If $\{f\in R[x]\:|\:f\text{ monic}\}$ is a right denominator set, is $\{f^i\:|\:i\geq 0\}$ a right denominator set also?

Let $R$ be a right (and left) Noetherian ring and $T=R[x]$ its polynomial ring. It was shown by Stafford that the set $S=\{f\in T\:|\:f\text{ monic}\}$ is a right denominator set. So my question is, if $g\in S$, then is $G=\{g^a\:|\:a\geq 0\}$ a right denominator set also?

If $t\in T$ and $g^i$ then I want to show $tG\cap g^i T\neq\emptyset$. As $g^i\in S$ then I know there exists some $t^\prime\in T$ and $f\in S$ such that $g^i t^\prime=tf$. So the question then becomes, does $f=g^j$ for some $j\geq 0$?

Edit: I thought of using the following result:

If $A\subset B$ is a multiplicative subset of a ring $B$ such that $aB$ is a two-sided ideal for any $a\in A$, then $A$ is a right permutable set.

Now, for $g^i\in G$ it is clear that $g^iT$ is right ideal of $T$. So, let $t\in T$ then $tg^iS\cap g^iT\neq\emptyset$, i.e. for some $s\in S$ and $t^\prime\in T$ we have $tg^is=g^it^\prime$. This seems promising but then I am unsure of how to proceed. I need to show $tg^i\bar{t}\in g^iT$ but do we need $\bar{t}=s\bar{t}^\prime$?

No, this is not true in general.

Let $g = x+\alpha$, let $t= \beta$ be an invertible element . Then we need to find $t'$ in $R[x]$ such that $(x+\alpha) t' = \beta (x+\alpha)^j = (x+ \beta \alpha \beta^{-1})^j \beta$, so $(x+\alpha) (t' \beta^{-1}) =(x+\beta \alpha \beta^{-1})^j$.

In particular, take $R$ to be a matrix algebra with $\alpha = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\beta = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, so $\beta \alpha \beta^{-1}$ is $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.

If $t' \beta^{-1}$ satisfies this identity, then it must be a diagonal matrix, as any non-diagonal matrix polynomial will have nonzero off-diagonal entries when multiplied by $x+\alpha$ (by looking only at the highest-degree non-diagonal entries and multiplying them by $x$ to get the highest-degree non-diagonal entries of the product, we can see that there is no cancellation in the top degree).

So $t'$ if it exists is a diagonal matrix, which implies that $x+0$ divides $(x+1)^j$ in the ring of polynomials in one variable, which is false, and thus $t'$ does not exist.

• Thanks for the answer, Will. You mention group algebras. What if $R$ is a group algebra over a finite group? Aug 3, 2017 at 22:08
• @SamWilliams: I can make this counter example work, by a similar argument, in a 2 x 2 matrix algebra, which of course occurs in finite group algebras. Aug 4, 2017 at 1:40
• @SamWilliams The counterexample is now simplified and applies to the group algebras of, e.g., $D_4$. Aug 5, 2017 at 0:37