Math 254A. Localization

This handout gives a description of the theory of localization. It is slightly different from the description in the textbook.

Localization generalizes the construction of fraction fields of an entire ring. It is more general in that one can restrict the set of denominators that can occur, or allow localizing rings that are not entire.

Definition. A multiplicative subset of a ring A is a subset S ⊆ A which contains 1 and is closed under multiplication. (Equivalently, it is a submonoid of the multiplicative monoid of A.)

Throughout this handout, A is a ring and S is a multiplicative subset of A.

We start with the basic construction of localization.

Proposition 1. Let ∼ be the relation on S × A defined by (s_{1}, a_{1}) ∼ (s_{2}, a_{2}) if there is an s′ ∈ S such that s′(s_{1}a_{2} − s_{2}a_{1}) = 0 . Then: (a). ∼ is an equivalence relation, and (b). ∼ is the smallest equivalence relation on S × A that satisfies

(s, a) ∼ (s′s, s′a)  for all s, s′ ∈ S and all a ∈ A .  (1)

Proof. (a). Exercise.

(b). Let ≈ be the smallest equivalence relation on S × A that satisfies (1). Clearly ∼ satisfies (1), so ≈ ⊆ ∼. Conversely, suppose (s_{1}, a_{1}) ∼ (s_{2}, a_{2}) . Then there is an s′ ∈ S such that

s′s_{1}a_{2} = s′s_{2}a_{1} . Therefore (since s′s_{1}, s′s_{2} ∈ S)

(s_{1}, a_{1}) ≈ (s′s_{2}s_{1}, s′s_{2}a_{1}) = (s′s_{1}s_{2}, s′s_{1}a_{2}) ≈ (s_{2}, a_{2}) ,

so ∼ ⊆ ≈. □

Definition 2. The localization S^{−1}A (also denoted A[S^{−1}] ) is the set of ∼ -equivalence classes in S × A. The equivalence class of a pair (s, a) is denoted s^{−1}a or a/s.

Proposition 1b then provides a more convenient way of checking whether certain types of formulas involving (s, a) lead to well-defined formulas involving a/s.

Corollary 3. Let B and C be sets, and let f: S × A × B → C be a function. If f(s, a, b) = f(s′s, s′a, b) for all s, s′ ∈ S, a ∈ A, and b ∈ B, then there is a unique well-defined function f̃: S^{−1}A × B → C such that f̃(a/s, b) = f(s, a, b) for all s, a, and b.

Proposition 4. The usual formulas for addition and multiplication of fractions give S^{−1}A a well-defined structure of (commutative) ring, such that the map a ↦ a/1 is a ring homomorphism A → S^{−1}A. Moreover, this homomorphism satisfies the following universal property:

Every ring homomorphism ψ: A → B in which ψ(S) ⊆ B^{∗} factors uniquely through the canonical map A → S^{−1}A.

Definition 5. Let M be an A-module. Let ∼ be the relation on S × M defined by (s_{1}, m_{1}) ∼ (s_{2}, m_{2}) if there exists s′ ∈ S such that s′s_{1}m_{2} = s′s_{2}m_{1} . This is an equivalence relation (exercise), and we let S^{−1}M denote the set of ∼ -equivalence classes on S × M . Then S^{−1}M is a module over S^{−1}A.

In fact, we have S^{−1}M ≅ M ⊗_{A} S^{−1}A. Moreover, for fixed A and S the map M ↦ S^{−1}M is a functor from the category of A-modules to the category of S^{−1}Amodules. In fact, it is an exact functor. This means that if

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

is an exact sequence of A-modules, then the resulting sequence

0 ⟶ S^{−1}M′ ⟶ S^{−1}M ⟶ S^{−1}M′′ ⟶ 0

of S^{−1}A-modules is also exact.

Examples 6.

• If A is entire, K is its fraction field, and S = A \ {0}, then S^{−1}A = K and A → S^{−1}A is the injection A → K.

• If S = {1} then A → S^{−1}A is an isomorphism.

• S^{−1}A = 0 ⟺ 0 ∈ S.

• If A is entire and 0 ∉ S, then A → S^{−1}A is the injection of A into a subring of its fraction field.

Proposition 7. Let ϕ: A → S^{−1}A be the canonical map. Then:

(a). ker ϕ = {a ∈ A : ann(a) ∩ S ≠ ∅}.

(b). Let 𝔞 and 𝔟 be ideals of A. Then S^{−1}𝔞 and S^{−1}𝔟 (defined as S^{−1}A-

modules) are ideals of S^{−1}A, and S^{−1}(𝔞𝔟) = (S^{−1}𝔞)(S^{−1}𝔟) .

(c). Let 𝔞 be an ideal in A. Then

ϕ^{−1}(S^{−1}𝔞) = {a ∈ A : as ∈ 𝔞 for some s ∈ S } .

(d). Let 𝔞′ be an ideal of S^{−1}A. Then S^{−1}(ϕ^{−1}(𝔞′)) = 𝔞′.

(e). Let 𝔞 be an ideal of A. Then S^{−1}A/S^{−1}𝔞 ≅ (a/𝔞) ⊗_{A} S^{−1}A ≅ \overline{S}^{−1}(A/𝔞)

(where \overline{S} ⊆ S^{−1}A is the image ϕ(S) ).

Proof. Exercise.

Corollary 8. The map 𝔞 ↦ S^{−1}A (from the set of ideals of A to the set of ideals of S^{−1}A) is surjective, and the map 𝔞′ ↦ ϕ^{−1}(𝔞′) is injective (and is its right inverse).

Proof. Exercise.

Corollary 9. If A is noetherian, then so is S^{−1}A.

Proof. Exercise.

Proposition 10. Let B and C be entire rings with A ⊆ B ⊆ C. Assume that 0 ∉ S. Then: (a). The injections A → B → C naturally induce injections

S^{−1}A → S^{−1}B → S^{−1}C .

(b). If β ∈ B is integral over A, then β/1 is integral over S^{−1}A.

(c). If β ∈ B and β/1 is integral over S^{−1}A, then sβ is integral over A for

some s ∈ S. (d). If B is the integral closure of A in C, then S^{−1}B is the integral closure of

S^{−1}A in S^{−1}C. (e). If A is integrally closed, then so is S^{−1}A.

Proof. Exercise.

Definition 11. The set of prime ideals of a ring A is denoted Spec A. Let ϕ: A → B be a ring homomorphism. For each ideal 𝔮 ∈ Spec B, the ideal ϕ^{−1}𝔮 is a prime ideal of A; hence ϕ induces a map f: Spec B → Spec A. In particular, if ϕ: A → S^{−1}A is as defined earlier, then we get an injective map

f: Spec S^{−1}A → Spec A .

We often identify Spec S^{−1}A with its image via f in Spec A.

Proposition 12. Let 𝔭 ∈ Spec A. Then:

(a). S^{−1}𝔭 = (1) ⟺ 𝔭 ∩ S ≠ ∅ (this holds if and only if 1 = s/s for some

s ∈ S ∩ 𝔭).

(b). If 𝔭 ∩ S = ∅, then a/s ∈ S^{−1}𝔭 ⟺ a ∈ 𝔭.

(c). If 𝔭 ∩ S = ∅ then S^{−1}𝔭 is prime. In that case, Spec S^{−1}A corresponds to {𝔭 ∈ Spec A : 𝔭 ∩ S = ∅}.

Proof. Exercise.

Proposition 13. Assume that the ring A is Dedekind and the 0 ∉ S. Then S^{−1}A is also Dedekind.

Proof. That S^{−1}A is entire, noetherian, and integrally closed follow from Example 6, Corollary 9, and Proposition 10e, respectively. If S^{−1}A is not Dedekind, then it would follow that there is a nonzero prime ideal 𝔭 of S^{−1}A which is not maximal. If so, then there would be a maximal ideal 𝔮 of S^{−1}A containing 𝔭, so 𝔭 ⫋ 𝔮 would be nonzero prime ideals of S^{−1}A. But then ϕ^{−1}(𝔭) ⫋ ϕ^{−1}(𝔮) would be nonzero prime ideals of A, contradicting the fact that A is Dedekind. □

Corollary 14. Assume that A is Dedekind, and let 𝔞 be a nonzero ideal of A. Let 𝔞 = 𝔭_{1}^{n_{1}} ⋯ 𝔭_{r}^{n_{r}} be the prime factorization of 𝔞 (with 𝔭_{1}, …, 𝔭_{r} distinct). Then S^{−1}A has prime factorization ∏_{i:𝔭_{i}∩S=∅}(S^{−1}𝔭_{i})^{n_{i}} , where all of the ideals S^{−1}𝔭_{i} in the above product are distinct.

Proof. By Propositions 7b and 12,

S^{−1}𝔞 = S^{−1}(𝔭_{1}^{n_{1}} ⋯ 𝔭_{r}^{n_{r}}) = (S^{−1}𝔭_{1})^{n_{1}} ⋯ (S^{−1}𝔭_{r})^{n_{r}} = ∏_{i:𝔭_{i}∩S=∅}(S^{−1}𝔭_{i})^{n_{i}} .

If 𝔭_{i} ∩ S = 𝔭_{j} ∩ S = ∅ and i ≠ j, then S^{−1}𝔭_{i} ≠ S^{−1}𝔭_{j} by Corollary 8. □