Localisation isomorphic to a quotient of polynomial ring [duplicate]

Solution 1:

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.

Solution 2:

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.