12 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES4. LocalizationToday I talked about localization. I explained the basic definitions,gave examples and I also tried to explain what the word “local” refersto. All rings are commutative here.4.1. basic definitions.Definition 4.1. A subset S ⊆ R is called a multiplicative set if S isclosed under multiplication, 0 /∈ S and 1 ∈ S.Some people leave out the assumption that 1 ∈ S. It is not necessarysince you can always add 1 to the set (take the union with {1}).Definition 4.2. If S is a multiplicative set in a ring R then S−1R isdefined to be the set of equivalence classes of symbolsxawhere x ∈R, a ∈ S andxa∼ybif (∃t ∈ S) xbt = aytThis is a clumsy equivalence relation. So, I reformulated it as follows.This equivalence relation is the transitive relation generated by thesymmetric relation given byxa≈txta(∀t ∈ S)The reason is that:xa≈xbtabt=aytabt≈ybProposition 4.3. S−1R is a ring with addition and multiplicationgiven byxa+yb=xb + ayabxa·yb=xyaband j(x) =x1gives a ring homomorphism j : R → S−1R.Proof. We need to show that addition and multiplication are well-defined. The rest is straightforward. To show that addition is well-defined, we can use the stronger relation ≈:sxsa+tytb=sxtb + satysatb=(st)(xb + ay)(st)ab≈xb + ayab.Similarly,sxsa·tytb=sxtysatb=(st)(xy)(st)ab≈xyab.So, multiplication is also well-defined. !MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 134.2. examples. I gave a bunch of examples but I delayed the expla-nations for the end of the lecture.Example 1. R = Z with multiplicative set S = {n ∈ Z | n (= 0}. ThenS−1Z = Q.This is a special case of the more general example:Example 2. Suppose that R is any domain and S = R\{0}. ThenQ(R) = S−1Ris the quotient field or field of fractions of R. This is a field since thenonzero elements of R are invertible:!ab"−1=baThis example required a discussion about which elements are zero andwhich are 1.Proposition 4.4. An elementxs∈ S−1R is equal to zero if and only ifthere is an element t ∈ S so that xt = 0.Proof. If such a t exists thenxs≈xtst=0st∼01= j(0)since 0 · 1 = 0 = st · 0. Conversely, ifxs∼01= j(0)then there is a t ∈ S so that xt = x1t = st0 = 0. !I forgot to do the same discussion with 1.Proposition 4.5. An elementxs∈ S−1R is equal to 1 =11= j(1) ifand only if there exists t ∈ S so that tx = ts.Proof. This is just the definition of the equivalence relation. !Example 3. Let R = C0(R). This is the ring of continuous functionsf : R → R with the usual pointwise addition and multiplication. LetS be the set of all functions f which are nonzero at 1:S = {f | f(1) (= 0}Then S is clearly a multiplicative set. I claim that S−1R is a “local ring”which means that the nonunits form an ideal. A fraction f (x)/g(x) isan element of S−1R if g(1) (= 0. It is invertible if f(1) (= 0. So the14 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESnonunits are given by the equation f (1) = 0 which clearly defines anideal. In fact it is the kernel of the evaluation mapev1: S−1R → R.This is a special case of the following more general example.Example 4. If P ⊆ R is a prime ideal then its complement S = R\Pis a multiplicative set. (In fact these conditions are equivalent.) So, wecan form the ring S−1R which is called the localization of R at P andwritten RP.Another special case of this is the following.Example 5. Suppose that p ∈ Z is irreducible (i.e. a prime number).Then (p) is a prime ideal and we can form the localization:Z(p)= S−1Z =#ab∈ Q | p( | b$This is a subring of Q.What is the relationship between Z(p)and Zp?4.3. universal property. In your homework you proved that if anyinteger n which is not divisible by p is uniquely invertible in Zp. There-fore, by the following universal property of the localization, there is aring homomorphismZ(p)→ Zp.Is it onto? Is it a monomorphism?Theorem 4.6. If S ⊂ R is a multiplicative set then S−1R has thefollowing universal property: Given any ring homomorphismφ : R → R"so that φ(S) ⊆ U(R") then there exists a unique ring homomorphismφ : S−1R → R"so that φ◦j = φ. I.e., the following diagram commutes.Rj!!!!!!!!!!!φ""R"S−1R∃!φ##""""Proof. (Existence) The ring homomorphism φ is given by φ(x/s) =φ(x)φ(s)−1. You actually need to show that this is a homomorphism.This follows from the fact that the definition of addition and multipli-cation in S−1R used the rules which hold for arbitrary elements of theform xs−1.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 15(Uniqueness) Suppose that ψ : S−1R → R"is a homomorphismwhich extends j. Then the equationxs=x1·1sgives the equationψ!xs"= ψ!x1"ψ%1s&= φ(x)φ(s)−1where ψ(1/s) = ψ(s)−1= φ(s)−1since ψ induces a group homomor-phismU(ψ) : U(R) → U(R").!4.4. local rings. Localization is used to produce local rings.Definition 4.7. A local ring is a ring R with a unique maximal idealm.Proposition 4.8. A ring R is local if and only if the complementR\U(R) of the set of units is an ideal.The proof used the followings two obvious properties of U(R).Lemma 4.9. An ideal in R cannot contain any units.This implies that any ideal is contained in R\U(R). So, if this is anideal, it must be maximal.Lemma 4.10. Any element a ∈ R which is not a unit generates anideal (a).So, if there is only one maximal ideal, it contains all such a. So,m = R\U(R).The term “localization” is justified by the following theorem whichI forgot to prove. So, you can do it for homework.Theorem 4.11. If P is a prime ideal in R then RPis a local ring.What I did explain is what the “local” means, at least topologically.4.5. germs of functions.Definition 4.12. Suppose that X, Y are topological spaces and f :X → Y is a mapping. Let x0∈ X. Then the germ of f at x0is definedto be the equivalence class of f under the equivalence relation f ∼ g ifthere exists an open neighborhood U of x0in X so thatf|U= g|U16 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESThis is an equivalence relation since, if g|V= h|V, thenf|U∩V= g|U∩V= h|U∩Vand U ∩ V is an open neighborhood of x0. Note that f need only bedefined in a neighborhood of x0.When the target Y is a ring, the set of map germs forms a ring. If weput some restriction on the functions f (e.g., continuous, differentiable,polynomial), we get a subring. For example, we have a ring of germsat 1 of continuous functions R → R. The notation is with a comma:gr1(f) : R,