Commutative Algebra 22

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.

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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.

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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 }\}.

warningUsually \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).

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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.

ideals_of_localization

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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