$\mathcal {Z}$-Armendariz Rings and Modules

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.

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 ann_r(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 a_ib_j = 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 M_R is called Armendariz ([1, Proposition 12]), if f(x)g(x) = 0 implies that m_ir_j = 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 M_R 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

$$ a_{0}b_{0} + (a_{0}b_{1}+a_{1}b_{0})x + (a_{0}b_{2} + a_{1}b_{1} + a_{2}b_{0})x^{2} + {\cdots} + a_{m}b_{n}x^{n+m} = 0. $$

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, M_n(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 = M_n(R) and E_ij ∈ 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 ann_r(E₁₂) ∩ E₂₂S = 0. A similar proof can be used for the ring of n × n upper (lower) triangular matrices over R.

Proposition 1

Let {R_i}_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 R_i 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 + I_kis 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 ∈ I_k 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,⋯ , I_t 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 a_ib_j + I_k 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 + ⋯ + f_ntⁿ, g(t) = g₀ + g₁t + ⋯ + g_mt^m ∈ R[x][t] and f(t)g(t) = 0, where f_i, g_j ∈ 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 + ⋯ + degf_n + degg₀ + ⋯ + degg_m. Then f(x^k) = f₀ + f₁x^k + ⋯ + f_nx^kn, g(x^k) = g₀ + g₁x^k + ⋯ + g_mx^km ∈ R[x] and f(x^k)g(x^k) = 0. So the product of each coefficient of f_i with every coefficient of g_j 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 f_i = a_ixⁱ and g_j = b_jx^j. 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[{x_α}_α∈A] is right$\mathcal {Z}$-Armendariz.

Proof

Let R be a right $\mathcal {Z}$-Armendariz ring and f, g ∈ R[{x_α}_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 ann_r(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 a_iis an arbitrary coefficient of f_ifor$i=1, 2, \dots ,n$. 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 a_i is an arbitrary coefficient of f_i 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 E_ij’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 ann_l(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 + ⋯ + c^n− 1x^n− 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 M_n(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 ann_l(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

$$ \mathcal{Z}(S_{S})= \left\{\left[\begin{array}{cc} a&m\\0&a \end{array}\right] \in S ~|~ a \in \mathcal{Z}(R_{R}) \right\}. $$

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

$$ S=\left\{ \left[\begin{array}{cccc} a_{1} & a_{2}&\dots&a_{n}\\ 0& a_{1} &\dots&a_{n-1}\\ \vdots& & & \vdots\\ 0 & 0&{\dots} &a_{1} \end{array}\right]~|~ a_{i} \in R, i=1,\dots,n \right\} $$

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, _RM_S ( 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, _RM_S be an (R, S)-bimodule and $T= \left [\begin {array}{cc} R&M\\0&S \end {array}\right ]$. If M_S 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

$$ \left( \left[\begin{array}{cc} 0&m\\0&0 \end{array}\right] + \left[\begin{array}{cc} 1&0\\0&0 \end{array}\right]x\right)\left( \left[\begin{array}{cc} 0&m\\0&0 \end{array}\right] - \left[\begin{array}{cc} 0&0\\0&1 \end{array}\right]x\right)=0. $$

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

$$ R\left[x_{1},x_{1}^{-1}, \dots, x_{n},x_{n}^{-1}\right] $$

is right$\mathcal {Z}$-Armendariz.

Proof

Consider the multiplicatively closed set

$$ S= \left\{x_{1}^{i_{1}}x_{2}^{i_{2}} {\dots} x_{n}^{i_{n}} ~|~i_{1},i_{2}, \dots, i_{n} \geq 0 \right\} $$

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

$$ S=\{(a,b) \in R \times R ~|~ a-b \in \mathcal{Z}(R_{R})\} $$

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

$$ F(x)= \sum\limits_{i=0}^{m} (a_{i},b_{i}) x^{i}, \quad G(x)= \sum\limits_{j=0}^{n} (a^{\prime}_{j}, b^{\prime}_{j}) x^{j} \in S[x] $$

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 M_R, the factor module $\frac {M}{\mathcal {Z}(M_{R})}$ is Armendariz too. But first we need a lemma.

Lemma 3

Let M_Rbe 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 M_Ris 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 M_R is Armendariz, m_ir_jcr = 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 M_R 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 M_R is a nonsingular $\mathcal {Z}$-Armendariz module, then M_R is Armendariz. A ring R is right $\mathcal {Z}$-Armendariz, if R_R 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 N_R ≤ M_R, then $\mathcal {Z}(N_{R}) = \mathcal {Z}(M_{R}) \cap N$ and for a family of right R-modules {M_i}_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 M_R 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, M_R 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 M_n 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 E_ij’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 M_Ris$\mathcal {Z}$-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 M_R be an R-module and K_R be a finitely generated submodule of M_R. Since K_R can be embedded in a free R-module, by Proposition 15, it is $\mathcal {Z}$-Armendariz. Now Proposition 16 implies that M_R 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 I_Ris 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 F_R and M_R 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 M_Rbe an Armendariz module, f ∈ M[x] and$g_{1}, g_{2}, \dots , g_{n} \in R[x]$. If fg₁g₂ ⋯ g_n = 0, then mb₁b₂ ⋯ b_n = 0, where mis an arbitrary coefficient of fand b_iis an arbitrary coefficient of g_ifor$i=1, 2, \dots , n$.

Proposition 19

Let M_R 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 ⋯ g_n ∈ K[x], then mb₁b₂ ⋯ b_n ∈ K, where mis any coefficient of fand b_iis any coefficient of g_ifor$i=1, 2, \dots , n$. In particular, M_Ris 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 M_Ris 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 M_R 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 M_Rbe an R-module. Then M_Ris$\mathcal {Z}$-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 M_Sis a$\mathcal {Z}$-Armendariz S-module, thenM_Ris$\mathcal {Z}$-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 M_Rbe 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 ann_r(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 ∈ann_r(mr) ∩ aR = 0. Hence, ab ∈ann_r(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

$$ m_{0}r_{k}+m_{1}r_{k-1}+ {\cdots} + m_{k}r_{0}=0 $$

(1)

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 M_R is Dedekind-finite if M≅M ⊕ N (for some R-module N_R) 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.