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For a multiplicatively closed subset $S$ of $A$, we have a functor $S^{-1}: A-Mod \rightarrow S^{-1}A-Mod$.

I am trying to understand this functor a little bit better and I was thinking about the following universal property:

Given a ring homomorphism $g: A \rightarrow B$ where every element of $S$ is sent to a unit, there is a unique map $\tilde{g}: S^{-1}A \rightarrow B$ where $\tilde{g} \circ f=g$. Here $f$ is the usual map $A \rightarrow S^{-1}A$ which sends $a \mapsto a/1$.

I learned about universal properties formally as initial objects in comma categories (MacLane). This is also stated as a universal arrow from some object $c$ in a category $C$ to some functor $T: D \rightarrow C$.

I was wondering if it was possible to view the above property in terms of this language. I've gotten a bit confused thinking about it now

Some Thoughts We can view $B$ as an $B$ module and hence as an $A$ module via $a \cdot b=g(a)b$. We can also probably view $B$ as an $S^{-1}A$ module via $(a/s) \cdot b= g(a)g(s)^{-1}b$.

I am a bit stuck from here.

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Well, that universal property is a statement about purely ring homomorphisms, so I describe a solution along these lines.

Consider the category ${\bf C}:={\bf Ring^{mlt}}$ of objects $(R,S)$ where $S$ is a multiplciatively closed subset of the ring $R$ and morphisms $\varphi:(R,S)\to(R',S')$ with $\varphi:R\to R'$ ring homomorphism such that $\varphi(s)\in S'$ for all $s\in S$.

Let $\bf D$ be its full subcategory on objects $(R,S)$ with $S\subseteq R^{-1}$, i.e. each $s\in S$ is invertible in $R$.

Then, the property translates as $(S^{-1}R,\,S)$ is just the reflection of $(R,S)$ on $\bf D$, that is, a universal arrow to the inclusion functor ${\bf D}\hookrightarrow{\bf C}$.

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  • $\begingroup$ Thanks, I was wondering along the lines of a restricted subcategory of commutative rings. What you have said seems to be exactly what I was looking for. Thanks for taking the time for writing this, it's of great help. $\endgroup$ – user135520 Mar 2 '15 at 4:58
  • $\begingroup$ Is that really correct? I think the reflective subcategory should consist of the objects $(R, S)$ where $S$ contains all the units, not the other way around. (Consider the extreme case where $S = \emptyset$.) $\endgroup$ – Zhen Lin Mar 2 '15 at 9:14
  • $\begingroup$ Well, isn't it the case that if $S\subseteq R^{-1}$ then $S^{-1}R=R$? $\endgroup$ – Berci Mar 3 '15 at 19:34

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