I've been reading Ravi Vakil's notes on algebraic geometry notes and am having a hard time connecting the standard definition of the localization of a module to the definition in terms of universal properties (page 24).

Can anybody either explain the connection or provide some kind of 'roadmap' of the material necessary to understand this connection?

I apologize if this question is vague or unclear.

EDIT: This is an attempt to reformulate my original question to make it more precise:

Suppose we did not know the definition of the localization of a module and we just saw the diagram presented on page 24, how do we recover the usual definition and how do we determine that anything else satisfying those properties is isomorphic to the usual definition?

  • $\begingroup$ Sorry, I edited the link to go directly to the pdf. it should work now $\endgroup$
    – WWright
    Sep 11, 2010 at 21:53
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    $\begingroup$ Isn't this the content of exercise 2.3.E? $\endgroup$ Sep 11, 2010 at 21:55
  • $\begingroup$ I guess. I think I might close this question until I can figure out what I'm trying to ask. I think I just need more time to let universal properties sit. I just don't see how to think of the regular definition as a universal property. $\endgroup$
    – WWright
    Sep 11, 2010 at 22:02
  • $\begingroup$ For much further motivation on the universal viewpoint I highly reccomend George Bergman's An Invitation to General Algebra and Universal Constructions, math.berkeley.edu/~gbergman/245 $\endgroup$ Sep 11, 2010 at 22:11
  • $\begingroup$ Imho it's better to start by first comprehending the universality of polynomial rings - where the essential ideas are much more familiar. $\endgroup$ Sep 11, 2010 at 22:14

3 Answers 3


In my opinion, the most enlightening (and the simplest) way to present the universal construction of localizations (and fractions) is to use instead of the pair construction the natural presentation in terms of generators and relations. This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (and to avoid the many tedious verifications required in the pair approach). Moreover, this approach is much more conceptual. Indeed, the pairs in the pair construction are nothing but normal forms for the polynomial terms in the presentation based approach. For details of this approach see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, and Voloch's: Rings of fractions the hard way. Note: presumably Voloch's title is a joke - since the presentation based approach is actually the easiest way - in fact both Rotman's and Voloch's expositions can be simplified.

If you're just beginning to understand universal constructions then I highly recommend that you peruse the beautiful exposition in George Bergman's An Invitation to General Algebra and Universal Constructions.

You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume [1].

[1] Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002


As the Rotman's, or Voloch's, constructions of Dubuque's answers show, you can't recover exactly "the usual definition" of the localization of a ring from its universal property. That is, a universal property won't tell you, in general, how to construct an object that verifies it (as you can see, there are at least, two constructions for the localization, both verifying the universal property: how are you going to choose among them?). What a universal property will tell you is that all the guys verifying it are necessarily isomorphic: for instance, the construction in Vakil's notes, and Rotman's are necessarily isomorphic.

Precisely, let's stay with rings and let $\gamma' : A \longrightarrow B$ be some guy that satisfies the universal property of the localization. Let $\gamma: A \longrightarrow S^{-1}A$ denote the construction of the localization explained in Vakil's notes, 2.3.3. Then, since both $S^{-1}A$ and $B$ satisfy the universal property, you'll have morphisms $\widetilde{\gamma} : S^{-1}A \longrightarrow B$ and $\widetilde{\gamma'}: B \longrightarrow S^{-1}A$ such that $\widetilde{\gamma'}\gamma = \gamma'$ and $\widetilde{\gamma}\gamma' = \gamma$. Let's show that both $\widetilde{\gamma}$ and $\widetilde{\gamma'}$ are inverses of each other. For instance, $\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B$ because

$$ (\widetilde{\gamma'} \widetilde{\gamma})\gamma' = \widetilde{\gamma'}\gamma = \gamma' \ . $$

But $\mathrm{id}_B$ verifies the same identity clearly:

$$ \mathrm{id}_B \gamma'= \gamma' \ . $$

Hence, because of the uniqueness of the universal property, you must have

$$ \widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B \ . $$


$$ \widetilde{\gamma}\widetilde{\gamma'} = \mathrm{id}_{S^{-1}A} \ . $$

What you can (and must) do is to verify that the usual definition (particular construction) of the localization verifies the universal property. That is, given some morphism of rings $\varphi : A \longrightarrow B$ such that $\varphi s \in B$ is a unit for every $s \in S$, then there is a unique morphism of rings $\widetilde{\varphi} : S^{-1}A \longrightarrow B$ such that $\widetilde{\varphi}\gamma = \varphi$.

Hint: you can easily check that

$$ \widetilde{\varphi} \left( \frac{a}{s} \right) = \frac{\varphi (a)}{\varphi (s)} $$

is well-defined (that is, if $\frac{a}{s} = \frac{b}{t}$, then $\widetilde{\varphi}\left( \frac{a}{s} \right) = \widetilde{\varphi}\left( \frac{b}{t} \right) $), is a morphism of rings, verifies $\widetilde{\varphi} \gamma = \varphi$ and is unique (that is, if you had another $\psi : S^{-1}A \longrightarrow B$ such that $\psi\gamma = \varphi$, then $\psi = \widetilde{\varphi}$).

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    $\begingroup$ The two ways are not different! The pairs in the pair construction are simply normal forms of the polynomials in the presentation by generators and relations. Using the presentation approach explains the genesis of the pair normal forms and, moreover, simplifies matters since one can use universal properties of polynomials rings and quotient constructions to eliminate the tedious details of the direct pair construction (checking the ring axioms etc). Note: the presentation approach is folklore - not original to Rotman or Voloch. $\endgroup$ Sep 12, 2010 at 16:07

Since you already obtained very good answers, let me add a short remark. In some cases, you can indeed quite easily derive a concrete construction from an universal property.

For instance, let's assume you already know the universal property of the limit of a functor $F$ (as a terminal $F$-cone) and you want to derive an explicit construction of the limit $\lim F$ in the special case that $F : \mathcal{C} \to \mathrm{Set}$ maps to the category of sets.

Then recall that a set $X$ is determined by its "global elements", i.e. by its maps $1 \to X$ where $1 = \{ \heartsuit \}$ is any one-element set ("the lonely heart"): $X \cong \mathrm{Hom}_\mathrm{Set}(1,X)$. Combining this observation and the universal property, we obtain

$$ \lim F \cong \mathrm{Hom}_\mathrm{Set}(1, \lim F) \cong \text{set of $F$-cones with apex $1$} \cong \{ (x_i)_{i \in \mathcal{C}}\ |\ F(f)(x_i) = x_j \text{ for all $f : i \to j$ in $\mathcal{C}$} \}.$$

A similar calculation can be used to determine the (underlying set of) the categorical product of vector spaces (use $\mathrm{Hom}_{\mathrm{Vect}(\mathbb{R})}(\mathbb{R}, \cdot)$ instead of $\mathrm{Hom}_\mathrm{Set}(1,\cdot)$). It does not work to derive a description of colimits in $\mathrm{Set}$.


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