \in \mathcal {Z}(M_{R})[x] = \mathcal {Z}(M[x]_{R[x]})\). We will show that for every \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\), \(m_{i}r_{j} \in \mathcal {Z}(M_{R})\). All coefficients of f(x)g(x) are in \(\mathcal {Z}(M_{R})\), so that for every nonzero c ∈ R there exists r ∈ R such that cr≠ 0 and f(x)g(x)cr = 0. Since MR is Armendariz, mirjcr = 0 for every \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\) and therefore, \(m_{i}r_{j} \in \mathcal {Z}(M_{R})\). □
A similar technique can be used to show that for any Armendariz ring R, the factor rings \(\frac {R}{\mathcal {Z}(R_{R})}\) and \(\frac {R}{\mathcal {Z}(_{R}R)}\) are Armendariz. Note that the converse of this statement is not true, for example, let R be a commutative ring. Then \(\frac {R}{\mathcal {Z}(R)}\) is reduced and so is Armendariz. However, commutative rings are not necessarily Armendariz.
In the rest of this section, we study \(\mathcal {Z}\)-Armendariz modules as a generalization of Armendariz modules.
Definition 2
A right R-module MR is called \(\mathcal {Z}\)-Armendariz, if the equation f(x)g(x) = 0 implies that \(m_{i}r_{j} \in \mathcal {Z}(M_{R})\) for every \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\), where \(f(x)= {\sum }_{i=0}^{m} m_{i}x^{i} \in M[x]\) and \(g(x)={\sum }_{j=0}^{n} r_{j}x^{j} \in R[x]\).
Clearly every Armendariz module (for example, every vector space over a division ring) is \(\mathcal {Z}\)-Armendariz. Also every singular right R-module is \(\mathcal {Z}\)-Armendariz and if MR is a nonsingular \(\mathcal {Z}\)-Armendariz module, then MR is Armendariz. A ring R is right \(\mathcal {Z}\)-Armendariz, if RR is a \(\mathcal {Z}\)-Armendariz module.
Proposition 15
The class of\(\mathcal {Z}\)-Armendariz modules over a ring R, is closed under submodules and arbitrary direct sums.
Proof
The proof follows from the fact that if NR ≤ MR, then \(\mathcal {Z}(N_{R}) = \mathcal {Z}(M_{R}) \cap N\) and for a family of right R-modules {Mi}i∈I, \(\mathcal {Z}(\oplus _{i \in I} M_{i}) = \oplus _{i \in I} \mathcal {Z} (M_{i})\). □
Corollary 6
A ring R is right\(\mathcal {Z}\)-Armendariz if and only if every submodule of a free right R-module is\(\mathcal {Z}\)-Armendariz.
Corollary 7
Every semisimple right module over a right\(\mathcal {Z}\)-Armendariz ring is\(\mathcal {Z}\)-Armendariz.
Proof
By Proposition 15, it is sufficient to prove the corollary for simple modules. Suppose that R is a right \(\mathcal {Z}\)-Armendariz ring and MR is a simple module. By [7, Exercise 7.12A], every simple module over an arbitrary ring is either singular or projective. According to Corollary 6, MR is a \(\mathcal {Z}\)-Armendariz module. □
In the next example we see that an infinite direct product of \(\mathcal {Z}\)-Armendariz modules is not necessarily \(\mathcal {Z}\)-Armendariz.
Example 7
Let \(R= \left [\begin {array}{cc} \mathbb {Z}& \mathbb {Z}\\ 0 & \mathbb {Z} \end {array}\right ]\). For every n ≥ 2, \(M_{n}= \left [\begin {array}{cc} \mathbb {Z}& \mathbb {Z}_{n}\\ 0 & \mathbb {Z} \end {array}\right ]\) is a right R-module with \(\mathcal {Z}(M_{n})\! =\! \left [\begin {array}{cc} 0& \mathbb {Z}_{n}\\ 0&0 \end {array}\right ]\). Since \(\mathbb {Z}\) is an Armendariz ring, one can show that Mn is a \(\mathcal {Z}\)-Armendariz R-module. Now consider \(M= {\prod }_{n \geq 2} M_{n}\). Put \(a= \left (\left [\begin {array}{cc}0& \bar {1}\\ 0&0 \end {array}\right ], \left [\begin {array}{cc} 0& \bar {1}\\ 0&0 \end {array}\right ], {\dots } \right )\), \(b= \left (\left [\begin {array}{cc} 1& 0\\ 0&0 \end {array}\right ], \left [\begin {array}{cc} 1& 0\\ 0&0 \end {array}\right ], \dots \right ) \in M\) and f(x) = a − bx ∈ M[x] and g(x) = E12 + E22x ∈ R[x], where Eij’s are those introduced in Example 3. We have f(x)g(x) = 0. But aE22 = a is not contained in \(\mathcal {Z}(M_{R})\), since \(\text {ann}_{r}(a)= \left [\begin {array}{cc} \mathbb {Z}& \mathbb {Z}\\0&0 \end {array}\right ]\), which is not an essential right ideal.
Proposition 16
A module MRis\(\mathcal {Z}\)-Armendariz if and only if every finitely generated submodule of M is\(\mathcal {Z}\)-Armendariz.
Proof
The only if part follows from Proposition 15. For the if part, note that for any f ∈ M[x] there exists a finitely generated submodule N of M such that f ∈ N[x]. □
Corollary 8
Let R be a ring such that every finitely generated right R-module can be embedded in a free module (for example, let R be a quasi-Frobenious ring). Then the following are equivalent:
-
(1)
R is a right\(\mathcal {Z}\)-Armendariz ring;
-
(2)
Every right R-module is\(\mathcal {Z}\)-Armendariz.
-
(3)
Every cyclic right R-module is\(\mathcal {Z}\)-Armendariz.
Proof
(1) ⇒ (2) Let MR be an R-module and KR be a finitely generated submodule of MR. Since KR can be embedded in a free R-module, by Proposition 15, it is \(\mathcal {Z}\)-Armendariz. Now Proposition 16 implies that MR is \(\mathcal {Z}\)-Armendariz. The proofs of (2) ⇒ (3) and (3) ⇒ (1) are clear. □
Proposition 17
Let I be a right ideal of a ring R such that I is not contained in\(\mathcal {Z}(R_{R})\)and\(\mathcal {Z}(R_{R})\)is a prime ideal of R. If IRis a\(\mathcal {Z}\)-Armendariz module, then R is a right\(\mathcal {Z}\)-Armendariz ring.
Proof
Let \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\), \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R[x]\) and f(x)g(x) = 0. For every a ∈ I and r ∈ R, we have af(x)g(x)r = 0. Obviously, af(x) ∈ I[x], so that \(aa_{i}b_{j}r \in \mathcal {Z}(I_{R}) \subseteq \mathcal {Z}(R_{R})\). Thus, \(Ia_{i}b_{j}R \subseteq \mathcal {Z}(R_{R})\) for any \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). As \(\mathcal {Z}(R_{R})\) is a prime ideal and \(I \not \subseteq \mathcal {Z}(R_{R})\), we have \(a_{i}b_{j} \in \mathcal {Z}(R_{R})\). Therefore, R is right \(\mathcal {Z}\)-Armendariz. □
Proposition 18
Every flat right R-module over a right\(\mathcal {Z}\)-Armendariz ring R is\(\mathcal {Z}\)-Armendariz.
Proof
In view of the fact that for any modules FR and MR and any R-homomorphism \(\varphi :F \rightarrow M\), \(\varphi (\mathcal {Z}(F_{R})) \subseteq \mathcal {Z}(M_{R})\), the proof is similar to the proof of [4, Theorem 2.15]. □
The proof of the following lemma is similar to the proof of [1, Proposition 1].
Lemma 4
Let MRbe an Armendariz module, f ∈ M[x] and\(g_{1}, g_{2}, \dots , g_{n} \in R[x]\). If fg1g2 ⋯ gn = 0, then mb1b2 ⋯ bn = 0, where mis an arbitrary coefficient of fand biis an arbitrary coefficient of gifor\(i=1, 2, \dots , n\).
Proposition 19
Let MR be a right R-module and\(\frac {M}{K}\)be an Amendariz module for some submodule K of\(\mathcal {Z}(M_{R})\). For any f ∈ M[x] and\(g_{1}, g_{2}, \dots , g_{n} \in R[x]\), if fg1g2 ⋯ gn ∈ K[x], then mb1b2 ⋯ bn ∈ K, where mis any coefficient of fand biis any coefficient of gifor\(i=1, 2, \dots , n\). In particular, MRis a\(\mathcal {Z}\)-Armendariz module.
Proof
Using Lemma 4, the proof is clear. □
Similar to the case for the Armendariz modules (Proposition 14), we have the following result.
Proposition 20
Let R be a ring. If MRis a\(\mathcal {Z}\)-Armendariz module, then so is the factor module\(\bar {M}=\frac {M}{\mathcal {Z}(M_{R})}\).
Proof
Suppose that \(\bar {f}(x)= {\sum }_{i=0}^{m} \bar {a_{i}} x^{i} \in \bar {M}[x]\) and \(g(x)={\sum }_{j=0}^{n} b_{j}x^{j} \in R[x]\) such that \(\bar {f}(x) g(x) = \bar {0}\) in \(\frac {M}{\mathcal {Z}(M_{R})}[x]\). We show that \(\bar {a_{i}}b_{j} \in \mathcal {Z}\left (\frac {M}{\mathcal {Z}(M_{R})}\right )\) for any \(i=0, 1, \dots ,m\) and \(j=0, 1, \dots , n\). We have \(f(x)g(x) \in \mathcal {Z}(M_{R})[x]\), where \(f(x)={\sum }_{i=0}^{m} a_{i}x^{i}\). Since every coefficient of f(x)g(x) is a singular element of M, for every nonzero element c ∈ R, there exists r ∈ R such that cr≠ 0 and f(x)g(x)cr = 0. Now \(a_{i}b_{j}cr \in \mathcal {Z}(M_{R})\) for \(i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\), since MR is \(\mathcal {Z}\)-Armendariz. Hence, \( \bar {a_{i}}b_{j} cr= \bar {0}\) in \(\frac {M}{\mathcal {Z}(M_{R})}\). Thus, \( \bar {a_{i}}b_{j} \in \mathcal {Z}\left (\frac {M}{\mathcal {Z}(M_{R})}\right )\). □
Corollary 9
Let R be a right nonsingular ring. Then for every\(\mathcal {Z}\)-Armendariz right R-module M, the factor module\(\frac {M}{\mathcal {Z}(M_{R})}\)is Armendariz.
Proof
By Proposition 20, \(\frac {M}{\mathcal {Z}(M_{R})}\) is \(\mathcal {Z}\)-Armendariz and by [7, Theorem 7.21], \(\mathcal {Z}\left (\frac {M}{\mathcal {Z}(M_{R})}\right )=0\). Therefore, \(\frac {M}{\mathcal {Z}(M_{R})}\) is an Armendariz module. □
The proof of the next result is similar to the proof of Proposition 2.
Proposition 21
Let MRbe an R-module. Then MRis\(\mathcal {Z}\)-Armendariz if and only if M[x]R[x]is so.
Note that if \(\theta : R \rightarrow S\) is a ring homomorphism and M is an S-module, then M is an R-module via mr = m𝜃(r).
Proposition 22
Let\(\theta : R \rightarrow S\)be a ring epimorphism. If MSis a\(\mathcal {Z}\)-Armendariz S-module, thenMRis\(\mathcal {Z}\)-Armendariz as an R-module.
Proof
Observe that \(\mathcal {Z}(M_{S}) \subseteq \mathcal {Z}(M_{R})\), now the rest of the proof is clear. □
In the next theorem, we show that over a right duo-ring, every right module is \(\mathcal {Z}\)-Armendariz. But first we state the following lemma.
Lemma 5
Let R be a right duo-ring and MRbe a right R-module. If\(mr^{2} \in \mathcal {Z}(M_{R})\)for some m ∈ M and r ∈ R, then\(mr \in \mathcal {Z}(M_{R})\).
Proof
Suppose that \(mr^{2} \in \mathcal {Z}(M_{R})\) and \(mr \notin \mathcal {Z}(M_{R})\). So there exists a ∈ R −{0} such that annr(mr) ∩ aR = 0. On the other hand, mr2ab = 0 for some b ∈ R such that ab≠ 0. Thus, mr(rab) = 0, which implies that rab ∈annr(mr) ∩ aR = 0. Hence, ab ∈annr(mr) ∩ aR = 0, which is a contradiction. □
Theorem 2
For a right duo-ring R, every right R-module is\(\mathcal {Z}\)-Armendariz.
Proof
Let \(f(x)= {\sum }_{i=0}^{m} m_{i} x^{i} \in M[x]\) and \(g(x)= {\sum }_{j=0}^{n} r_ j x^{j} \in R[x]\) such that f(x)g(x) = 0. We will show that \(m_{i}r_{j} \in \mathcal {Z}(M_{R})\) for every \( i=0, 1, \dots , m\) and \(j=0, 1, \dots , n\). We prove by induction on i + j. Clearly \(m_{0}r_{0}=0 \in \mathcal {Z}(M_{R})\). Suppose that the statement is true when i + j < k. If i + j = k, we multiply the equation
by r0. Since R is a right duo-ring, for each \(i=0, 1, \dots , (k-1)\), we have \(m_{i}r_{k-i}r_{0}= m_{i}r_{0}r^{\prime }_{i}\) for some \(r^{\prime }_{i} \in R\). By the induction hypotheses, \(m_{i}r_{0} \in \mathcal {Z}(M_{R})\) for i < k. Thus, \(m_{k}{r_{0}^{2}} \in \mathcal {Z}(M_{R})\). By Lemma 5, \(m_{k}r_{0} \in \mathcal {Z}(M_{R})\). Now multiplying (1), by r1, we deduce that \(m_{k-1}{r_{1}^{2}} \in \mathcal {Z}(M_{R})\) and again by Lemma 5, \(m_{k-1}r_{1} \in \mathcal {Z}(M_{R})\). By continuing this proses, we have \(m_{i}r_{k-i} \in \mathcal {Z}(M_{R})\) for every \(i=0, 1, \dots , k\). □
Remark 5
(1) We show that the converse of Theorem 2 is not true. Recall that a ring is right distributive if its lattice of right ideals is distributive. By [10, Corollary 7], over a right distributive ring, any right module is Armendariz (and hence \(\mathcal {Z}\)-Armendariz). But there is a right distributive ring which is not right duo ([9, Example 7.1.6]).
(2) Recall that an R-module MR is Dedekind-finite if M≅M ⊕ N (for some R-module NR) implies that N = 0. We show that a \(\mathcal {Z}\)-Armendariz module is not necessarily Dedekind-finite. For example, let R be a commutative ring and \(M= R^{(\mathbb {N})}\). By Theorem 2, M is a \(\mathcal {Z}\)-Armendariz module but clearly it is not Dedekind-finite.
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Nejadzadeh, A., Amini, A., Amini, B. et al. \(\mathcal {Z}\)-Armendariz Rings and Modules. Vietnam J. Math. 48, 131–143 (2020). https://doi.org/10.1007/s10013-019-00380-4
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DOI: https://doi.org/10.1007/s10013-019-00380-4
Keywords
- Right singular ideal
- Armendariz ring
- Right \(\mathcal {Z}\)-Armendariz ring
- \(\mathcal {Z}\)-Armendariz module