# Ring of fractions mapping property

Let $R$ be a commutative ring with unit and let $S \subseteq R$ be a multiplicative closed subset. Consider the ring of fractions $S^{-1}R$ with the homomorphism $f\colon R \to S^{-1}R$ which sends $r$ to $(rs)/s$. If $T$ is a commutative ring with unit for which there exists a homomorphism $\alpha$ s.t. $\alpha(s)$ is invertible in $T$ for any $s$ in $S$, then there exists a unique homomorphism $g\colon S^{-1}R \to T$. This happens for all $T$ and for all $\alpha$ satisfying the previous properties.

Now, let $A$ be a commutative ring with unit such that there exists a homomorphism $\beta\colon R \to A$ with $\beta(s)$ invertible in $T$ for every $s \in S$. Suppose that for all commutative rings with unit $T$ and for all homomorphisms $\alpha\colon R \to T$ with $\alpha(s)$ invertible in $T$ for all $s\in S$ there exists a unique homomorphism $\gamma\colon A \to T$ such that $\gamma \circ \beta = \alpha$, then

is it true that there exists an isomorphism between $S^{-1}R$ and $A$ ?

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Yes, by the standard argument that objects defined by universal properties are unique up to unique isomorphism. Namely, take $\alpha: R \to S^{-1}R$ to be the obvious map. This yields $\gamma: A \to S^{-1}R$ which will be an isomorphism. – Tom Bachmann Mar 10 '12 at 17:05

You are asking whether the ring of fractions $S^{-1}R$ is uniquely defined (up to isomorphism) by its universal property. This is indeed true.
Let $A$ and $B$ be two rings which both satisfy the universal property of $S^{-1}R$. Then there exist homomorphisms $\alpha: R \to A$ and $\beta: R \to B$ such that $\alpha(s)$ and $\beta(s)$ are invertible for all $s\in S$. Since both $A$ and $B$ have the universal property there are unique homomorphisms $\varphi: A \to B$ and $\psi: B \to A$ such that $\beta = \varphi \circ \alpha$ and $\alpha = \psi \circ \beta$. (You might want to draw a diagram to visualize the homomorphisms). Since $\alpha(s)$ is invertible for all $s \in S$ there is also a unique homomorphism $f: A \to A$ with $\alpha = f \circ \alpha$. Since the identity on $A$ is such a homomorphism, by uniqueness we must have $f = \mathrm {id}_A$. But $\psi \circ \varphi : A \to A$ is another homomorphism with $\alpha = (\psi \circ \varphi) \circ \alpha$ (by commutativity of the diagrams), so $\psi \circ \varphi$ must also be the identity on $A$. By interchanging the roles of $A$ and $B$ we see similarly that $\varphi \circ \psi$ is the identity on $B$, so $\varphi$ and $\psi$ are isomorphisms.