3. Fractions in Rings and Modules Atiyah and Macdonald Define Multiplicative Sets to Be Subsets S ⊆ a of a Ring So That

MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 19

3. Fractions in rings and modules Atiyah and MacDonald define multiplicative sets to be subsets S A of a ring so that 1 S and S is closed under multiplication. They don’t⊆ assume 0 / S (as ∈in my Def 1.19). So, we will now allow 0 to be in S. ∈ 1 1 (But, if 0 S then S− A = 0 and S− M = 0!) ∈ 1 If S A is a multiplicative set then define S− A to be the ring whose elemen⊆ ts are equivalence of pairs (a, s) where a A, s S with equivalence relation: ∈ ∈ (a, s) (b, t) atu = bsu for some u S. ∼ ⇔ ∈ a Equivalence classes are written a/s or s . Letting (b, t) = (0, 1) we get: a 1 = 0 in S− A au = 0 for some u S s ⇔ ∈ 1 Putting a = 1 we see that S− A = 0 iff 0 S. Addition and multipli- cation are given in the usual way by ∈ a b at + bs a b ab + = , = s t st s t st 1 There is a ring homomorphism f : A S− A given by f(a) = a/1. → f(a) = 0 as = 0 for some s S. ⇔ ∈ Exercise 3.1. (1) Let A = Z, S = Z (3), T = 1 + (3). Show that 1 1 − S− Z = T − Z. (2) More generally, show that if m is a maximal ideal, S = A m 1 1 − and T = 1 + m then S− A = T − A. One standard example is when A is an integral domain and S = 1 A 0 . Then S− A is a field called the field of fractions of A. − { } 1 1 3.1. characterization of S− A. The ring S− A is characterized by a universal property which can be stated abstractly and more concretely.

1 Proposition 3.2 (A-M 3.1). If S A is a multiplicative set then S− A has the following universal property:⊂ Given any ring homomorphism g : A B → so that g(S) U(B) then there exists a unique ring homomorphism 1 ⊆ h : S− A B so that h f = g. I.e., the following diagram commutes. → ◦ g A / B EE y< EE y EE y f EE y !h E" y ∃ 1 S− A 20 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA

This categorical language is not as useful as the concrete interpreta- tion: Corollary 3.3 (A-M 3.2). Suppose that g : A B is a ring homo- 1 → morphism. Then B ∼= S− A iff it has the following properties. (1) s S g(s) is a unit in B. (2) g(∈a) =⇒0 as = 0 for some s S. ⇒ ∈ 1 (3) Every element of B has the form g(s)− g(a) for some s S, a A. ∈ ∈

1 Proof. (1) says that g factors through S− A. (3) says the induced map 1 h : S− A B is onto. (2) says that h is 1-1. → ! 3.2. localization. The most important case of this construction is when S = A p is the complement of a prime ideal p. But first, we need to do −the following easy exercise: Exercise 3.4. For any multiplicative set S and ideal a in A show that

1 a S− a = a a, s S s | ∈ ∈ 1 ! e " is an ideal in S− A equal to the extension a . Theorem 3.5. If S = A p is the complement of a prime ideal p then 1 − 1 S− A is a local ring with unique maximal ideal S− p.

1 This local ring is denoted Ap = S− A and called the localization of A at p.

3.3. localization of modules. If M is an A-modules and S A is a 1 ⊆ multiplicative set then let S− M be the set of all equivalence classes of pairs (x, s) M S where ∈ × (x, s) (y, t) xtu = ysu for some u S. ∼ ⇔ ∈ x The equivalence class of (x, s) is written x/s or s . Taking (y, t) = (0, 1) we get x 1 = 0 in S− M xu = 0 for some u S s ⇔ ∈ 1 1 1 Then S− M is an S− A-module with action of S− A given by a x ax = s t st 1 Exercise 3.6. If M is finitely generated then show that S− M = 0 iff there exists an s S so that sM = 0. ∈ MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 21

In the special case S = A p where p is prime, we write − 1 S− M = Mp If f : M N is a homomorphism of A-modules, we get an induced → 1 1 1 1 homomorphism of S− A-modules S− f : S− M S− N given by 1 → S− f(x/s) = f(x)/s. We get the following important theorem with an easy proof. 1 Theorem 3.7 (A-M 3.3). The functor S− is exact, i.e. given any exact sequence M # M M ##, we get an exact sequence →f →g 1 1 1 S− f 1 S− g 1 S− M # S− M S− M ## −−−→ −−−→ 1 The module S− M satisfies a universal property similar to the one 1 for S− A. That universal property can be expressed in the following terms. 1 Proposition 3.8. Given an A-module M and an S− A-module N we have a natural isomorphism 1 Hom (M, N) = Hom 1 (S− M, N) A A ∼ S− A where AN is N considered as an A-module by restriction of scalars. To see that this is a universal property, look at the following diagram: g M / N GG w; GG w GG w f GG w !h G# w ∃ 1 S− M 1 Given any g : M N we get a unique h : S− M N. → A → 1 1 Corollary 3.9 (A-M 3.5). S− M = S− A M. ∼ ⊗A Proof. The adjunction formula 1 Hom (M, N) = Hom 1 (S− A M, N) A A ∼ S− A ⊗A 1 tells us that S− A M satisfies the same universal formula! ⊗A ! 1 Corollary 3.10 (A-M 3.6). S− A is a flat A-module. Exercise 3.11. If S = 1 + a then show that 1 S− (A/a) = A/a