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Given a ring $R$, suppose we have two multiplicative subsets $S,U \subseteq R$ (which contain 1). Write $US = \{us \mid u \in U, s \in S\}$ also a multiplicative subset, containing both $S$ and $U$

Writing $\lambda: R \to R_S$ the localisation map, so that $\lambda(U)$ is also a multiplicative subset. I'm trying to find a sleek proof that $$ \left( R_S \right)_{ \lambda(U)} \cong R _ {SU} $$ which uses universal properties of localisation.

I'm currently trying to show that $\left( R_S \right)_{ \lambda(U)}$ also has the universal property:

Given $\phi: R \to T$ such that $SU$ maps to units in $T$, then $S$ maps to units in $T$ so we get a unique ring homomorphism $\phi_1$ such that the following commutes: $$\begin{array}{cccc} R & \xrightarrow{\phi} & T \\ & \searrow & \uparrow & \phi_1 \\ & & R_S \end{array}$$ and then $\phi_1$ maps $\lambda(U)$ into units of $T$ (since $\phi_1 \circ \lambda(U) = \phi(U)$ by the diagram) yielding unique $\phi_2$ such that $$\begin{array}{cccc} R_S & \xrightarrow{\phi_1} & T \\ & \searrow & \uparrow & \phi_2 \\ & & \left( R_S \right)_{ \lambda(U)} \end{array}$$ commutes.

This yields that we get a commutative diagram with unique $\phi_2$ $$\begin{array}{cccccc} R & \xrightarrow{\phi} & T & \xleftarrow{\phi_2} & \left( R_S \right)_{ \lambda(U)}\\ & \searrow & \uparrow & \nearrow \\ & & R_S \end{array}$$

So existence follows. My problem is that this contruction depends on $\phi_1$. This is where I'm struggling: the universal property would need uniqueness with $$\begin{array}{cccc} R & \xrightarrow{\phi_1} & T \\ & \searrow & \uparrow & \phi_2 \\ \mu \circ \lambda& & \left( R_S \right)_{ \lambda(U)} \end{array}$$

where $\mu: R_S \to \left( R_S \right)_{ \lambda(U)}$ is the natural localisation map. (and $\mu \circ \lambda$ is the diagonal arrow. I'm sorry about the label placement - I wouldn't normally use arrays but apparently AMScd doesn't have diagonal arrows)

It feels like something along the lines of this would be sufficient, but I'm not sure: given a commutative $$\begin{array}{cccccc} &R & \xrightarrow{\phi} & T \\ \lambda &\downarrow & & \uparrow & \phi_2 \\ &R_S & \xrightarrow{\mu} & \left( R_S \right)_{ \lambda(U)} \end{array}$$ then can we simply define $\phi_1 := \phi_2 \circ \mu$ which yields the diagram above, which is unique by our first argument, and then uniqueness of $\phi_2$ follows?

Finishing off from there if this is true, we use universal property of $R_{SU}$ and recently shown one for $\left( R_S \right)_{ \lambda(U)}$ that there are unique $\phi$,$\psi$ such that:

$$ \begin{array}{ccccc} && R \\ & \swarrow & \downarrow & \searrow \\ R _ {SU} & \xrightarrow{\phi} & \left( R_S \right)_{ \lambda(U)} & \xrightarrow{\psi} & R _ {SU} \end{array}$$

and so again using universal property of $R_{SU}$, we have uniqueness of the identity of $R_{SU}$ so that $1 = \psi \circ \phi$ and similarly $1 = \phi \circ \psi$ yielding the isomorphism.

(I also realise showing $$ \dfrac{r}{s} / \dfrac{u}{1} \mapsto \dfrac{r}{su} $$ is an isomorphism is relatively quick, but showing well-definedness and injectivity gets quite element heavy, and I'm hoping for something different.)

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Your approach is correct. You have shown that for $(R_S)_{\lambda(U)}$ and $\phi:R\rightarrow T$ satisfying $\phi(US) \subseteq R^\times$ you find an extension $\phi_2$.

Now suppose that $\psi:(R_S)_{\lambda_U}\rightarrow T$ is another extension. Note that the composite $\sigma:R_S \overset{\mu}\rightarrow (R_S)_{\lambda(U)} \overset{\psi}\rightarrow T$ satisfies $\sigma \circ \lambda = \phi$ by assumption, so by the universal property of $R_S$ you have $\sigma = \phi_1$. But now you are done, since by the universal property of $(R_S)_{\lambda(U)}$ we obtain $\psi = \phi_2$.

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