Let $R$ be a right Noetherian ring and $S=\{f\in R[x]\;|\;f\text{ monic}\}$. It is a result of Stafford that $S$ is a right denominator set in $R[x]$, so in particular we can localize $R[x]$ at any $f\in S$. My question is this, can we localize $R[x,x^{-1}]$ at any $f\in S$? I know,


$R[x,x^{-1}]\_f\cong R[x]\_{xf}\cong R[x]\_{fx}$,


and of course $R[x]\_f$ and $(R[x]\_f)\_x$ both exist, but is this enough?


I suppose the question is this, does the above guarantee $\{f^n\:|\:n\geq 0\}$ is a right denominator set in $R[x,\,x^{-1}]$? As $R[x,x^{-1}]$ is a Laurent polynomial ring over a Noetherian ring, then it is also right Noetherian, so I know that I only need to check $\{f^n\:|\:n\geq 0\}$ is right Ore, but does the above guarantee this?



Right localization of $R[x,x^{-1}]$ at monic $f\in R[x]$
I may be wrong here but does this not follow from the fact $S=\{f^n\:|\:n\geq 0\}$ is right Ore? Consider $f^{m}, h$ for some $m\geq 0$ and $h\in R[x,\,x^{-1}]$. Let $y=x^{-1}$ and note that if $h\in R[x]$ then there is nothing to prove. For simplicity, suppose $h\in R[y]$ and has degree $k$, say. Then choose $h^\prime\in R[y]$ that is also of degree $k$. Then,


$f^{m} h^\prime=f^{m} y^k g=y^k f^{m} g=y^k g^{\prime}f^{l}$,


for $l\geq 0,\; g, g^\prime\in R[x]. Now, h=y^k \tilde{h}$ so we need to choose $g$ such that $g^\prime=\tilde{h}$. However, we can always do this because $S$ is right Ore. A similar argument exists if $h\in R[x,x^{-1}]$ (as opposed to just $R[y]$.


