Localization of a ring

Note: The definition of “ring” used here does not require an identity for multiplication.

Let R be a commutative ring, and suppose S \subseteq R is some subset with the property that 0 \notin S, S contains no zero divisors, and S is closed under multiplication. Then by considering R \times S and defining and quotienting out an equivalence relation \sim (to “ignore” different representations of fractions), we can form something known as the ring of quotients, or localization, of elements of R by elements of S, denoted Q(R,S).

Let us assume we have this situation, and write Q = Q(R,S). Then the ring Q has a couple of properties:

  • R can be embedded into Q, that is, it can be viewed as a subring of Q.
  • Every s \in S, under the above identification, becomes a unit in Q.
  • Up to isomorphism, Q is the unique smallest ring containing R which also satisfies the previous property.

This is not a precise statement, so let us try to flesh out what we really want to say. In the first property, what is it exactly that we mean by “embedding” R into Q? Well, in algebra, the word embedding is usually used as a synonym for injective homomorphism (to be concise: monomorphism). So indeed, with the first property, we really mean to say that there exists a ring monomorphism \iota : R \to Q. You can think of this map as taking the structure of R and carrying it into Q in a rigid manner, so it is indeed justified to term it an embedding (I say “rigid” due to the map’s injectivity). This map \iota, when instead viewed as a map from R to \mathrm{Image}(\iota), then becomes a ring isomorphism between R and some subring of Q.

Now that we have properly articulated the first property, let us move on to the second. It is quite simple to restate this one: if we take any s \in S and consider it as an element of Q by way of our embedding \iota, the claim is that it will be invertible. In other words, \iota(s) is a unit of Q for all s \in S.

The third property is where things begin to get interesting, because it is what is known as a universal mapping property (in disguise). If you have seen an abstract treatment of tensor products, for example, you will recognize its form. It turns out that it takes quite a bit of effort to really say what we mean here. So let’s reason this out. What do we mean by “Q is the unique smallest ring containing R satisfying the second property”? Well, of course Q does not, strictly speaking, “contain” R, but rather, R is embeddable into Q. Also, we want to get across the idea that Q can always be embedded in any ring Q' to which R admits a ring monomorphism with the second property, and this embedding is moreover uniquely determined by such a monomorphism. So our statement becomes “Q is the unique smallest ring T to which R admits a ring monomorphism \varphi such that \varphi(s) is a unit in T for every s \in S“. We are getting closer, but still not quite there yet. To make the “smallest ring” bit formal, we have to observe that this means the following:

For any commutative, unital ring Q' and ring monomorphism \varphi : R \to Q' such that \varphi(s) is a unit in Q' for every s \in S, there is a unique ring monomorphism \psi : Q \to Q' with the property that \psi \circ \iota = \varphi.

By now, you are almost certainly rolling your eyes at how long this is getting, but hilariously enough, we are still not quite finished. Here’s the subtlety: we really want to provide the assurance that indeed, Q is characterized by this property. That is, we want to say that there do not exist two structurally different (non-isomorphic) “smallest” rings Q and \tilde{Q} “containing” R with the property that every s \in S becomes a unit in that ring. After all, if we’re going to use words like “smallest”, then one usually speaks of the smallest object in a collection, not a smallest object in a collection. We need to communicate that this is, indeed, the case. So, the penultimate revision of our condition is:

For any commutative, unital ring Q' and ring monomorphism \varphi : R \to Q' such that \varphi(s) is a unit in Q' for every s \in S, there is a unique ring monomorphism \psi : Q \to Q' with the property that \psi \circ \iota = \varphi. Moreover, this property characterizes Q up to isomorphism.

What do we mean by this “characterization”? Well, we mean this:

For any commutative, unital ring Q' and ring monomorphism \varphi : R \to Q' such that \varphi(s) is a unit in Q' for every s \in S, there is a unique ring monomorphism \psi : Q \to Q' with the property that \psi \circ \iota = \varphi.

Moreover, suppose Q_1 and Q_2 are two rings, both having the above property. (That is, suppose that the above property holds, when every occurrence of Q is replaced by Q_1, and also when every occurrence is replaced by Q_2). Then there exists a ring isomorphism \beta : Q_1 \overset{\cong}{\to} Q_2.

The first part is illustrated by this diagram:

And the second part is (perhaps poorly) illustrated by this one:

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About mlbaker

just another guy trying to make the diagrams commute.
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