Hungerford: Algebra III.4. Rings of Quotients and Localization 1. Determine the complete ring of quotients of the ring Z n for each n 2. Proof: Denote Z n = {0, 1,, n 1}, and the set of all non zero divisors of Z n by S(Z n ). (1.1) Claim: S(Z n ) = {m : (m, n) = 1}. Suppose that 0 < m < n and (m, n) = d. If m S(Z n ). Then there exists an x with 0 < x < n such that mx = 0. Thus n (mx). If (m, n) = 1, then n x, contrary to the assumption that 0 < x < n. Thus d > 1, and so S(Z n ) {m : (m, n) = 1}. Conversely, assume that d > 1. Then for some positive integers m and n with 0 < m < n and 0 < n < n, we have m = m d and n = n d. Therefore mn = mn = nm = 0 Z n, and so m S(Z n ). Thus S(Z n ) {m : (m, n) = 1}. This proves the claim. By the claim, we can write S 1 Z n = {m/s : (n, s) = 1}. Example: For n = 4, S 1 Z 4 = {1/3, 2/3} Z 4. 2. Let R be a multiplication subset of a commutative ring with identity and let T be a multiplicative subset of the ring S 1 R. Let S = {r R : r/s T for some s S}. Then S is a multiplicative subset of R and there is a ring isomorphism S 1 R = T 1 (S 1 R). Proof: We may assume that 0 S. For any r, r S, there exist s, s S such that r/s, r /s T. Since T is multiplicative, (rr )/ss ) T ; and since S is multiplicative, ss S. Thus rr S, and so S is also multiplicative. We now define a map f : S 1 R T 1 (S 1 R). For any r/w S 1 R with r R and w S, there exists s S such that w/s T. Thus we define f(r/w) = (r/s)/(w/s) T 1 (S 1 R). (2.1) f is well-defined. Suppose that r/w = r /w in S 1 R. As w, w S, there exist s, s S such that w/s, w /s T. Since r/w = r /w in S 1 R, we have, for some r S, r (rw r w) = 0 in R, and for some s S, 1
r /s T. It follows that in R, we also have r (rw ss r wss ) = 0. Hence in S 1 R, we have (r /s )((r/s)(w /s ) (rw )/(ss )) = 0, and so (r/s)(w /s ) = (rw )/(ss ) = (r w)/(ss ) = (r /s )(w/s). This, in turn, implies that in T 1 (S 1 R), f(r/w) = (r/s)/(w/s) = (r /s )/(w /s ) = f(r /w ). (2.2) f : S 1 R T 1 (S 1 R) is a ring homomorphism. Let r 1 /w 1, r 2 /w 2 S 1 R. Then s 1, s 2 S such that w 1 /s 1, w 2 /s 2 T, which implies that (w 1 w 2 )/(s 1 s 2 ) T. f(r 1 /w 1 + r 2 /w 2 ) = f((r 1 w 2 + r 2 w 1 )/(w 1 w 2 )) (addition in S 1 R) = ((r 1 w 2 + r 2 w 2 )/(s 1 s 2 ))/((w 1 w 2 )/(s 1 s 2 )) (definition of f) = ((r 1 /s 1 )(w 2 /s 2 ) + (r 2 /s 2 )(w 1 /s 1 ))/((w 1 /s 1 )(w 2 /s 2 )) (addition and multiplication in T 1 (S 1 R) = (r 1 /s 1 )/(w 1 /s 1 ) + (r 2 /s 2 )/(w 2 /s 2 ) (addition in T 1 (S 1 R) = f(r 1 /w 1 ) + f(r 2 /w 2 ) (definition of f) Let 1 denote the identity of R. Since S is multiplicative, 1/1 = s/s in S 1 R for any s S. For any w/s T, (w/s)((1/1)(w/s) (w/s)(1/1)) = 0/s and so (1/1)/(1/1) = (w/s)/(w/s) in T 1 (S 1 R). (2.3) For any w S and for any s S, sw S. In fact, if w S, then s S such that w/s T. For any s S, since 1/1 = s/s S 1 R, (sw)/(ss ) = (1/1)(w/s) T, and so as ss S, sw S. (2.4) Let r R. If for some w S, rw = 0, then for any w S, r/w = (rw w)/(ww ) = 0/(ww ) = 0/w. In particular, in S 1 R, {0/w} = {r/w : r R, w S, and for one w S, rw = 0 in R}. (2.5) f is a bijection. For any (r/s)(w/s) T 1 (S 1 R), we have r R, s S and w/s T. Thus w S and so f(r/w) = (r/s)(w/s). Thus f is surjective. By (2.3) and (2.4) Kerf = {r/w S 1 R : f(r/w) = 0 T 1 (S 1 R)} = {r/w S 1 R : for some w R, s S with w/s T, (r/s)/(w/s) = (0/s)/(w/s) T 1 (S 1 R)} 2
= {r/w S 1 R : for some w R, s S with w/s T, (rw)/s = 0/s S 1 R} = {r/w S 1 R : for some w R, s S with w/s T, rws = 0 R} = {0/w S 1 R}. 3. (a) The set E of positive even integers is a multiplicative subset of Z such that E 1 Z is the field of the rational numbers. (b) State and prove condition(s) on a multiplicative subset S of Z which insure that S 1 Z is the field of rational numbers. Proof: (a) Since the product of two positive even numbers is still a positive even number, E is a multiplicative subset of Z. Let Q = (Z {0}) 1 Z denote the field of the rational numbers. Then p/q Q, p, q Z and q 0. Without loss of generality, we may assume that q > 0, and so p/q = (2p)/2q) E 1 bfz. This proves that Q E 1 bfz, and so equality must hold. (b) We claim that for a multiplicative subset S Z {0}, S 1 Z = Q if and only if for any prime p, S pz. First assume that S 1 Z = Q. For any prime p, as 1/p Q = S 1 Z, a Z and b S such that a/b = 1/p, or in Z, ap = b S. HenceS pz =. Conversely, we assume that for any prime p, S pz. Let a/b Q. Then as Z is a PID (and so a UFD, and every irreducible element is a prime), for some primes p 1, p 2,, p m, and positive integers n 1, n 2,, n m, b = p n1 1 pn2 2 pnm m. Since p i Z S, for each i, s i Z such that s i p i S. Let s = s 1 s 2 s m. Then sb = (s 1 p 1 )(s 2 p 2 ) (s m p m ) S, and so a/b = (sa)/sb) S 1 Z. This proves Q S 1 bfz, and completes the proof. 4. If S = {2, 4} and R = Z 6, then S 1 R = Z 3. Proof: For the simplicity of notation, we denote Z 6 = {0, 1, 2, 3, 4, 5}. Therefore, S R = 12. As 3 2 = 3 4 = 0 2 = 0 4, and so the equivalence class in S 1 R represented by 0 2 contains 4 elements; as 1 4 = 5 2 = 2 2 = 4 4, and so the equivalence class in S 1 R represented by 2 2 contains 4 elements; as 5 4 = 1 2 = 2 4 = 4 2, and so the equivalence class in S 1 R represented by 4 2 contains 4 elements. Thus S R has exactly 3 equivalence classes. Define φ : S 1 R Z 3 by φ( 0 2 ) 0, φ( 2 2 ) = 1 and φ( 4 2 ) = 2. Then one can verify that this is a ring epimorphism (verification is routine and so omitted here). Since S 1 R = Z 3 = 3, φ is also a monomorphism and so φ is an isomorphism. 5. Let R be an integral domain with quotient field F. It T is an integral domain such that R T F, then F is (isomorphic to) the quotient field of T. Proof: Let E be the field of quotients of T. Then there is an monomorphism f : T E. By Corollary III-4.6 (Let R be an integral domain considered as a subring of its quotient field F. If 3
E is a field and f : R E is a monomorphism of rings, then there is a unique monomorphism of fields f : F E such that the restriction of f to R is f. We here apply Corollary III-4.6 with T replacing R.), there is a field monomorphism f : E F extending f. View this isomorphism as an embedding, we may assume that T E F. Now apply Corollary III-4.6 to R with E replacing F in Corollary III-4.6, then we have R F E. 8. Let R be a commutative ring with identity, I an ideal of R and π : R R/I the canonical projection. (a) If S is a multiplicative subset of R, then πs = π(s) is a multiplicative subset of R/I. (b) The mapping θ : S 1 R (πs) 1 (R/I) given by r/s π(r)/π(s) is a well defined function. (c) θ is a ring epimorphism with kernel S 1 I and hence induces a ring isomorphism S 1 R/S 1 I = (πs) 1 (R/I). Proof: (a) s, s S, as S is a multiplicative subset, ss S. Thus π(s)π(s ) = π(ss ) π(s) = πs, and so πs is a multiplicative subset. (b) Suppose that r/s = r /s S 1 R. Then rs = r s. Since π : R R/I is a homomorphism, π(r)π(s ) = π(rs ) = π(r s) = π(r)π(s ). It follows that θ(r/s) = π(r)/π(s) = π(r )/π(s ) = θ(r /s ) in (πs) 1 (R/I). (c) For any r/s, r /s S 1 R, θ(r/s + r /s ) = θ((rs + r s)/ss ) = π(rs + r s)/π(ss ) = (π(r)π(s ) + π(r )π(s))/(π(s)π(s )) = π(r)/π(s) + π(r )/π(s ) = θ(r/s) + θ(r /s ). θ((r/s)(r /s )) = θ((rr)/(ss )) = π(rr )/π(ss ) = (π(r)π(r ))/)π(s)π(s )) = (π(r)/π(s))(π(r )/π(s )) = θ(r/s)θ(r /s ). For any a/b (πs) 1 (R/I) with a R/I and b πs, since π is an epimorphism, r R, s S such that a = π(r) and b = π(s). Therefore, θ(r/s) = a/b and so θ is also an epimorphism. Kerθ = {r/s S 1 R : θ(r/s) = 0 (πs) 1 (R/I)} = {r/s S 1 R : π(r)/π(s) = 0/π(s) (πs) 1 (R/I)} = {r/s S 1 R : π(r) = 0 R/I and π(s) π(s)} = {r/s S 1 R : r I and s S} = S 1 I. 4
9. Let S be a multiplicative subset of a commutative ring with identity. If I is an ideal in R, then S 1 (Rad I) = Rad (S 1 (I)). Proof: It is routine to show that S 1 ( Rad I) Rad (S 1 I). In fact, if a b S 1 ( Rad I), then a Rad (I) and b S. Thus for some integer n > 0, a n I. As S is multiplicative, b n S. It follows that ( a b )n = an b n S 1 I, and so S 1 ( Rad I) Rad (S 1 I). Conversely, let a b Rad(S 1 I). Then for some integer m > 0, am b m for some c I and d S, am b m = c d. Thus s S, such that s(a m d b m c) = 0, or equivalently, a m sd = b m sc. = ( a b )m S 1 I. Therefore, Since c I and since I is an ideal, a m sd = b m sc I. Since S is a multiplicative set, and since b, s, d S, we have bsd S and so (asd) m = a m sd(sd) m 1 I, (bsd) m S, and so ( ) m asd S 1 I. bsd Thus asd bsd Rad(S 1 I). But then for any s S, s (absd absd) = 0, and so This proves that Rad (S 1 (I)) S 1 (Rad(I)). a b = asd bsd Rad(S 1 I). 12. A commutative ring with identity is local if and only if for all r, s R, r + s = 1 R implies r or s is a unit. Proof: Let R be a commutative ring with identity 1 R. Suppose first that R is local, and so R has exactly one maximal ideal M. Suppose that r, s R with r + s = 1. If neither r nor s are units, then (r) and (s) are principal ideals. Let M 1 and M 2 be maximal ideals of R containing (r) and (s), respectively. Then since M is the only maximal ideal of R, we have M 1 = M 2 = M, and so r, s M. Since r + s = 1 R, we must have 1 R M, and so M = R, contrary to the assumption that M R. Conversely, assume that r, s R, that r +s = 1 R implies that r or s is a unit. By contradiction, we assume that R has at least two distinct maximal ideals M 1 and M 2. Since M 1 and M 2 are distinct maximal ideals, the ideal M 1 + M 2 must be R itself. Therefore, since 1 R R, r M 1 and s M 2 such that r + s = 1 R. Therefore, either r or s is a unit, and so either M 1 or M 2 equals R, contrary to the fact that a maximal ideal is not equal to R. 13. The ring R consisting of all rational numbers with denominators not divisible by some (fixed) prime p is a local ring. Proof: Consider the ring Z of the integers. For a fixed prime p, (p) is a prime ideal of Z and so S = Z (p) is a multiplicative subset. By Theorem III-4.11(ii) (Let P be a prime idea in a commutative ring R with identity, the ideal P P = S 1 P is the unique maximal ideal of the ring 5
R P = S 1 R, the localization of R at P ), Z (p), having S 1 (p) as its only maximal ideal, is a local ring. Fact: The structure of Z (p). As (p) = pz, S = Z (p) = {n Z {0} : p n}. Thus Z (p) = {n/m Q : p m}. In particular. Z (2) = {n/m Q : m is odd }. 6