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

Here's another answer using the universal property in another way (I know it's a bit late, but is it ever too late ?)
As for universal properties in general, the ring satisfying the universal property described by Arturo Magidin in his answer is unique up to isomorphism. Thus to show that $R[x]/(ax-1) \simeq R_D$, it suffices to show that $R[x]/(ax-1)$ has the same universal property ! 
But that is quite easy: let $\phi: R\to T$ be a ring morphism such that $\phi(a) \in T^{\times}$. 
Using the universal property of $R[x]$, we get a unique morphism $\overline{\phi}$ extending $\phi$ with $\overline{\phi}(x) = \phi(a)^{-1}$. 
Quite obviously, $ax -1 \in \operatorname{Ker}\overline{\phi}$. Thus $\overline{\phi}$ factorizes uniquely through $R[x]/(ax-1)$. 
Thus we get a unique morphism $\mathcal{F}: R[x]/(ax-1) \to T$ with $\mathcal{F}\circ \pi = \phi$, where $\pi$ is the canonical map $R\to R[x]/(ax-1)$. This shows that $\pi: R \to R[x]/(ax-1)$ has the universal property of the localization, thus it is isomorphic to the localization. 
This is essentially another way of seeing Arturo Magidin's answer
