2 主要结果
定义2.1 环$R$称为$N$ -弱拟Armendariz环, 如果$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in R[x],
$满足$f(x)R[x]g(x)=0, $则有$a_{i}Rb_{j}\subseteq N(R), i, j=0,
1.$
引理2.2 [8, n2.1] 设$f(x), g(x)\in R[x],
$则$f(x)Rg(x)=0$当且仅当$f(x)R[x]g(x)=0.$
设$R$为环, $M$为$(R, R)$ -双模, $R$对于$M$的平凡扩张$T(R,
M)=R\bigoplus M, $其中运算为
| $(r_{1}, m_{1})+(r_{2}, m_{2})=(r_{1}+r_{2},
m_{1}+m_{2}), (r_{1}, m_{1})(r_{2}, m_{2})=(r_{1}r_{2},
r_{1}m_{2}+m_{1}r_{2}).$ |
易知$T(R, M)\cong
\left\{\left(\begin{array}{ccccc}r&m\\0&r\end{array}\right)|r\in
R, m\in M\right\}.$
记
| $S_{n}(R)=\left\{\left(\begin{array}{ccccc}a&a_{12}&\cdots&a_{1n}\\0&a&\cdots&a_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&a\end{array}\right)|a, a_{ij}\in R\right\}(n\ge 2). $ |
由文[3, 例2.5]可知, 若$R$是弱拟-Armendariz环,
$S_{n}(R)$不是弱拟-Armendariz环, 但对于$N$
-弱拟Armendariz环却有以下结论.
定理2.3 设$R$是环, 则下列命题等价:
(1) $R$是$N$ -弱拟Armendariz环;
(2) $T_{n}(R)$是$N$ -弱拟Armendariz环, 对任意$n\geq 2$;
$(2^\prime)$ $T_{n}(R)$是$N$ -弱拟Armendariz环, 对某个$n\geq
2$;
(3) $S_{n}(R)$是$N$ -弱拟Armendariz环, 对任意$n\geq 2$;
$(3^\prime)$ $S_{n}(R)$是$N$ -弱拟Armendariz环, 对某个$n\geq
2$;
(4) $T(R, R)$是$N$ -弱拟Armendariz环.
证 $(1)\Rightarrow(2)$设$f(x)=A_{0}+A_{1}x,
g(x)=B_{0}+B_{1}x\in T_{n}(R)[x]$, 有$f(x)T_{n}(R)[x]g(x)=0, $其中
| $A_{i}={\left(\begin{array}{ccccc}a_{11}^{i}&a_{12}^{i}&\cdots&a_{1n}^{i}\\0&a_{22}^{i}&\cdots&a_{2n}^{i}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&a_{nn}^{i}\end{array}\right)}, \ B_{j}={\left(\begin{array}{ccccc}b_{11}^{j}&b_{12}^{j}&\cdots&b_{1n}^{j}\\0&b_{22}^{j}&\cdots&b_{2n}^{j}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&b_{nn}^{j}\end{array}\right)}, i, j=0, 1.$ |
易证存在环同构$T_{n}(R)[x]\rightarrow T_{n}(R[x])$;
| $\sum\limits_{i=0}^{n}{\left(\begin{array}{ccccc}a_{11}^{i}&a_{12}^{i}&\cdots&a_{1n}^{i}\\0&a_{22}^{i}&\cdots&a_{2n}^{i}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&a_{nn}^{i}\end{array}\right)}x^{i}\mapsto {\left(\begin{array}{ccccc}\sum\limits_{i=0}^{n}a_{11}^{i}x^{i}&\sum\limits_{i=0}^{n}a_{12}^{i}x^{i}&\cdots&\sum\limits_{i=0}^{n}a_{1n}^{i}x^{i}\\0&\sum\limits_{i=0}^{n}a_{22}^{i}x^{i}&\cdots&\sum\limits_{i=0}^{n}a_{2n}^{i}x^{i}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&\sum\limits_{i=0}^{n}a_{nn}^{i}x^{i}\end{array}\right)}.
$ |
则可令
| $f(x)={\left(\begin{array}{ccccc}f_{11}&f_{12}&\cdots&f_{1n}\\0&f_{22}&\cdots&f_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&f_{nn}\end{array}\right)}, \ g(x)={\left(\begin{array}{ccccc}g_{11}&g_{12}&\cdots&g_{1n}\\0&g_{22}&\cdots&g_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&g_{nn}\end{array}\right)}, $ |
其中$f_{st}=a_{st}^{o}+a_{st}^{1}x, g_{vw}=b_{vw}^{o}+b_{vw}^{1}x\
\ (1\leq s, t, v, w\leq n).$故对任意的$r\in R, $由$f(x)T_{n}(R)[x]g(x)=0$，得
| $\begin{array}{rcl}
&&{\left(\begin{array}{ccccc}f_{11}&f_{12}&\cdots&f_{1n}\\0&f_{22}&\cdots&f_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&f_{nn}\end{array}\right)}{\left(\begin{array}{ccccc}r&0&\cdots&0\\0&r&\cdots&0\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&r\end{array}\right)}{\left(\begin{array}{ccccc}g_{11}&g_{12}&\cdots&g_{1n}\\0&g_{22}&\cdots&g_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&g_{nn}\end{array}\right)}\\
&=&{\left(\begin{array}{ccccc}f_{11}rg_{11}&*&\cdots&*\\0&f_{22}rg_{22}&\cdots&*\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&f_{nn}rg_{nn}\end{array}\right)}=0.\end{array}$ |
因此$f_{ss}Rg_{ss}=0, $即$f_{ss}R[x]g_{ss}=0$, $s=1, 2, \cdots, n.$由于$R$是$N$ -弱拟Armendariz环, 所以$a_{ss}^{i}Rb_{ss}^{j}\subseteq N(R).$
下证$A_{i}T_{n}(R)B_{j}\subseteq N(T_{n}(R)).$任取
| $C={\left(\begin{array}{ccccc}c_{11}&c_{12}&\cdots&c_{1n}\\0&c_{22}&\cdots&c_{2n}\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&c_{nn}\end{array}\right)}\in T_{n}(R),
$ |
则
| $\begin{array}{rcl}
A_{i}CB_{j}&=&{\left(\begin{array}{ccccc}a_{11}^{i}c_{11}b_{11}^{j}&*&\cdots&*\\0&a_{22}^{i}c_{22}b_{22}^{j}&\cdots&*\\
\cdots&\cdots&\cdots&\cdots\\
0&0&\cdots&a_{nn}^{i}c_{nn}b_{nn}^{j}\end{array}\right)},
\end{array}$ |
$a_{ss}^{i}Rb_{ss}^{j}\subseteq N(R)$, 所以对任意的$s$及$i, j$, 存在$m_{ijs}^{c}\in\mathbb{N}^{+}$, 使得$(a_{ss}^{i}c_{ss}b_{ss}^{j})^{m_{ijs}^{c}}=0.$令$m_{ij}^{c}=\max\{m_{ij1}^{c}, m_{ij2}^{c}, \cdots,
m_{ijn}^{c}\}$, 则$(a_{tt}^{i}c_{tt}b_{tt}^{j})^{m_{ij}^{c}}=0,
t=1, 2, \cdots, n, $因此$\Big((A_{i}CB_{j})^{m_{ij}^{c}}\Big)^{n}=0$.所以$T_{n}(R)$是$N$ -弱拟Armendariz环, 对任意$n\geq 2$.
$(2)\Rightarrow(2^\prime)$显然成立.
$(2^\prime)\Rightarrow(1)$设$f(x)=a_{0}+a_{1}x,
g(x)=b_{0}+b_{1}x\in R[x], $有$f(x)R[x]g(x)=0.$则
| $\begin{eqnarray*}&&\left(\begin{array}{ccccc}f(x)&&&\\
&f(x)&\\
&&\ddots\\
&&&f(x)\end{array}\right)T_{n}(R[x])\left(\begin{array}{ccccc}g(x)&&&\\
&g(x)&\\
&&\ddots\\
&&&g(x)\end{array}\right)=0, \\
&& T_{n}(R)[x]\cong T_{n}(R[x]), \end{eqnarray*}$ |
于是
| $\left(\left(\begin{array}{ccccc}a_{0}&&&\\
&a_{0}&\\
&&\ddots\\
&&&a_{0}\end{array}\right )+\left(\begin{array}{ccccc}a_{1}&&&\\
&a_{1}&\\
&&\ddots\\
&&&a_{1}\end{array}\right)x\right)T_{n}(R)[x]$ |
| $\left(\left(\begin{array}{ccccc}b_{0}&&&\\
&b_{0}&\\
&&\ddots\\
&&&b_{0}\end{array}\right)+\left(\begin{array}{ccccc}b_{1}&&&\\
&b_{1}&\\
&&\ddots\\
&&&b_{1}\end{array}\right)x\right)=0.$ |
由于$T_{n}(R)$是$N$
-弱拟Armendariz环, 所以
| $\left(\begin{array}{ccccc}a_{i}&&&\\
&a_{i}&\\
&&\ddots\\
&&&a_{i}\end{array}\right)T_{n}(R)\left(\begin{array}{ccccc}b_{j}&&&\\
&b_{j}&\\
&&\ddots\\
&&&b_{j}\end{array}\right)\subseteq N(T_{n}(R)), i, j=0, 1. $ |
特别地, 对任意的$r\in R$, 存在$n_{ij}^{r}\in \mathbb{N}^{+}$, 使得
| $\left(\left(\begin{array}{ccccc}a_{i}&&&\\
&a_{i}&\\
&&\ddots\\
&&&a_{i}\end{array}\right)\left(\begin{array}{ccccc}r&&&\\
&r&\\
&&\ddots\\
&&&r\end{array}\right)\left(\begin{array}{ccccc}b_{j}&&&\\
&b_{j}&\\
&&\ddots\\
&&&b_{j}\end{array}\right)\right)^{n_{ij}^{r}}=0, $ |
则$(a_{i}rb_{j})^{n_{ij}^{r}}=0$, 于是$a_{i}Rb_{j}\subseteq N(R).$所以$R$是$N$ -弱拟Armendariz环.
$(1)\Rightarrow
(3)\Rightarrow(3^\prime)\Rightarrow(1)$与上述证明过程类似.
$(1)\Rightarrow (3)\Rightarrow(4)\Rightarrow(1)$易证, 从而得证.
注 由定义可知, 弱拟-Armendariz环一定是$N$
-弱拟Armendariz环.
下面的例子说明了反之是不成立的.
例2.4 设$W$为约化环, 由文[3, 例2.5]可知$R=T(W,
W)$为弱拟-Armendariz环, 且$S_{3}(R)$不是弱拟-Armendariz环, 但由定理2.3知$S_{n}(R)$为$N$ -弱拟Armendariz环, 对任意$n\geq 2$.
引理2.5 [3] 若$R$是半素环, 则$aRb=0$当且仅当$bRa=0$, 且若$aRbRa=0, $则$aRb=0$, 任意的$a, b\in
R.$
引理2.6 若$R$是半素环, 则$R[x]$是$N$ -弱拟Armendariz环.
证 由文[3, 定理2.7]可以直接得出该结论, 下面给出该引理的另一种证法.
设$F(y)=f_{0}+f_{1}y, G(y)=g_{0}+g_{1}y\in R[x][y], f_{i},
g_{j}\in R[x], i, j=0, 1, $有$F(y)R[x][y]G(y)=0, $即$F(y)R[x]G(y)=0$.则对任意的$f(x)\in R[x], $
| $f_{0}fg_{0}=0, \
f_{0}fg_{1}+f_{1}fg_{0}=0, \ f_{1}fg_{1}=0.$ |
由文[9, 定理10.19]知, 环$R$是半素的当且仅当环$R[x]$是半素的.
$f_{0}fg_{0}=0$, 由引理2.5知$g_{0}ff_{0}=0$, 即$g_{0}R[x]f_{0}=0.$任取$h(x)\in R[x],
$在$f_{0}fg_{1}+f_{1}fg_{0}=0$两边同时右乘$hf_{0}$, 则有
| $f_{0}fg_{1}hf_{0}=-f_{1}fg_{0}hf_{0}=0.$ |
由引理2.5知$f_{0}fg_{1}=0, $故$f_{i}fg_{j}\in N( R[x]),
$即$f_{i}R[x]g_{j}\subseteq N( R[x]), i, j=0, 1.$所以$R[x]$是$N$
-弱拟Armendariz环.
引理2.7 若$R[x]$是$N$ -弱拟Armendariz环, 则$R$是$N$ -弱拟Armendariz环.
证 设$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in R[x]$, 有$f(x)R[x]g(x)=0, $即$f(x)Rg(x)=0.$则对任意的$r\in R,
$
| $a_{0}rb_{0}=0, \ a_{0}rb_{1}+a_{1}rb_{0}=0, \
a_{1}rb_{1}=0.$ |
令$p(y)=a_{0}+a_{1}y, q(y)=b_{0}+b_{1}y\in
R[x][y]$.
下证$p(y)R[x]q(y)=0.$对任意的$r\in R$, $t\in \mathbb{N}$, 由上面的三个等式可知,
| $p(y)(rx^{t})q(y)=(a_{0}rb_{0})x^{t}+(a_{0}rb_{1}+a_{1}rb_{0})x^{t}y+(a_{1}rb_{1})x^{t}y^{2}=0. $ |
于是$p(y)R[x]q(y)=0$, 即$p(y)R[x][y]q(y)=0.$由于$R[x]$是$N$-弱拟Armendariz环, 所以$a_{i}R[x]b_{j}\subseteq
N(R[x]). $特别地, $a_{i}Rb_{j}\subseteq N(R), i, j=0, 1.$所以$R$是$N$ -弱拟Armendariz环.
由引理2.6, 引理2.7可即得,
定理2.8 若$R$是半素环, 则$R$是$N$ -弱拟Armendariz环.
注 1) 由文[3, 定理2.7]也可推出该定理成立.
2) 例2.4说明了该定理的逆命题不成立.
事实上, 令$a=\left(\begin{matrix} 0&1\\0&0\end{matrix}\right)\in
R$, 任意的
| $\left(\begin{matrix}
w_{11}&w_{12}\\0&w_{11}\end{matrix}\right)\in R,
\left(\begin{matrix}
0&1\\0&0\end{matrix}\right)\left(\begin{matrix}
w_{11}&w_{12}\\0&w_{11}\end{matrix}\right)\left(\begin{matrix}
0&1\\0&0\end{matrix}\right)=0, $ |
即$aRa=0$, 但$a\neq 0, $所以$R$不是半素环.
命题2.9 设$e$是环$R$的中心幂等元.则$R$是$N$
-弱拟Armendariz环当且仅当$eR$和$(1-e)R$都是$N$ -弱拟Armendariz环.
证 $(\Rightarrow )$设$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in (eR)[x]$, 有$f(x)(eR)[x]g(x)=0, $即$f(x)(eR)$ $g(x)=0.$又$f(x)e=f(x).$故对任意的$r\in R, f(x)rg(x)=f(x)(er)g(x)=0, $即$f(x)R[x]g(x)=0.$由于$R$是$N$ -弱拟Armendariz环, 所以$a_{i}Rb_{j}\subseteq N(R), i, j=0, 1, $则对任意的$er\in eR, r\in R, a_{i}(er)b_{j}=(a_{i}e)rb_{j}=a_{i}rb_{j}\in N(R)\bigcap eR=N(eR)$.所以$eR$是$N$ -弱拟Armendariz环, 类似可证$(1-e)R$是$N$ -弱拟Armendariz环.
$(\Leftarrow)$设$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in R[x], $有$f(x)R[x]g(x)=0, $即$f(x)Rg(x)=0$.故对任意的$r\in R, $
| $ef(x)(er)eg(x)=ef(x)rg(x)=0,
(1-e)f(x)(1-e)r(1-e)g(x)=(1-e)f(x)rg(x)=0.$ |
即
| $ef(x)(eR)[x]eg(x)=0, (1-e)f(x)((1-e)R)[x](1-e)g(x)=0.$ |
由于$eR$和$(1-e)R$都是$N$ -弱拟Armendariz环, 所以对任意的$r\in R$, 存在$n_{ij}^{r}, m_{ij}^{r}\in\mathbb{N}^{+}, $使得
| $[(ea_{i})(er)(eb_{j})]^{n_{ij}^{r}}
=e(a_{i}rb_{j})^{n_{ij}^{r}}=0, [(1-e)a_{i}((1-e)r)(1-e)b_{j}]^{m_{ij}^{r}}=(1-e)(a_{i}rb_{j})^{m_{ij}^{r}}=0.$ |
令$k_{ij}^{r}=\max\{n_{ij}^{r}, m_{ij}^{r}\}, $则
| $e(a_{i}rb_{j})^{k_{ij}^{r}}=0, (1-e)(a_{i}rb_{j})^{k_{ij}^{r}}=0,
(a_{i}rb_{j})^{k_{ij}^{r}}=e(a_{i}rb_{j})^{k_{ij}^{r}}+(1-e)\\(a_{i}rb_{j})^{k_{ij}^{r}}=0,
$ |
于是$a_{i}Rb_{j}\subseteq N(R), i, j=0, 1.$所以$R$是$N$
-弱拟Armendariz环.
环$R$称为nil-半交换的[10], 若对任意的$a, b\in
R,$ $ab\in N(R),$有$aRb\subseteq N(R).$ $I\lhd R$称为nil -半交换的, 若$I$满足上述条件.
命题2.10 设$R$是环, 则有
(1) 若$R$是$N$ -弱拟Armendariz环, $L$为$C(R)$的零化子, 则$R/L$是$N$ -弱拟Armendariz环.
(2) $I\lhd R, $若$R/I$是$N$ -弱拟Armendariz环, $I$是nil-半交换的, 则$R$是$N$ -弱拟Armendariz环.
证 (1) $\varnothing \neq C(R)=\{s\in R|sr=rs, r\in R\}, $令$\bar{a}=a+L, a\in R,$ $\bar{R}=R/L.$设$\bar{f}(x)=\bar{a_{0}}+\bar{a_{1}}x, \bar{g}(x)=\bar{b_{0}}+\bar{b_{1}}x\in \bar{R}[x], $有$\bar{f}(x) \bar{R}[x]\bar{g}(x)=\bar{0}.$
下证$\bar{a_{i}}\bar{R}\bar{b_{j}}\subseteq N( \bar{R}), i, j=0, 1, $ $\bar{f}(x) \bar{R}[x]\bar{g}(x)=\bar{0}, $故对任意的$\bar{r}\in \bar{R}, \bar{f}(x) \bar{r}\bar{g}(x)=\bar{0}, $则$a_{0}rb_{0}, \ a_{0}rb_{1}+a_{1}rb_{0}, \ a_{1}rb_{1}\in L,
$所以对任意的$s\in C(R), $
| $s(a_{0}rb_{0})=0, \
s(a_{0}rb_{1}+a_{1}rb_{0})=0, \ s(a_{1}rb_{1})=0.$ |
则$(sa_{0}+sa_{1}x)R(b_{0}+b_{1}x)=0, $即$(sa_{0}+sa_{1}x)R[x](b_{0}+b_{1}x)=0.$由于$R$是$N$
-弱拟Armendariz环, 所以对任意的$r\in R$, 存在$n_{ij}^{r}\in
\mathbb{N}^{+}, $使得$(sa_{i}rb_{j})^{n_{ij}^{r}}=s^{n_{ij}^{r}}(a_{i}rb_{j})^{n_{ij}^{r}}=0,
$则$(a_{i}rb_{j})^{n_{ij}^{r}} \in L(s\in C(R)\Rightarrow
s^{n_{ij}^{r}}\in C(R)), $于是$\bar{a_{i}}\bar{R}\bar{b_{j}}\subseteq N( \bar{R}), i, j=0,
1.$所以$R/L$是$N$ -弱拟Armendariz环.
(2) 设$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in R[x]$, 有$f(x)R[x]g(x)=0, $即$f(x)Rg(x)=0.$则对任意的$r\in R$,
| $a_{0}rb_{0}=0, \ a_{0}rb_{1}+a_{1}rb_{0}=0, \ a_{1}rb_{1}=0.$ |
另一方面$\bar{f}(x) \bar{R}\bar{g}(x)=\bar{0}, $由于$\bar{R}$是$N$ -弱拟Armendariz环, 所以$\bar{a_{i}}\bar{R}\bar{b_{j}}\subseteq N(\bar{R}), i, j=0, 1. $故对任意的$r\in R, $存在$n_{ij}^{r}\in \mathbb{N}^{+}, $使得$(a_{i}rb_{j})^{n_{ij}^{r}}\in I, $令$(a_{0}rb_{1})^{p}\in I, p\in \mathbb{N}^{+}.$
下证$a_{0}rb_{1}\in N(R).$由$a_{0}rb_{0}=0$得$(b_{0}a_{0}r)^{2}=0$, 于是
| $(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}a_{1}r(b_{0}a_{0}r)^{2}(b_{0}(a_{0}rb_{1})^{p+1})=0.$ |
因为
| $(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}a_{1}r(b_{0}a_{0}r)\in I, (b_{0}a_{0}r)(b_{0}(a_{0}rb_{1})^{p+1})\in I, b_{1}(a_{0}rb_{1})^{p}a_{1}r\in I, $ |
且$I$是nil -半交换的, 所以
| $(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}a_{1}r(b_{0}a_{0}r)b_{1}(a_{0}rb_{1})^{p}a_{1}r(b_{0}a_{0}r)(b_{0}(a_{0}rb_{1})^{p+1})\in N(I).$ |
即
| $[(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}]^{2}a_{1}r(b_{0}a_{0}r)(b_{0}(a_{0}rb_{1})^{p+1})\in N(I).$ |
继续此过程可得, $[(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}]^{4}\in N(I), $于是$(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}\in N(I).$在等式$ a_{0}rb_{1}+a_{1}rb_{0}=0$两边同时右乘$(a_{0}rb_{1})^{p+1}$, 则有
| $(a_{0}rb_{1})^{p+2}=-(a_{1}rb_{0})(a_{0}rb_{1})^{p+1}\in N(I).$ |
这样$ a_{0}rb_{1}\in N(I)\subseteq N(R)$, 则$a_{i}Rb_{j}\subseteq N(R), i, j=0, 1.$所以$R$是$N$ -弱拟Armendariz环.
推论2.11 若$R$是nil -半交换环, 则$R$是$N$ -弱拟Armendariz环.
$x\in R$称为正则的[11], 若对任意的$0\neq r\in R, rx\neq 0, xr\neq 0.$
命题2.12 设$\triangle$为$R$中中心正则元构成的乘法封闭子集.则$R$是$N$ -弱拟Armendariz环当且仅当$\triangle ^{-1}R$是$N$ -弱拟Armendariz环.
证 $(\Rightarrow)$设$F(x)=\alpha _{0}+\alpha _{1}x, G(x)=\beta _{0}+\beta _{1}x\in (\triangle ^{-1}R)[x]$, 有$F(x)(\triangle ^{-1}R)[x]G(x)=0, $即
| $F(x)(\triangle ^{-1}R)G(x)=0, $ |
其中$\alpha _{i}=u^{-1}a_{i}, \beta _{j}=v^{-1}b_{j}, u, v\in \triangle, a_{i}, b_{j}\in R$.故对任意的$w^{-1}r\in \triangle ^{-1}R, $
| $0=u^{-1}(a_{0}+a_{1}x)(w^{-1}r)v^{-1}(b_{0}+b_{1}x)=(uwv)^{-1}(a_{0}+a_{1}x)r(b_{0}+b_{1}x).$ |
令$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x, $则$f(x), g(x)\in R[x]$, 且$f(x)Rg(x)=0$, 即$f(x)R[x]g(x)=0$.由于$R$是$N$ -弱拟Armendariz环, 所以对任意的$r\in R, $存在$n_{ij}^{r}\in \mathbb{N}^{+}, $使得$(a_{i}rb_{j})^{n_{ij}^{r}}=0$, 则
| $[(uwv)^{-1}a_{i}rb_{j}]^{n_{ij}^{r}}=0,$ |
即$(uwv)^{-1}a_{i}rb_{j}\in N(\triangle ^{-1}R), i, j=0, 1, $则$\alpha _{i}(\triangle ^{-1}R)\beta _{j}\subseteq N(\triangle ^{-1}R).$所以$\triangle ^{-1}R$是$N$
-弱拟Armendariz环.
$(\Leftarrow)$设$f(x)=a_{0}+a_{1}x, g(x)=b_{0}+b_{1}x\in
R[x], $有$f(x)R[x]g(x)=0, $即$f(x)Rg(x)=0.$故对任意的$r\in R,
w\in \triangle, 0=w^{-1}f(x)rg(x)=f(x)(w^{-1}r)g(x), $则
| $f(x)(\triangle ^{-1}R)g(x)=0,$ |
即$f(x)(\triangle
^{-1}R)[x]g(x)=0$.由于$\triangle ^{-1}R$是$N$ -弱拟Armendariz环, 所以对任意的$w^{-1}r\in \triangle ^{-1}R, $存在$n_{ij}^{wr}\in\mathbb{N}^{+} $, 使得$[a_{i}(w^{-1}r)b_{j}]^{n_{ij}^{wr}}=0,
$则$(a_{i}rb_{j})^{n_{ij}^{wr}}=0, $于是$a_{i}Rb_{j}\subseteq N(
R), i, j=0, 1.$所以$R$是$N$ -弱拟Armendariz环.
环$R[x;x^{-1}]=\{\sum\limits_{i=k}^{n}m_{i}x^{i}|m_{i}\in R, k, n\in \mathbb{Z}\}$, 其中加法和乘法运算与$R[x]$中相同.
推论2.13 设$R$为环.则$R[x]$是$N$ -弱拟Armendariz环当且仅当$R[x;x^{-1}]$是$N$ -弱拟Armendariz环.
证 令$\triangle =\{1, x, x^{2}, \cdots\}, $则$\triangle$为$R[x]$中中心正则元构成的乘法封闭子集.
$R[x;x^{-1}]=\triangle ^{-1}R[x], $由命题2.12即证.
2015, Vol. 35 
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