# Localization

Recall that given an integral domain, there is a canonical way to construct the “smallest field containing it”, its field of fractions. Here, we will generalize this construction to arbitrary rings.

We let A be a fixed ring throughout.

Definition.

A subset $S\subseteq A$ is said to be multiplicative if

• $1 \in S$;
• if $s, s'\in S$, then $ss'\in S$.

Our objective is to construct a ring by inverting all elements of S. We will, in fact, define this via its universal property.

Definition.

For a multiplicative $S\subseteq A$, its localization comprises of a pair

$(S^{-1}A, \phi : A \to S^{-1}A)$

where $S^{-1}A$ is a ring and $\phi$ is a ring homomorphism such that $\phi(s)$ is a unit for any $s\in S$, with the following property.

• For any pair $(B, \psi : A \to B)$, where $B$ is a ring and $\psi$ is a homomorphism such that $\psi(s)$ is a unit for any $s\in S$, there is a unique ring homomorphism $f : S^{-1}A \to B$ such that $f\circ \phi = \psi$.

In summary, any ring homomorphsim $A\to B$ which maps S into the units of B must factor through the localization.

By definition, we obtain a bijection for any ring B

\begin{aligned} \mathrm{hom}_{\mathbf{Ring}}(S^{-1}A, B) &\leftrightarrow \{ \psi \in \mathrm{hom}_{\mathbf{Ring}}(A, B) : \psi(S) \subseteq U(B)\}, \\ f &\mapsto f\circ \phi.\end{aligned}

Intuitively, we may imagine $S^{-1}A$ as the “smallest” ring extension of A in which every element of S becomes a unit. [Technically, this is wrong since $A \to S^{-1}A$ is not injective in general, but this is just for intuition.]

Exercise A

Prove that if $(A_1, \phi_1 : A\to A_1)$ and $(A_2, \phi_2 : A\to A_2)$ are both localizations, there is a unique ring isomorphism $f:A_1 \to A_2$ such that $f\circ \phi_1 = \phi_2$.

Prove that if A is a domain and $S = A -\{0\}$, then $S^{-1}A$ is the field of fractions of A.

# Concrete Definition

We will prove that $S^{-1}A$ exists by construction.

Proposition 1.

Take $A\times S$ with the equivalence relation:

• $(a, s) \sim (a', s')$ if there exists $t\in S$ such that $t(sa' - s'a) = 0$.

and $S^{-1}A$ be the set of its equivalence classes; write $\frac a s$ for the equivalence class containing $(a,s)$. Then $S^{-1}A$ is a ring under the following operations.

$\frac a s + \frac {a'} {s'} = \frac{s'a + sa'} {ss'}, \quad \frac a s \times \frac{a'}{s'} = \frac{aa'} {ss'}$.

Proof

The proof is tedious but straight-forward. For example, for transitivity, if $(a,s) \sim (a',s')$ and $(a',s') \sim (a'', s'')$, then there exist $t, t'\in S$ such that

\begin{aligned} &tsa' = ts'a,\ t's'a'' = t's''a' \\ \implies &tt's'(sa'') = ts(t's'a'') = ts(t's''a') = t's''(tsa') = t's''(ts'a) = tt's'(s''a).\end{aligned}

To prove that addition and product are well-defined, suppose $\frac {a'}{s'} = \frac{a''}{s''}$ so that $ts'a'' = ts''a'$ for some $t\in S$; we wish to show $(s'a + sa', ss') \sim (s''a + sa'', ss'')$ and $(aa', ss') \sim (aa'', ss'')$. Now

$s^2(ts'a'') = s^2 (ts''a') \implies\begin{cases} &t(ss')(s''a + sa'') = t(ss'')(s'a + sa'), \\&ts(ss')(aa'') = ts(ss'')(aa'). \end{cases}$

and we are done. It remains to show that these operations turn $S^{-1}A$ into a ring, which we will leave to the reader. ♦

Proposition 2.

The ring $S^{-1}A$ together with the homomorphism

$\phi :A\to S^{-1}A, \quad a\mapsto \frac a 1$

gives the localization.

Proof

Note that $\phi(s) = \frac s 1$ is a unit for each $s\in S$ since $\frac s 1 \times \frac 1 s = \frac s s = \frac 1 1$. Now for any ring B and homomorphism $\psi : A\to B$ such that $\psi(s)$ is a unit for each $s\in S$, we set

$f : S^{-1}A \longrightarrow B, \quad \frac a s \mapsto \psi(a)\psi(s)^{-1} \in B.$

This map is well-defined since if $\frac a s = \frac {a'}{s'}$ then $tsa' = ts'a$ for some $t\in S$ so $\psi(t)\psi(s)\psi(a') = \psi(t)\psi(s')\psi(a)\in B$ and since $\psi(s), \psi(s'), \psi(t)$ are all units we have $\psi(a)\psi(s)^{-1} = \psi(a')\psi(s')^{-1}$.

It is also straight-forward to show that f is a ring homomorphism and $f\circ \phi = \psi$.

To show that f is unique, we have $f(\frac a 1) = f(\phi(a)) = \psi(a)$ for all $a\in A$. Hence for all $a\in A$, $s\in S$ we have

$f(\frac a s) \psi(s) = f(\frac a s) f(\frac s 1) = f(\frac {as}{s}) = f(\frac a 1) = \psi(a).$

Note

From this concrete description of $S^{-1}A$, we see that

• if S has no zero-divisors of A, then $\phi : A \to S^{-1}A$ is injective;
• in particular if A is an integral domain and $0\not\in S$, then $A\subseteq S^{-1}A \subseteq \mathrm{Frac} A$, the field of fractions of A;
• as a consequence, if A is an integral domain so is $S^{-1}A$.

# Some Properties of Localizations

Lemma 1.

If $S\subseteq A$ and $T\subseteq B$ are multiplicative subsets and $f:A\to B$ is a ring homomorphism such that $f(S) \subseteq T$, we obtain a ring homomorphism

$A_S \longrightarrow B_T, \quad \frac a s \mapsto \frac{f(a)}{f(s)}$.

Proof

Easy exercise: can be done directly or by universal property. ♦

Some common instances of localizations are as follows.

### Example 1: Inverting One Element

The easiest way to get a multiplicative set is to pick $f\in A$ and set $S = \{1, f, f^2, \ldots\}$. We write $A_f$ for the resulting $A_S$.

For example if $f = 2 \in \mathbb Z = A$, then

$A_f = \{ \frac a {2^k} \in \mathbb Q : a \in \mathbb Z,\ k \in \mathbb Z_{\ge 0}\} = \mathbb Z[\frac 1 2]$,

the $\mathbb Z$-subalgebra of $\mathbb Q$ generated by $\frac 1 2$.

Exercise B

Prove that $A_f \cong A[X]/(f\cdot X - 1)$, where $A[X]$ is the polynomial ring.

### Example 2: Localization at a Prime

Let $\mathfrak p \subset A$ be a prime ideal. Then $S := A-\mathfrak p$ is multiplicative by the definition of prime ideals. We write $A_{\mathfrak p} := S^{-1} A$, the localization of A at the prime ideal $\mathfrak p$.

For example, suppose $\mathfrak p = (2) \subset \mathbb Z = A$. Then

$A_{\mathfrak p} = \{ \frac a b \in \mathbb Q : a, b\in \mathbb Z, \ b\text{ odd }\}.$

Usually $\mathbb Z_{2}$ denotes the ring of 2-adic integers (which we will cover much later). Hence we will write $\mathbb Z[\frac 1 2]$ for the first example and $\mathbb Z_{(2)}$ for the second (note the brackets).

# Ideals of Localized Ring

### General Case

Given a general ring homomorphism $f:A\to B$, we can “push” an ideal $\mathfrak a\subseteq A$ through f to obtain an ideal $f(\mathfrak a) B$ of $B$. Recall that the product notation means we take the set of all finite sums

$f(a_1)b_1 + f(a_2) b_2 + \ldots + f(a_k)b_k, \quad a_1, \ldots, a_k \in \mathfrak a, b_1, \ldots, b_k \in B.$

One can also think of it as the ideal of B generated by $f(\mathfrak a)$.

Conversely, an ideal $\mathfrak b$ of B gives us an ideal $f^{-1}(\mathfrak b)$ of A. For convenience we write $\mathfrak a B$ for $f(\mathfrak a)B$ and $\mathfrak b \cap A$ for $f^{-1}(\mathfrak b)$ if f is implicit.

Note that this definition of $\mathfrak a B$ is consistent with our earlier definition of $\mathfrak a M$ where M is an A-module.

### Main Case of Interest

Now we apply this to our map $\phi : A \to S^{-1}A$, $a\mapsto \frac a 1$.

Lemma 2.

$\mathfrak a (S^{-1}A)$ is exactly the set of elements of $S^{-1}A$ which can be written as $\frac a s$ for some $a \in \mathfrak a, s\in S$.

Note

It is possible for $\frac a s = \frac {a'}{s'}$ to hold in $S^{-1} A$ with $a\in \mathfrak a$ but $a'\not\in \mathfrak a$.

Proof

Let $\mathfrak b$ be the set of elements as described. It suffices to show:

• $\phi(\mathfrak a)\subseteq \mathfrak b \subseteq \mathfrak a(S^{-1}A)$,
• $\mathfrak b$ is an ideal of $S^{-1}A$.

For the first claim, let $a\in \mathfrak a, s\in S$. Then $\frac a 1 \in \mathfrak b$ so $\phi(\mathfrak a)\subseteq \mathfrak b$. Also $\frac a s = \frac a 1 \cdot \frac 1 s = \phi(a) \cdot \frac 1 s$. Hence $\mathfrak b \subseteq \mathfrak a(S^{-1}A)$.

For the second, let $x, y \in \mathfrak a$, $s, s'\in S$ and $\frac a t \in S^{-1} A$. Then

$\frac x s - \frac a t \cdot\frac y {s'} = \frac{s'tx - say}{sts'} \in \mathfrak b$

since $s'tx - say \in \mathfrak a$ and $sts' \in S$. So $\mathfrak b$ is an ideal of $S^{-1}A$. ♦

The following gives a relationship between ideals of $A$ and those of $S^{-1}A$.

Proposition 3.

For an ideal $\mathfrak b\subseteq S^{-1}A$, let $\mathfrak a = \mathfrak b \cap A$, an ideal of A. Then

$\mathfrak a (S^{-1}A) = \mathfrak b$.

Proof

(⊆) Suppose $a \in \mathfrak a = \mathfrak b \cap A$, i.e. $\frac a 1 \in \mathfrak b$. Since $\mathfrak b$ is an ideal of $S^{-1}A$ we have $\mathfrak a (S^{-1}A) \subseteq \mathfrak b$.

(⊇) Pick $\frac a s \in \mathfrak b$ where $a\in A, s\in S$. Then $\frac a 1 \in \mathfrak b$ so $a \in \mathfrak a$. We see that $\frac a s = \frac a 1 \cdot \frac 1 s \in \mathfrak a (S^{-1}A)$. ♦

Note

Hence we obtain the following correspondence.

One can imagine this as: the ideals of $S^{-1}A$ form a subset of those of A. Clearly this subset is proper in general (e.g. take $A = \mathbb Z$ and $S^{-1}A = \mathbb Q$).

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