Proof. Using the fact that $\sigma \left(B\right)=B$ and if $c\in C\left(N\right)$, then $\sigma \left(c\right)\in C\left(N\right)$, it can be easily proved that $\sigma \left(T_N\left(B\right)\right)=T_N\left(B\right)$. □

Proof. $\sigma \left(N\right)\subseteq N$ because $\sigma \left(N\right)$ is a nilpotent ideal of $B$. Let $n\in N$. Then $n=\sigma \left(a\right)$ for some $a\in B$. Therefore ${\sigma }^{-1}\left(n\right)=a$, and ${\sigma }^{-1}\left(N\right)=\left\{a\in B\right.$, such that $a={\sigma }^{-1}\left(n\right)$, some $\left.n\in N\right\}$ = I (say) is an ideal of $B$. Now $\sigma \left(I\right)\subseteq \sigma \left(N\right)$. Hence $\sigma \left(N\right)=N$. □

Proof. Note that $ax=x\sigma \left(a\right)$ for all $a\in B$. Let $S=\left\{x^m{a}_m+...+a_0\in B\left[x,\sigma \right]f\right.$, some m$\left.\right\}\cup \left\{0\right\}$. Let $a_m\in S$ and $c\in R$. Then we have some $g=\sum x^i{a}_i\in B\left[x,\sigma \right]$, (i = 1, 2, ..., n) and $f=\sum x^j{d}_j$, (j = 1, 2, ..., t) regular in $B\left[x,\sigma \right]$ such that leading coefficient of $gf$ is $a_m$; i.e. ${\sigma }^t\left(a_n\right)d_t=a_m$. Now $h=\sum x^j{\sigma }^{-t}\left(c\right)a_j\in B\left[x,\sigma \right]$, (j = 1, 2, ..., t), and therefore $hf$ has leading coefficient in $S$; i.e. $c{\sigma }^t\left(a_n\right)d_t\in S$. Thus $S$ is a left ideal. We now show that $S$ is essential. Let $0\ne I\subseteq B$ be a left ideal of B. Then it is easy to see that $I\left[x,\sigma \right]$ is a left ideal of $B\left[x,\sigma \right]$. Now $f\in B\left[x,\sigma \right]$ is regular, therefore $B\left[x,\sigma \right]f$ is an essential left ideal of $B\left[x,\sigma \right]$; i.e. $I\left[x,\sigma \right]\cap B\left[x,\sigma \right]f\ne \left(0\right)$. Let $\sum x^i{a}_i\in I\left[x,\sigma \right]\cap B\left[x,\sigma \right]f$, (i = 1, 2, ..., k). Then $a_k\in I$ and $a_k\in S$, and therefore $S$ is essential as a left ideal. So $S$ contains a left regular element by Goldie‘s Theorem, see for example [1, Theorem (1.37)]. Now $B$ is semiprime Noetherian implies that $S$ contains a regular element. Hence there exists $g\in B\left[x,\sigma \right]$ such that $gf$ has leading coefficient regular in $B$. □

Proof. By 2.1 above $\sigma \left(T_N\left(B\right)\right)=T_N\left(B\right)$, therefore $\left(T_N\left(B\right)\right)\left[x,\sigma \right]$ is a right ideal of $B\left[x,\sigma \right]$. Now on the same lines as in [6, Proposition (1.1)] with some manipulations on $\sigma$, it can be easily proved that $T_N\left(B\left[x,\sigma \right]\right)\subseteq \left(T_N\left(B\right)\right)\left[x,\sigma \right]$. □

Proof. Let $F_{\alpha }$ = Sum of all submodules of $A$ of Krull dimension less than $\alpha$. Then as in [6, Theorem(1.2)], $F_{\alpha }=\cap A_{\alpha }$, where $A_{\alpha }$ are ideals of $A$ such that $A∕A_{\alpha }$ is $\alpha$-homogeneous. Also observe that $A∕\sigma \left(F_{\alpha }\right)$ is isomorphic to $A∕F_{\alpha }$. Therefore $A∕\sigma \left(F_{\alpha }\right)$ is $\alpha$-homogeneous. So $\sigma \left(F_{\alpha }\right)\subseteq F_{\alpha }$. Now let $F_{\alpha }^{\ast }$ = Sum of submodules of $A\left[x,\sigma \right]$ having Krull dimension less than $\alpha +1$. Then $F_{\alpha }\left[x,\sigma \right]\subseteq F_{\alpha }^{\ast }$ and $F_{\alpha }^{\ast }=\cap A_{\alpha }^{\ast }$, where $A\left[x,\sigma \right]∕A$ is $\alpha +1$-homogeneous. Let $Ann\left(M\right)=B$, where $Ann\left(M\right)$ denotes the annihilator of $M$. Then $M$ is a critical $A\left[x,\sigma \right]∕B$ module, which is also faithful since $\mid M\mid =\alpha +1$, therefore $\mid A\left[x,\sigma \right]∕B\mid =\alpha +1$. Now $M$ is critical with $\mid M\mid =\alpha +1$, so by [2] $A\left[x,\sigma \right]∕B$ is isomorphic to a submodule of direct sum of n copies of $M$. This easily yields that $A\left[x,\sigma \right]∕B$ is $\alpha +1$-homogeneous. Hence $F_{\alpha }\left[x,\sigma \right]\subseteq B$. Thus $M$ is an $A\left[x,\sigma \right]∕F_{\alpha }\left[x,\sigma \right]$ module. If $N\left(A∕F_{\alpha }\right)$ is the prime radical of $A∕F_{\alpha }$, then since $A∕F_{\alpha }$ is $\alpha$-homogeneous, $\left(A∕F_{\alpha }\right)∕N\left(A∕F_{\alpha }\right)$ is also $\alpha$-homogeneous, because $A$ is a commutative ring. Now using [6, Theorem (1.2)] and the fact that every critical module is compressible over $A∕F_{\alpha }$ by [4, Theorem (2.5)], we get by [2, Proposition (3.6)] that $\left(A∕F_{\alpha }\right)∕T_N\left(A∕F_{\alpha }\right)$ has an artinian quotient ring, therefore by [5, Theorem (3.1)] $\left(A∕F_{\alpha }\right)\left[x,\sigma \right]∕\left(T_N\left(A∕F_{\alpha }\right)\right)\left[x,\sigma \right]$ has an artinian quotient ring. Now 2.5 implies that

Now since

is $\alpha +1$-homogeneous as $\left(A∕F_{\alpha }\right)∕N\left(A∕F_{\alpha }\right)$ is $\alpha$-homogeneous, so again by an application of [2, Proposition (3.6)], we get that $M$ as a module over $\left(A∕F_{\alpha }\right)\left[x,\sigma \right]$ is compressible. Hence $M$ has a prime annihilator. □