[proofplan]
We construct explicit $R$-algebra homomorphisms in both directions and verify they are mutually inverse. The map $R_u \to R[T]/(uT-1)$ is obtained from the universal property of localisation by observing that the image of $u$ in $R[T]/(uT-1)$ is a unit. The map $R[T]/(uT-1) \to R_u$ is obtained from the universal property of the polynomial ring by sending $T \mapsto u^{-1} \in R_u$, which kills the relation $uT - 1$. The mutual inverse property follows from the explicit formulas on generators.
[/proofplan]
[step:Construct $\varphi: R_u \to R[T]/(uT-1)$ via the universal property of localisation]
Let $\overline{R} := R[T]/(uT-1)$ and let $\pi: R[T] \to \overline{R}$ be the canonical projection. Define the $R$-algebra structure map $f: R \to \overline{R}$ by $f(r) = \pi(r)$ (the composition of the inclusion $R \hookrightarrow R[T]$ with $\pi$). We verify that $f(u)$ is a unit in $\overline{R}$: indeed, $f(u) = \pi(u)$ and $\pi(u) \cdot \pi(T) = \pi(uT) = \pi(1) = 1_{\overline{R}}$, since $uT - 1 \in (uT - 1)$. So $f(u)$ is a unit with inverse $\pi(T)$.
Since $R_u$ is the localisation of $R$ at the multiplicative set $\{1, u, u^2, \dots\}$, and $f(u)$ is a unit in $\overline{R}$, the [universal property of localisation](/theorems/???) yields a unique $R$-algebra homomorphism
\begin{align*}
\varphi: R_u &\to \overline{R} \\
\frac{r}{u^n} &\mapsto f(r) \cdot f(u)^{-n} = \pi(r) \cdot \pi(T)^n = \pi(rT^n).
\end{align*}
That is, $\varphi\!\left(\frac{r}{u^n}\right) = rT^n + (uT - 1)$.
[guided]
The localisation $R_u$ inverts the single element $u$, so its universal property states: for any ring homomorphism $f: R \to A$ such that $f(u)$ is a unit in $A$, there is a unique extension to $R_u$. We apply this with $A = \overline{R} = R[T]/(uT-1)$.
Why is $f(u)$ a unit? The defining relation in $\overline{R}$ is $uT \equiv 1 \pmod{(uT-1)}$, so $\pi(u) \cdot \pi(T) = 1$. The inverse of $\pi(u)$ is $\pi(T)$.
The explicit formula $\varphi(r/u^n) = \pi(rT^n)$ arises from $r/u^n = r \cdot u^{-n}$ in $R_u$, which maps to $f(r) \cdot f(u)^{-n} = \pi(r) \cdot \pi(T)^n = \pi(rT^n)$.
[/guided]
[/step]
[step:Construct $\psi: R[T]/(uT-1) \to R_u$ via the universal property of the polynomial ring]
By the [Universal Property of Polynomial Algebra](/theorems/2820), the assignment $T \mapsto u^{-1} := \frac{1}{u} \in R_u$ extends uniquely to an $R$-algebra homomorphism
\begin{align*}
\Phi: R[T] &\to R_u \\
p(T) &\mapsto p(u^{-1}),
\end{align*}
where $R_u$ is regarded as an $R$-algebra via the canonical map $\iota: R \to R_u$, $r \mapsto r/1$. We verify that $(uT - 1) \subseteq \ker \Phi$: $\Phi(uT - 1) = u \cdot u^{-1} - 1 = 1 - 1 = 0$ in $R_u$.
By the universal property of quotient rings, $\Phi$ descends to a unique $R$-algebra homomorphism
\begin{align*}
\psi: \overline{R} &\to R_u \\
p + (uT-1) &\mapsto p(u^{-1}).
\end{align*}
[/step]
[step:Verify $\psi \circ \varphi = \operatorname{id}_{R_u}$ and $\varphi \circ \psi = \operatorname{id}_{\overline{R}}$]
**$\psi \circ \varphi = \operatorname{id}_{R_u}$:** For an arbitrary element $\frac{r}{u^n} \in R_u$,
\begin{align*}
(\psi \circ \varphi)\!\left(\frac{r}{u^n}\right) = \psi(\pi(rT^n)) = (rT^n)\big|_{T = u^{-1}} = r \cdot (u^{-1})^n = \frac{r}{u^n}.
\end{align*}
**$\varphi \circ \psi = \operatorname{id}_{\overline{R}}$:** It suffices to check on a generating set. Since $\overline{R}$ is generated as an $R$-algebra by the element $\pi(T) = T + (uT-1)$, and both $\varphi \circ \psi$ and $\operatorname{id}_{\overline{R}}$ are $R$-algebra homomorphisms, it suffices to check on $\pi(T)$:
\begin{align*}
(\varphi \circ \psi)(\pi(T)) = \varphi(u^{-1}) = \varphi\!\left(\frac{1}{u}\right) = \pi(T).
\end{align*}
Since both maps agree on the $R$-algebra generator $\pi(T)$ and are $R$-algebra homomorphisms, $\varphi \circ \psi = \operatorname{id}_{\overline{R}}$.
Hence $\varphi$ and $\psi$ are mutually inverse $R$-algebra isomorphisms, and $R_u \cong R[T]/(uT-1)$.
[guided]
For the first composition, we trace an element $r/u^n$ through both maps: $\varphi$ sends it to the coset $rT^n + (uT-1)$, and then $\psi$ evaluates this polynomial at $T = u^{-1}$, recovering $r \cdot u^{-n} = r/u^n$.
For the second composition, rather than checking on every coset $p + (uT-1)$, we use the fact that both $\varphi \circ \psi$ and $\operatorname{id}_{\overline{R}}$ are $R$-algebra endomorphisms of $\overline{R}$. An $R$-algebra endomorphism of $\overline{R}$ is determined by where it sends the generator $\pi(T)$ (since every element of $\overline{R}$ is a polynomial in $\pi(T)$ with $R$-coefficients). We compute: $\psi(\pi(T)) = u^{-1} \in R_u$, and then $\varphi(u^{-1}) = \varphi(1/u) = \pi(T)$. So $\varphi \circ \psi$ fixes $\pi(T)$, and therefore $\varphi \circ \psi = \operatorname{id}_{\overline{R}}$.
This is a standard verification pattern for universal property arguments: construct maps in both directions using universal properties, then check mutual inverse on generators.
[/guided]
[/step]
