TITLE: Localisation isomorphic to a quotient of polynomial ring
QUESTION [7 upvotes]: Let $R$ be a commutative ring and $A=\{1,a,a^2,\dots\}$ for some $a\in R$. Prove that $A^{-1}R$ is isomorphic to $R[T]/(aT-1)$. 

I guess I'm meant to find a surjective homomorphism between $A^{-1}R$ and $R[T]$ and then use first isomorphism theorem. What homomorphism should I use?

REPLY [3 votes]: This question becomes even easier if you know the universal properties (which some people would call the "right definitions") of localization, of polynomial rings, and of quotients.
$A^{-1}R$ is the universal example of a commutative ring with a homomorphism from $R$ that sends all elements of $A$ to invertible elements. That's equivalent to just sending $a$ to an invertible element.
$R[T]/(aT-1)$ is the universal example of a commutative ring with a homomorphism from $R$ and an element $T$ that serves as an inverse for the image of  $a$.
Since the two universal properties are equivalent, the rings (and homomorphisms) they define are isomorphic.