Prologue~
"We shall not cease from exploration
And the end of all our exploring
Will be to arrive where we started
And know the place for the first time."
— T. S. Eliot, Four Quartets
In the previous post, we approached localization through its universal property. Given a ring $R$ and a multiplicative subset $S$, we sought a ring $S^{-1}R$ together with a morphism $ \iota: R \longrightarrow S^{-1}R$ such that every element of $S$ becomes invertible, and such that $S^{-1}R$ is universal with respect to this property.
This point of view is extremely powerful: rather than constructing an object explicitly and then studying its properties, we first describe what the object ought to do and only later worry about how it is built. In many ways, this is one of the recurring themes of modern mathematics. There is, however, another perspective on localization that is equally illuminating.
Verse I: The Architecture~
The answer is surprisingly elegant: localization can be realized as a colimit. Colimits are usually introduced as a way of gluing together diagrams of objects. Localization, on the other hand, appears to be a process of adjoining inverses. The common link between the two is--when we form the localization $S^{-1}R$, every element $s\in S$ is required to become invertible. Informally, this means that multiplication by $s$, $R\xrightarrow{\cdot s}R$, should become an isomorphism after localization.
Rather than forcing all these inverses to exist simultaneously, we examine the collection of maps induced by the elements of $S$. The multiplicative subset $S$ naturally carries a preorder. Given $s,t\in S$, we write $ s\leq t $ if there exists some $u\in S$ such that $ t=su. $This relation reflects the idea that $t$ is a refinement of the denominator $s$. Since $S$ is multiplicatively closed, any two elements $s$ and $t$ admit a common upper bound, namely $st$. Thus $S$ becomes a filtered preorder.
So essentially, to each $s\in S$, we associate a copy of the $R$-module $R$. Whenever $t=su$, we obtain a morphism $R\xrightarrow{\cdot u}R$. In this way, $S$ determines a filtered diagram of $R$-modules. The remarkable fact is that the localization is precisely the colimit of this diagram. $$ S^{-1}R \cong \varinjlim_{s\in S} R$$
Verse II: A Necessary Interlude~
Before proving the above theorem, let us recall the notion of a filtered diagram and its colimit. Roughly speaking, a filtered diagram is a directed system of objects in which any finite collection of objects can be compared at some later stage. The colimit of such a diagram may be thought of as the object obtained by identifying elements that eventually become equal.
Given a poset $(P, \leq)$, we construct a category whose objects are elements of $P$ and morphisms are $\operatorname{Hom}(x,y)= \{\ast\}$ when $ x\le y$ and $\operatorname{Hom}(x,y)= \varnothing$ otherwise. In this setting, the notion of a filtered category becomes particularly transparent.
A poset $P$ is said to be filtered if it is nonempty and every pair of elements admits a common upper bound. That is, for any $x,y \in P$, there exists some $z \in P$ such that $x \leq z$ and $y \leq z$. Moreover, since $S$ is multiplicatively closed, any two elements $s,t \in S$ possess a common upper bound, namely $st$, because $s \leq st$ and $t \leq st$. Thus $S$ forms a filtered preorder and naturally serves as an indexing category for the diagram whose colimit will produce the localization $S^{-1}R$.
Verse III: The Limit of the Journey~
A filtered category is very useful because it provides an indexing system for directed collections of objects and morphisms. Let $\mathscr{C}$ be a category and let $I$ be a filtered category. A diagram indexed by $I$ is simply a functor $D: I \to \mathscr{C}$. For each object $i \in I$, the functor assigns an object $D(i) \in \mathscr{C}$, and for each morphism $i \to j$, in $I$, it assigns a morphism $D(i) \to D(j)$ in $\mathscr{C}$. When $I$ is a filtered preorder, we can visualise this data as a directed system $D(i_1)\longrightarrow D(i_2)\longrightarrow D(i_3)\longrightarrow\cdots$ in which any finite collection of objects eventually maps into a common stage. The colimit of such a diagram is an object that records all the information contained in the system while identifying elements that become equal at sufficiently large stages.
More precisely, let $D:I\to R-\mathbf{Mod} $ be a filtered diagram of $R$-modules. The colimit of $D$, denoted $\varinjlim_{i\in I}D(i), $ is characterized by the following universal property: for every $R$-module $M$, giving a morphism, $\varinjlim_{i\in I}D(i)\to M$ is equivalent to giving a compatible family of morphisms $D(i) \to M$ for all $i \in I$.
Concretely, filtered colimits of modules admit a particularly simple description. Consider the direct sum $ \bigoplus_{i\in I}D(i). $ Whenever $ f_{ij}:D(i)\to D(j) $ is a morphism in the diagram and $x \in D(i)$, we impose the relation $ x\sim f_{ij}(x)$. The colimit is obtained by quotienting all such relations. $$\varinjlim_{i\in I}D(i) = \left(\bigoplus_{i\in I}D(i)\right)\Big/ \left\langle x-f_{ij}(x) \right\rangle$$
Verse IV: The Familiar Object ~
Now, keeping in mind the above discussions, we delve into the concept of localization. The multiplicative subset $S$ determines a filtered category, and to each $s \in S$ we associate a copy of the $R$-module $R$. The resulting filtered diagram will have colimit $ \varinjlim_{s\in S}R,$ and our goal is to show that this colimit is naturally isomorphic to the localization $S^{-1}R.$
We return to the multiplicative subset . Recall that $S$ carries a filtered preorder given by $ s\leq t \iff t=su$ for some $u\in S.$ To each object $s \in S$, assign the $R$-module $R$. Thus, $ D(s)=R$ whenever $s\leq t, $ say $t=su, $ we define the corresponding morphism $D(s)\longrightarrow D(t) $ to be multiplication by $u: r \longmapsto ru. $
This assignment defines a functor $D:S\to R\text{-}\mathbf{Mod}.$ The resulting diagram therefore has the form $R \xrightarrow{\cdot u} R \xrightarrow{\cdot v} R \xrightarrow{\cdot w} \cdots.$ The colimit $\varinjlim_{s\in S}R$ is obtained by identifying an element $r \in D(s)$ with its image $ru\in D(su).$ In other words, $(r,s)\sim (ru,su)$ for every $r\in R,\quad u\in S.$ And this is precisely the relation which appears in the construction of fractions.
Verse V: The Identification~
The discussion above culminates in the following theorem.
Theorem. Let $R$ be a commutative ring and $S$ a multiplicative subset of $R$. Consider the filtered diagram $D:S\to R\text{-}\mathbf{Mod}$ defined by $D(s)=R$ and $D(s\le t)=\cdot u:R\to R$ whenever $t=su.$. Then localization of $R$ at $S$ is naturally isomorphic to the colimit of this diagram. That is, $S^{-1}R \cong \varinjlim_{s\in S}R. $
Essentially, localization may be realized as the filtered colimit of copies of $R$connected by multiplication maps coming from the elements of $S$.
Proof. Recall that the colimit of a diagram of $R$-modules may be described as the quotient $\left(\bigoplus_{s\in S}R\right)\Big/\sim,$ where the equivalence relation is generated by $(r,s)\sim (ru,su)$ for every $r\in R,\qquad u\in S.$
We define a map $\Phi:\varinjlim_{s\in S}R\longrightarrow S^{-1}R$ by $(r,s) \mapsto \frac{r}{s}$.
We first check that $\Phi$ is well-defined. Suppose $(r,s) \sim (ru,su).$ Then $\Phi(ru,su)=\frac{ru}{su}.$ Since $R$ is commutative, $\frac{ru}{su}=\frac{r}{s},$ and therefore $ \Phi(r,s)=\Phi(ru,su).$
To show surjectivity, observe that every element of $S^{-1}R$ is represented by a fraction $\frac{r}{s}$, which is the image of the class of $(r,s)$.
for injectivity, suppose $ \Phi(r,s)=\Phi(r',t).$. Then $\frac{r}{s}=\frac{r'}{t}$ in $S^{-1}R$. By the definition of localization, there exists $u \in S$ such that $ u(rt-r's)=0.$
Multiplying both pairs by suitable elements of $S$, both $(r,s)$ and $(r',t)$ map to the stage indexed by $stu$. Their images there are $rtu$ and $r'su,$ which are equal by the previous relation. Hence the two classes coincide in the colimit.
Therefore, $\Phi$ is bijective. Since both constructions carry natural $R$-module structures and $\Phi$ preserves them, $\Phi$ is an isomorphism. Consequently, $S^{-1}R \cong \varinjlim_{s\in S}R.$
Epilogue~
"What we call the beginning is often the end
And to make an end is to make a beginning."
— T. S. Eliot, Four Quartets
This post was inspired by the categorical description of localization mentioned in The Rising Sea by Ravi Vakil. In the next post, we will look at the Ore condition and localization for non-commutative rings (NCRs).