Localisation is isomorphic to a quotient of polynomial ring I am having trouble with the following problem. 

Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$. 

I just want a hint.
Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks

In the quotient $R[x]/(ax-1)$, the class of $x$ is invertible. What is its inverse?
This will tell you where $x$ must be mapped under $R[x]\to R_D$ if this must will induce an isomorphism $R[x]/(ax-1)\to R_D$.
