For $m \in \text{max} \ R[X]$ and $f \in R[X] \setminus m, \ R[X]_f$ is not a local ring Let $R$ be a commutative ring with $1$ and $m$ be a maximal ideal of $R[X]$. Let $f \in R[X]$ such that $f \notin m$. Then I want to show that $R[X]_f$ is not a local ring, where $R[X]_f$ is the localization of $R[X]$ with respect to the multiplicative closed set $\{ 1 , f ,f ^{2}, \ldots \}$.


My effort: On the contrary if we assume that $R[X]_f$ is local then clearly $R[X]_f \cong R[X]_{m}.$ 

But then I couldn't draw any contradiction. Please don't give me a reference where it was proved that there is another maximal ideal $n$ of $R[X]$ different from $m$ such that $f \notin n$ as well. To be honest I want to prove the previously stated fact in a different way.
We refer to the following commutative diagram:



Let $\text{loc}_f:R[X]\to R[X]_f$ be the canonical localization and $\eta:R[X]\to R[X]/m$ be the canonical quotient. Note that $R[X]/m$ is a field with $\eta(f)\neq 0$ and thus by the universal property of localization that there exists a unique ring morphism $k:R[X]_f\to R[X]/m$ with $\eta=k\circ \text{loc}$. Clearly $\text{ker}(k)\subsetneq R[X]_f$ is maximal. 

Now denote by $\pi:R[X]\to R[X]$ the unique automorphism fixing $R\subset R[X]$ and taking $x\mapsto x+1$ and by $\pi^{-1}$ its inverse. From the U.P. of localization this induces an automorphism $\text{loc}(\pi):R[X]_f\to R[X]_f$ such that the "obvious" square $\text{loc}_f\circ \pi= \text{loc}(\pi)\circ \text{loc}_f$ commutes.

Observe that $\text{ker}(k\circ \text{loc}(\pi)^{-1})\subsetneq R[X]_f$ is like $\text{ker}(k)$ maximal. But $$k\circ\text{loc}_f=\eta=k\circ \text{loc}(\pi)^{-1}\circ \text{loc}_f\circ \pi$$ From which it follows that $$\text{loc}_f(p)\in \text{ker}(k)\iff p\in \text{ker}(\eta)\iff \text{loc}_f(p+1)\in \text{ker}(k\circ \text{loc}(\pi)^{-1})$$ And thus that $\text{ker}(k)$ and $\text{ker}(k\circ \text{loc}(\pi)^{-1})$ are disjoint maximal ideals of $R[X]_f$, as desired $\blacksquare$
We have $R[X]_{m} /(m\cdot R[X]_m)$ is a field. Whereas, 
$B = (R[x]_f)/(m \cdot R[x]_f) = (R/m) [X]_f$ is not a field as the image $\bar{f}$, of $f$ in $(R/m) [X]$ which is a principal ideal domain with infinitely many prime ideals, has only finitely many prime
divisors and all the prime ideal to which $\bar{f}$ does not belong survive in $B$.