# Localisation isomorphic to a quotient of polynomial ring [duplicate]

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?

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• There is no morphism from $A^{-1}R$ to $R[T]$ – Georges Elencwajg Apr 24 '14 at 19:21

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 $h(T)=T^na^nh(T)+r(aT-1)h(T)\in (aT-1)$.

• +1 very nice and elementary solution. Thank you for writing. – Mojojojo Sep 29 '18 at 20:59

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.

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.

What about $\Phi: R[T]\to A^{-1}R$ with $T\mapsto 1/a$?