Localisation isomorphic to a quotient of polynomial ring 
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?
 A: The homomorphism $A^{-1}R \to R[T]/(aT - 1)$ sends $\frac{r}{a^i} \mapsto rT^i$.  The way to define it is to first define a homomorphism $R \to R[T]/(aT - 1)$ and then use the universal property of localizations.
I think the easiest way to prove it's an isomorphism is to just define a map in the other direction and prove that the composition either way is the identity.
A: 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.
A: Define a ring homomorphism:$$\alpha: R[T] \longrightarrow A^{-1}R$$ $$r\longmapsto r/1, \mbox{ for $r \in R$} $$ $$T\longmapsto1/a.$$
Clearly $\alpha$ is surjective.
Let's prove that $\ker(\alpha)=(Ta-1)$.
$(\supseteq)$ Clear.
$(\subseteq)$ $h\in \ker(\alpha)\Rightarrow h \in(Ta-1)?$
$h=h(T)$ satisfies $h(1/a)=0 \in A^{-1}R$, that is, $a^nh(1/a)=0 \in R$, for some $n \geq \deg h$. Then $a^nh(T)=G(aT)$, where $G=G(Y)\in R[Y]$ satisfies $G(1)=0$. So $G(Y)=(Y-1)G_1(Y)$ for some polynomial $G_1(Y)$, so that $a^nh(T)=G(aT)=(aT-1)G_1(aT)\Rightarrow a^nh(T)\in (aT-1)$ for some $n$.
Notice $a$ and $aT-1$ are coprime, so that $$a^nh(T)\in (aT-1)\Rightarrow h(T)\in (aT-1).$$
Indeed, $1=aT-(aT-1)$, so that by taking $n^{th}$-powers and using the binomial theorem
$$1=a^nT^n+r(aT-1) \mbox{ for some $r \in R[T]$}.$$
Therefore $h(T)=T^na^nh(T)+r(aT-1)h(T)\in (aT-1)$.
I hope this help you.
A: What about $\Phi: R[T]\to A^{-1}R$ with $T\mapsto 1/a$?
