Throughout this paper, $S$ always stands for a monoid. A nonempty set $A$ is called a right $S$-act, if there exists a mapping $A\times S\rightarrow A, (a, s)\mapsto as$, satisfying the conditions $(as)t=a(st)$ and $a1=a$, for all $a\in A$ and all $s, t\in S$. Let $S$ be a monoid and $A$ be a right $S$-act. $A$ is called principally weakly flat if the functor $A\otimes-$ preserves embeddings of principal left ideals of $S$ into $S$. In this article, we generalize regular monoids, and define a new class of monoids, which is called $n$-regular monoids. Using a generalization of principal weak flatness, we give some characterizations of these monoids, and some known results are easily obtained.

Definition 1.1 Let $S$ be a monoid and $n$ a positive integer. A right $S$-act $A$ is called principally weakly $n$-flat if for every $s\in S$, $A\otimes Ss^n\rightarrow A\otimes S$ is monic.

Remark 1.2  It is clear that in the above definition, when $n=1$, then a right $S$-act $A$ is principally weakly $n$-flat if and only if $A$ is principally weakly flat. When $n\geq 2$, we will show that there exists a right $S$-act $A$ which is principally weakly $n$-flat but not principally weakly flat.

Suppose $I$ is a proper right ideal of $S$, by [1] the amalgam of two copies of $S$ with the core $I$ is defined as follows. If $x,y$ and $z$ denote elements not belonging to $S$, define $A(I)=(\{x,y\}\times (S-I))\cup (\{z\}\times I),$ and define a right $S$-action on $A(I)$ by

$\begin{eqnarray*} (x,u)s&=&\left\{ \begin{array}{ccc} (x,us), & \mbox{if $us \notin I,$} & \\ (z,us), & \mbox{ if $us \in I$,} & \end{array} \right.\\ (y,u)s&=&\left\{ \begin{array}{ccc} (y,us), & \mbox{if $us \notin I,$} & \\ (z,us), & \mbox{ if $us \in I$, } & \end{array} \right.\\ (z,u)s&=&(z,us). \end{eqnarray*}$

Then $A(I)$ is a right $S$-act.

Lemma 1.3 [3] Let $S$ be a monoid, $a,a^{\prime}\in A_S,b,b^{\prime}\in_{S}\mspace{-6mu}B$. Then $a\otimes b=a^{\prime}\otimes b^{\prime}$ in $A_S\otimes _{S}\!B$ if and only if there exist a positive integer $m$ and elements $a_1,\cdots,a_m\in A_S, b_2,\cdots,b_m\in _{S}\mspace{-6mu}B$, $s_1,t_1,\cdots,s_m,t_m\in S$ such that

$\begin{alignat*}{2} &a=a_1s_1,\ \\ &a_1t_1= a_2s_2, &\hspace{30pt} &s_1b=t_1b_2,\\ &a_2t_2= a_3s_3, & &s_2b_2=t_2b_3,\\ &\ \ \hspace{6.2mm} \vdots & &\ \ \hspace{6.6mm} \vdots\\ &a_mt_m= a^{\prime}, & &s_mb_m=t_mb^{\prime}. \end{alignat*}$

Lemma 1.4  Let $S$ be a monoid, $A$ an $S$-act and $n$ a positive integer. The following conditions are equivalent:

(1) $A$ is principally weakly $n$-flat.

(2) For every $s\in S$, any $a,a^{\prime}\in A$, $a\otimes s^n=a^{\prime}\otimes s^n$ in $A\otimes S$ implies that there exist $m\in N$, elements $a_1,\cdots,a_m\in A_S$, and $s_1,t_1,\cdots,s_m,t_m\in S$ such that

$\begin{alignat*}{2} &a=a_1s_1,\ \\ &a_1t_1=a_2s_2, &\hspace{30pt} &s_1s^n=t_1s^n,\\ &a_2t_2=a_3s_3, & &s_2s^n=t_2s^n,\\ &\ \ \hspace{6.2mm} \vdots & &\ \ \hspace{6.6mm} \vdots\\ &a_mt_m=a^{\prime}, & &s_ms^n=t_ms^n. \end{alignat*}$

Proof (1)$\Rightarrow$(2) Since $A$ is principally weakly $n$-flat, for every $s\in S$, any $a,a^{\prime}\in A$, $a\otimes s^n=a^{\prime}\otimes s^n$ in $A\otimes S$ implies that $a\otimes s^n=a^{\prime}\otimes s^n$ in $A\otimes Ss^n$. By Lemma 1 there exists a positive integer $m$ and elements $a_1,\cdots,a_m\in A_S, q_2,\cdots,q_m\in Ss^n$, $u_1,v_1,\cdots,u_m,v_m\in S$ such that

$\begin{alignat*}{2} &a=a_1u_1,\ \\ &a_1v_1= a_2u_2, &\hspace{30pt} &u_1s^n=v_1q_2,\\ &a_2v_2= a_3u_3, & &u_2q_2=v_2q_3,\\ &\ \ \hspace{6.2mm} \vdots & &\ \ \hspace{6.6mm} \vdots\\ &a_mv_m= a^{\prime}, & &u_mq_m=v_ms^n. \end{alignat*}$

Since $q_2,q_3,\cdots q_m\in Ss^n$, there exist $c_2,c_3,\cdots,c_m\in S$ such that $q_i=c_is^n(i=2,\cdots,m)$. Hence we have

$\begin{alignat*}{2} &a=a_1u_1,\ \\ &a_1v_1c_2= a_2u_2c_2, &\hspace{30pt} &u_1s^n=v_1c_2s^n,\\ &a_2v_2c_3= a_3u_3c_3, & &u_2c_2s^n=v_2c_3s^n,\\ &\ \ \hspace{6.2mm} \vdots & &\ \ \hspace{6.6mm} \vdots\\ &a_{m-1}v_{m-1}c_m= a_mu_mc_m, & &u_{m-1}c_{m-1}s^n=v_{m-1}c_ms^n,\\ &a_mv_m= a^{\prime}, & &u_mc_ms^n=v_ms^n. \end{alignat*}$

Denote $u_1=s_1, v_m=t_m, u_ic_i=s_i, v_{i-1}c_i=t_{i-1}, i=2,\cdots,m.$ Then the result follows.

(2)$\Rightarrow$(1) For every $s\in S$, any $a,a^{\prime}\in A$, $a\otimes s^n=a^{\prime}\otimes s^n$ in $A\otimes S$. By (2) there exist $m\in N$, elements $a_1,\cdots,a_m\in A_S$, and $s_1,t_1,\cdots,s_m,t_m\in S$ such that

$\begin{alignat*}{2} &a=a_1s_1,\ \\ &a_1t_1=a_2s_2, &\hspace{30pt} &s_1s^n=t_1s^n,\\ &a_2t_2=a_3s_3, & &s_2s^n=t_2s^n,\\ &\ \ \hspace{6.2mm} \vdots & &\ \ \hspace{6.6mm} \vdots\\ &a_mt_m=a^{\prime}, & &s_ms^n=t_ms^n. \end{alignat*}$

Now

$\begin{alignat*}{2} a\otimes s^n&=a_1s_1\otimes s^n=a_1\otimes s_1s^n=a_1\otimes t_1s^n=a_1t_1\otimes s^n\\ &=a_2s_2\otimes s^n=a_2\otimes s_2s^n=a_2\otimes t_2s^n=a_2t_2\otimes s^n=\cdots=a^{\prime}\otimes s^n \end{alignat*}$

in $A\otimes Ss^n$.

A monoid $S$ is called regular, if for every $s\in S$, there exists $x\in S$ such that $s=sxs$. Now we have the following:

Definition 2.1  Let $n$ be a positive integer. A monoid $S$ is called a $n$-regular monoid, if for every $s\in S$, there exists $x\in S$ such that $s^n=s^nxs^n$.

Lemma 2.2  Let $S$ be a monoid, $I$ a proper right ideal of $S$ and $n$ a positive integer. Then the following conditions are equivalent:

(1) $A(I)$ is principally weakly $n$-flat.

(2) For every $s\in S$, if there exists $r\in S-I$ such that $rs^n\in I$, then there exists $j\in I$ such that $rs^n=js^n$.

Proof (1)$\Rightarrow$(2) Since $A(I)$ is principally weakly $n$-flat, for every $s\in S$, if there exists $r\in S-I$ such that $rs^n\in I$, then $(x,r)\otimes s^n=(y,r)\otimes s^n$ in $A(I)\otimes S$. By Lemma 1.4 there exist $m\in N$, $p_1,\cdots,p_m\in S$, $s_1,t_1,\cdots,s_m,t_m\in S,$ and $w_1,\cdots,w_m\in \{x,y,z\}$ such that

$\begin{alignat*}{2} (x,r)&=(w_1,p_1)s_1,\ \\ (w_1,p_1)t_1&=(w_2,p_2)s_2, &\hspace{45pt} s_1s^n&=t_1s^n,\\ &\ \ \vdots & &\ \ \vdots\\ (w_{m-1},p_{m-1})t_{m-1}&=(w_m,p_m)s_m, & s_{m-1}s^n&=t_{m-1}s^n,\\ (w_m,p_m)t_m&=(y,r), &s_ms^n&=t_ms^n. \end{alignat*}$

There exists $k\in \{1,\cdots,m-1\}$ such that $w_k\neq w_{k+1}$, and so, there exists $j\in I$ such that $p_kt_k=p_{k+1}s_{k+1}=j$. So we have

$rs^n=p_1s_1s^n=p_1t_1s^n=p_2s_2s^n=\cdots=p_kt_ks^n= js^n.$

(2)$\Rightarrow$(1) Let for every $s\in S$, any $a,a^{\prime}\in A(I)$, $a\otimes s^n=a^{\prime}\otimes s^n$ in $A(I)\otimes S$. Since $(x,1)S\cong S\cong (y,1)S$(a free hence principally weakly $n$-flat $S$-act), without loss of generality we may restrict ourselves to the case in which $a=(x,r_1), a^{\prime}=(y,r_2), r_1,r_2\in S-I$. Since $(x,r_1)\otimes s^n=(y,r_2)\otimes s^n$, then $r_1s^n=r_2s^n\in I.$ By (2) there exists $j\in I$ such that $r_1s^n=js^n=r_2s^n$. Hence

$(x,r_1)\otimes s^n=(x,1)\otimes r_1s^n=(x,1)\otimes js^n=(y,1)\otimes js^n=(y,r_2)\otimes s^n$

in $A(I)\otimes Ss^n.$

Corollary 2.3  [3] Let $S$ be a monoid and $I$ be a proper right ideal of $S$. Then the following conditions are equivalent:

(1) $A(I)$ is principally weakly flat.

(2) For every $i\in I$, there exists $j\in I$ such that $i=ji$.

Proposition 2.4  Let $S$ be a monoid, $I$ a proper right ideal of $S$ and $n$ a positive integer. Then the following conditions are equivalent:

(1) $S/I$ is principally weakly $n$-flat.

(2) For every $s\in S$, if there exists $r\in S-I$ such that $rs^n\in I$, then there exist $j\in I$ such that $rs^n=js^n$.

Proof (1)$\Rightarrow$(2) Since $S/I$ is principally weakly $n$-flat, for every $s\in S$, if there exists $r\in S-I$ such that $rs^n\in I$, then for every $j^{\prime}\in I$, we have $[r]\otimes s^n=[j^{\prime}]\otimes s^n$ in $S/I\otimes S$. By Lemma 1.4 there exist $m\in N$, $u_1,\cdots,u_m\in S$, and $s_1,t_1,\cdots,s_m,t_m\in S$ such that

$\begin{alignat*}{2} [r]&=[u_1]s_1,\ \\ [u_1]t_1&=[u_2]s_2, &\hspace{45pt} s_1s^n&=t_1s^n,\\ &\ \ \vdots & &\ \ \vdots\tag{$*$}\\ [u_{m-1}]t_{m-1}&=[u_m]s_m, & s_{m-1}s^n&=t_{m-1}s^n,\\ [u_m]t_m&=[j^{\prime}], &s_ms^n&=t_ms^n. \end{alignat*}$

Since $j^{\prime}\in I$, we have $u_mt_m\in I$. Let $k$ be the least number such that $k\in \{1,2,\cdots,m\}$ and $u_kt_k\in I$. If $j=u_kt_k$, then $u_{k-1}t_{k-1}\in S-I$. Since $[u_{k-1}]t_{k-1}=[u_k]s_k$, we have $u_{k-1}t_{k-1}=u_ks_k$ and so

$rs^n=u_1s_1s^n=u_1t_1s^n=u_2s_2s^n=u_2t_2s^n=\cdots=\\ u_{k-1}t_{k-1}s^n=u_ks_ks^n=u_kt_ks^n= js^n.$

(2)$\Rightarrow$(1) For every $s\in S$, any $u,u^{\prime}\in S$, if $[u]\otimes s^n=[u^{\prime}]\otimes s^n$ in $S/I\otimes S$, we have the following four cases to consider:

Case 1 $u,u^{\prime}\in I$. Then it is clear that $[u]\otimes s^n=[u^{\prime}]\otimes s^n$ in $S/I\otimes Ss^n$.

Case 2 $u\in I$. $u^{\prime}\in S-I$. By assumption there exists $j\in I$ such that $u^{\prime}s^n=js^n$. Then

$[u]\otimes s^n=[j]\otimes s^n=[1]\otimes js^n=[1]\otimes u^{\prime}s^n=[u^{\prime}]\otimes s^n$

in $S/I\otimes Ss^n$.

Case 3 $u\in S-I$. $u^{\prime}\in I$. It is similar to Case 2.

Case 4 $u,u^{\prime}\in S-I$. By $[u]\otimes s^n=[u^{\prime}]\otimes s^n$ in $S/I\otimes S$, we have $us^n=u^{\prime}s^n$ or $us^n,u^{\prime}s^n\in I$. If $us^n=u^{\prime}s^n$, the result follows. Otherwise by assumption there exists $j_1,j_2\in I$ such that $us^n=j_1s^n$ and $u^{\prime}s^n=j_2s^n$. So

$[u]\otimes s^n=[1]\otimes us^n=[1]\otimes j_1s^n=[j_1]\otimes s^n=[j_2]\otimes s^n=[1]\otimes j_2s^n=[1]\otimes u^{\prime}s^n=[u^{\prime}]\otimes s^n$

in $S/I\otimes Ss^n$.

Corollary 2.5 [4] Let $S$ be a monoid and $I$ be a proper right ideal of $S$. Then the following conditions are equivalent:

(1) $S/I$ is principally weakly flat.

(2) For every $i\in I$, there exists $j\in I$ such that $i=ji$.

Theorem 2.6 Let $n$ be a positive integer. The following conditions on a monoid $S$ are equivalent:

(1) All right $S$-acts are principally weakly $n$-flat.

(2) All finitely generated right $S$-acts are principally weakly $n$-flat.

(3) $S$ is a $n$-regular monoid.

Proof (1)$\Rightarrow$(2) is clear.

(2)$\Rightarrow$(3) Let $s\in S$. If $s^nS=S$, it is clear that there exists $x\in S$ such that $s^n=s^nxs^n$. Otherwise $s^nS$ is a proper right ideal of $S$. Denote $I=s^nS$, then $A(I)$ is a finitely generated $S$-act, by assumption $A(I)$ is principally weakly $n$-flat. Since $s^n\in I$ and by Lemma 2.2 there exists $j\in I$ such that $s^n=js^n$. Hence there exists $x\in S$ such that $j=s^nx$, that is $s^n=s^nxs^n$.

(3)$\Rightarrow$(1) Suppose $A$ is a right $S$-act. Since $S$ is $n$-regular, for every $s\in S$, there exists $x\in S$ such that $s^n=s^nxs^n$. For any $a,a^{\prime}\in A$, if $a\otimes s^n=a^{\prime}\otimes s^n$ in $A\otimes S$. Then

$a\otimes s^n=a\otimes s^nxs^n=as^n\otimes xs^n=a^{\prime}s^n\otimes xs^n=a^{\prime}\otimes s^nxs^n=a^{\prime}\otimes s^n$

in $A\otimes Ss^n.$

Corollary 2.7 [5] The following conditions on a monoid $S$ are equivalent:

(1) All right $S$-acts are principally weakly flat.

(2) All finitely generated right $S$-acts are principally weakly flat.

(3) $S$ is a regular monoid.

Theorem 2.8  Let $n$ be a positive integer. The following conditions on a monoid $S$ are equivalent:

(1) All cyclic right $S$-acts are principally weakly $n$-flat.

(2) All Rees factor right $S$-acts are principally weakly $n$-flat.

(3) $S$ is a $n$-regular monoid.

Proof (1)$\Rightarrow$(2) is clear.

(2)$\Rightarrow$(3) By Proposition 2.4, the proof is similar to Theorem 2.6.

(3)$\Rightarrow$(1) By Theorem 2.6, it is clear.