Definition 10.9.1. Let $R$ be a ring, $S$ a subset of $R$. We say $S$ is a *multiplicative subset of $R$* if $1\in S$ and $S$ is closed under multiplication, i.e., $s, s' \in S \Rightarrow ss' \in S$.

## 10.9 Localization

Given a ring $A$ and a multiplicative subset $S$, we define a relation on $A \times S$ as follows:

It is easily checked that this is an equivalence relation. Let $x/s$ (or $\frac{x}{s}$) be the equivalence class of $(x, s)$ and $S^{-1}A$ be the set of all equivalence classes. Define addition and multiplication in $S^{-1}A$ as follows:

One can check that $S^{-1}A$ becomes a ring under these operations.

Definition 10.9.2. This ring is called the *localization of $A$ with respect to $S$*.

We have a natural ring map from $A$ to its localization $S^{-1}A$,

which is sometimes called the *localization map*. In general the localization map is not injective, unless $S$ contains no zerodivisors. For, if $x/1 = 0$, then there is a $u\in S$ such that $xu = 0$ in $A$ and hence $x = 0$ since there are no zerodivisors in $S$. The localization of a ring has the following universal property.

Proposition 10.9.3. Let $f : A \to B$ be a ring map that sends every element in $S$ to a unit of $B$. Then there is a unique homomorphism $g : S^{-1}A \to B$ such that the following diagram commutes.

**Proof.**
Existence. We define a map $g$ as follows. For $x/s\in S^{-1}A$, let $g(x/s) = f(x)f(s)^{-1}\in B$. It is easily checked from the definition that this is a well-defined ring map. And it is also clear that this makes the diagram commutative.

Uniqueness. We now show that if $g' : S^{-1}A \to B$ satisfies $g'(x/1) = f(x)$, then $g = g'$. Hence $f(s) = g'(s/1)$ for $s \in S$ by the commutativity of the diagram. But then $g'(1/s)f(s) = 1$ in $B$, which implies that $g'(1/s) = f(s)^{-1}$ and hence $g'(x/s) = g'(x/1)g'(1/s) = f(x)f(s)^{-1} = g(x/s)$. $\square$

Lemma 10.9.4. The localization $S^{-1}A$ is the zero ring if and only if $0\in S$.

**Proof.**
If $0\in S$, any pair $(a, s)\sim (0, 1)$ by definition. If $0\not\in S$, then clearly $1/1 \neq 0/1$ in $S^{-1}A$.
$\square$

Lemma 10.9.5. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. The category of $S^{-1}R$-modules is equivalent to the category of $R$-modules $N$ with the property that every $s \in S$ acts as an automorphism on $N$.

**Proof.**
The functor which defines the equivalence associates to an $S^{-1}R$-module $M$ the same module but now viewed as an $R$-module via the localization map $R \to S^{-1}R$. Conversely, if $N$ is an $R$-module, such that every $s \in S$ acts via an automorphism $s_ N$, then we can think of $N$ as an $S^{-1}R$-module by letting $x/s$ act via $x_ N \circ s_ N^{-1}$. We omit the verification that these two functors are quasi-inverse to each other.
$\square$

The notion of localization of a ring can be generalized to the localization of a module. Let $A$ be a ring, $S$ a multiplicative subset of $A$ and $M$ an $A$-module. We define a relation on $M \times S$ as follows

This is clearly an equivalence relation. Denote by $m/s$ (or $\frac{m}{s}$) be the equivalence class of $(m, s)$ and $S^{-1}M$ be the set of all equivalence classes. Define the addition and scalar multiplication as follows

It is clear that this makes $S^{-1}M$ an $S^{-1}A$-module.

Definition 10.9.6. The $S^{-1}A$-module $S^{-1}M$ is called the *localization* of $M$ at $S$.

Note that there is an $A$-module map $M \to S^{-1}M$, $m \mapsto m/1$ which is sometimes called the *localization map*. It satisfies the following universal property.

Lemma 10.9.7. Let $R$ be a ring. Let $S \subset R$ a multiplicative subset. Let $M$, $N$ be $R$-modules. Assume all the elements of $S$ act as automorphisms on $N$. Then the canonical map

induced by the localization map, is an isomorphism.

**Proof.**
It is clear that the map is well-defined and $R$-linear. Injectivity: Let $\alpha \in \mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, N)$ and take an arbitrary element $m/s \in S^{-1}M$. Then, since $s \cdot \alpha (m/s) = \alpha (m/1)$, we have $ \alpha (m/s) =s^{-1}(\alpha (m/1))$, so $\alpha $ is completely determined by what it does on the image of $M$ in $S^{-1}M$. Surjectivity: Let $\beta : M \rightarrow N$ be a given R-linear map. We need to show that it can be "extended" to $S^{-1}M$. Define a map of sets

Clearly, this map respects the equivalence relation from above, so it descends to a well-defined map $\alpha : S^{-1}M \rightarrow N$. It remains to show that this map is $R$-linear, so take $r, r' \in R$ as well as $s, s' \in S$ and $m, m' \in M$. Then

and we win. $\square$

Example 10.9.8. Let $A$ be a ring and let $M$ be an $A$-module. Here are some important examples of localizations.

Given $\mathfrak p$ a prime ideal of $A$ consider $S = A\setminus \mathfrak p$. It is immediately checked that $S$ is a multiplicative set. In this case we denote $A_\mathfrak p$ and $M_\mathfrak p$ the localization of $A$ and $M$ with respect to $S$ respectively. These are called the

*localization of $A$, resp. $M$ at $\mathfrak p$*.Let $f\in A$. Consider $S = \{ 1, f, f^2, \ldots \} $. This is clearly a multiplicative subset of $A$. In this case we denote $A_ f$ (resp. $M_ f$) the localization $S^{-1}A$ (resp. $S^{-1}M$). This is called the

*localization of $A$, resp. $M$ with respect to $f$*. Note that $A_ f = 0$ if and only if $f$ is nilpotent in $A$.Let $S = \{ f \in A \mid f \text{ is not a zerodivisor in }A\} $. This is a multiplicative subset of $A$. In this case the ring $Q(A) = S^{-1}A$ is called either the

*total quotient ring*, or the*total ring of fractions*of $A$.If $A$ is a domain, then the total quotient ring $Q(A)$ is the field of fractions of $A$. Please see Fields, Example 9.3.4.

Lemma 10.9.9. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module. Then

where the preorder on $S$ is given by $f \geq f' \Leftrightarrow f = f'f''$ for some $f'' \in R$ in which case the map $M_{f'} \to M_ f$ is given by $m/(f')^ e \mapsto m(f'')^ e/f^ e$.

**Proof.**
Omitted. Hint: Use the universal property of Lemma 10.9.7.
$\square$

In the following paragraph, let $A$ denote a ring, and $M, N$ denote modules over $A$.

If $S$ and $S'$ are multiplicative sets of $A$, then it is clear that

is also a multiplicative set of $A$. Then the following holds.

Proposition 10.9.10. Let $\overline{S}$ be the image of $S$ in $S'^{-1}A$, then $(SS')^{-1}A$ is isomorphic to $\overline{S}^{-1}(S'^{-1}A)$.

**Proof.**
The map sending $x\in A$ to $x/1\in (SS')^{-1}A$ induces a map sending $x/s\in S'^{-1}A$ to $x/s \in (SS')^{-1}A$, by universal property. The image of the elements in $\overline{S}$ are invertible in $(SS')^{-1}A$. By the universal property we get a map $f : \overline{S}^{-1}(S'^{-1}A) \to (SS')^{-1}A$ which maps $(x/s')/(s/1)$ to $x/ss'$.

On the other hand, the map from $A$ to $\overline{S}^{-1}(S'^{-1}A)$ sending $x\in A$ to $(x/1)/(1/1)$ also induces a map $g : (SS')^{-1}A \to \overline{S}^{-1}(S'^{-1}A)$ which sends $x/ss'$ to $(x/s')/(s/1)$, by the universal property again. It is immediately checked that $f$ and $g$ are inverse to each other, hence they are both isomorphisms. $\square$

For the module $M$ we have

Proposition 10.9.11. View $S'^{-1}M$ as an $A$-module, then $S^{-1}(S'^{-1}M)$ is isomorphic to $(SS')^{-1}M$.

**Proof.**
Note that given a $A$-module M, we have not proved any universal property for $S^{-1}M$. Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.

We define the maps as follows

We have to check that these homomorphisms are well-defined, that is, independent the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other. $\square$

If $u : M \to N$ is an $A$ homomorphism, then the localization indeed induces a well-defined $S^{-1}A$ homomorphism $S^{-1}u : S^{-1}M \to S^{-1}N$ which sends $x/s$ to $u(x)/s$. It is immediately checked that this construction is functorial, so that $S^{-1}$ is actually a functor from the category of $A$-modules to the category of $S^{-1}A$-modules. Moreover this functor is exact, as we show in the following proposition.

Proposition 10.9.12. Let $L\xrightarrow {u} M\xrightarrow {v} N$ be an exact sequence of $R$-modules. Then $S^{-1}L \to S^{-1}M \to S^{-1}N$ is also exact.

**Proof.**
First it is clear that $S^{-1}L \to S^{-1}M \to S^{-1}N$ is a complex since localization is a functor. Next suppose that $x/s$ maps to zero in $S^{-1}N$ for some $x/s \in S^{-1}M$. Then by definition there is a $t\in S$ such that $v(xt) = v(x)t = 0$ in $M$, which means $xt \in \mathop{\mathrm{Ker}}(v)$. By the exactness of $L \to M \to N$ we have $xt = u(y)$ for some $y$ in $L$. Then $x/s$ is the image of $y/st$. This proves the exactness.
$\square$

Lemma 10.9.13. Localization respects quotients, i.e. if $N$ is a submodule of $M$, then $S^{-1}(M/N)\simeq (S^{-1}M)/(S^{-1}N)$.

**Proof.**
From the exact sequence

we have

The corollary then follows. $\square$

If, in the preceding Corollary, we take $N = I$ and $M = A$ for an ideal $I$ of $A$, we see that $S^{-1}A/S^{-1}I \simeq S^{-1}(A/I)$ as $A$-modules. The next proposition shows that they are isomorphic as rings.

Proposition 10.9.14. Let $I$ be an ideal of $A$, $S$ a multiplicative set of $A$. Then $S^{-1}I$ is an ideal of $S^{-1}A$ and $\overline{S}^{-1}(A/I)$ is isomorphic to $S^{-1}A/S^{-1}I$, where $\overline{S}$ is the image of $S$ in $A/I$.

**Proof.**
The fact that $S^{-1}I$ is an ideal is clear since $I$ itself is an ideal. Define

where $\overline{x}$ and $\overline{s}$ are the images of $x$ and $s$ in $A/I$. We shall keep similar notations in this proof. This map is well-defined by the universal property of $S^{-1}A$, and $S^{-1}I$ is contained in the kernel of it, therefore it induces a map

On the other hand, the map $A \to S^{-1}A/S^{-1}I$ sending $x$ to $\overline{x/1}$ induces a map $A/I \to S^{-1}A/S^{-1}I$ sending $\overline{x}$ to $\overline{x/1}$. The image of $\overline{S}$ is invertible in $S^{-1}A/S^{-1}I$, thus induces a map

by the universal property. It is then clear that $\overline{f}$ and $g$ are inverse to each other, hence are both isomorphisms. $\square$

We now consider how submodules behave in localization.

Lemma 10.9.15. Any submodule $N'$ of $S^{-1}M$ is of the form $S^{-1}N$ for some $N\subset M$. Indeed one can take $N$ to be the inverse image of $N'$ in $M$.

**Proof.**
Let $N$ be the inverse image of $N'$ in $M$. Then one can see that $S^{-1}N\supset N'$. To show they are equal, take $x/s$ in $S^{-1}N$, where $s\in S$ and $x\in N$. This yields that $x/1\in N'$. Since $N'$ is an $S^{-1}R$-submodule we have $x/s = x/1\cdot 1/s\in N'$. This finishes the proof.
$\square$

Taking $M = A$ and $N = I$ an ideal of $A$, we have the following corollary, which can be viewed as a converse of the first part of Proposition 10.9.14.

Lemma 10.9.16. Each ideal $I'$ of $S^{-1}A$ takes the form $S^{-1}I$, where one can take $I$ to be the inverse image of $I'$ in $A$.

**Proof.**
Immediate from Lemma 10.9.15.
$\square$

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