## 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: