36 Rings of fractions

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1 36 Rings of fractions Recall. If R is a PID then R is a UFD. In particular Z is a UFD if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R F where F is a field. 2) If p(x) R[x] then p(x) F[x] and since F[x] is a UFD thus p(x) has a unique factorization into irreducibles in F[x]. 3) Use the factorization in F[x] and the fact that R is a UFD to obtain a factorization of p(x) in R[x] Definition. If R is a ring then a subset S R is a multiplicative subset if 1) 1 S 2) if a, b S then ab S 36.2 Example. Multiplicative subsets of Z: 1) S = Z 137

2 2) S 0 = Z {0} 3) S p = {n Z p n} where p is a prime number. Note: S p = Z p Proposition. If R is a ring, and I is a prime ideal of R then S = R I is a multiplicative subset of R. Proof. Exercise Construction of a ring of fractions. Goal. For a ring R and a multiplicative subset S R construct a ring S 1 R such that every element of S becomes a unit in S 1 R. Consider a relation on the set R S: (a, s) (a, s ) if s 0 (as a s) = 0 for some s 0 S Check: is an equivalence relation. Elements of S 1 R = (equivalence classes of R S under the relation ). Notation. a/s := the equivalence class represented by (a, s) Note. 1) If 0 S then (a, s) (a, s ) for all (a, s), (a, s ) R S, and so S 1 R consists of only one element. 2) If R is an integral domain and 0 S then (a, s) (a, s ) iff as = a s. 138

3 Multiplication in S 1 R: (a/s)(a /s ) = (aa )/ss Addition in S 1 R: a/s + a /s = (as + a s)/ss Check: 1) The operations of addition and multiplication in S 1 R are well defined. 2) These operations define a commutative ring structure on S 1 R. 3) The map ϕ S : R S 1 R, ϕ S (r) = r/1 is a ring homomorphism. Note. If s S then ϕ S (s) = s/1 is a unit in S 1 R: (s/1)(1/s) = s/s = 1/1 = 1 S 1 R 36.5 Definition. If S is a multiplicative subset of a ring R then S 1 R is called the ring of fractions of R with respect to S Theorem. If S is a multiplicative subset of a ring R and f : R R is a ring homomorphism such for ever s S the element f(s) R is a unit, then there exists a unique homomorphism f : S 1 R R such that the following diagram commutes: ϕ S R S 1 R f f 139 R

4 Proof. Define f(r/s) = f(r)f(s) 1. Check that 1) f is a well defined ring homomorphism 2) f is the only homomorphism such that f = fϕ S Examples. 1) If S Z, S = Z then S 1 Z = {0}. 2) If S 0 Z, S 0 = Z {0} then S 1 0 Z = Q. 3) If S 0 Z, S 0 = Z p then Sp 1 Z is isomorphic to the subring of Q consisting of all fractions m such that p n. n 4) If R = Z/6Z, S = {1, 3} R then S 1 R = Z/2Z (check!) Proposition. If S is a multiplicative subset of a ring R then Ker(ϕ S ) = {a R sa = 0 for some s S} Proof. We have ϕ S (r) = 0/1 iff s(r 1 0 1) = 0 for some s S Corollary. If R is an integral domain and S R is a multiplicative subset such that 0 S then the homomorphism ϕ S : R S 1 R is 1-1. As a consequence in this case we can identify R with a subring of S 1 R. 140

5 36.10 Note. 1) If R is an integral domain and S = R {0} then the ring S 1 R is a field. In this case S 1 R is called the field of fractions of R. 2) If I is a prime ideal of a ring R then the set S = R I is a multiplicative subset of R. In this case the ring S 1 R is called the localization of R at I and it is denoted R I Definition. A ring R is a local ring if R has exactly one maximal ideal Examples. 1) If F is a field then it is a local ring with the maximal ideal I = {0}. 2) Check: if F is a field then F[[x]] is a local ring with the maximal ideal x. 3) Check: if F is a field then F[x] is not a local ring since for every a F the ideal x a is a maximal ideal in F[x]. 4) Z is not a local ring Proposition. If R is a local ring then the maximal ideal J R consists of all non-units of R. Proof. Since J R thus J does not contain any units. Conversely, if a R is a non-unit then 1 a so a R. Therefore by Theorem 29.1 a is contained in some maximal ideal of R. Since J is the only maximal ideal we obtain a J, and so a J. 141

6 36.14 Proposition. If R is a ring and I R is a prime ideal then the ring R I is a local ring, and the maximal ideal J R I is given by J := {a/s a I, s I} Proof. Exercise. 142

7 37 Factorization in rings of polynomials Goal: 37.1 Theorem. If R is a UFD then so is R[x] Lemma. If R is an integral domain then p(x) R[x] is a unit in R[x] iff deg p(x) = 0 and p(x) = a 0 where a 0 is a unit in R. Proof. Exercise Lemma. If R is an integral domain and p(x) = a 0 is a polynomial of degree 0 in R[x] then p(x) is irreducible in R[x] iff a 0 is irreducible in R. Proof. Exercise Definition. If R is a UFD and p(x) = a 0 + a 1 x +... a n x n R[x], then p(x) is a primitive polynomial if gcd(a 0,..., a n ) Lemma. If R is a UFD and p(x) R[x] is an irreducible polynomial such that deg p(x) > 0 then p(x) is a primitive polynomial. Proof. If p(x) = a 0 + a 1 x +... a n x n is not primitive then p(x) = gcd(a 0,... a n )q(x) for some q(x) R[x], deg q(x) > 0. Since both gcd(a 0,..., a n ) and q(x) are non-units in R[x] the polynomial p(x) is not irreducible. 143

8 37.6 Lemma (Gauss). If R is a UFD and p(x), q(x) R[x] are primitive polynomials then p(x)q(x) is also primitive. Proof. Assume that r(x) = p(x)q(x) is not primitive. Then we have r(x) = c r(x) where c R is an irreducible element, and r(x) R[x]. Take the canonical epimorphism π : R R/ c This defines a homomorphism of rings of polynomials given by Note: π : R[x] (R/ c )[x] π(a 0 + a 1 x a n x m ) := π(a 0 ) + π 1 (a)x π(a n )x n 1) Since c is irreducible element thus c is a prime ideal and so R/ c is a domain. As a consequence (R/ c )[x] is a domain. 2) We have π(p(x)) π(q(x)) = π(c r(x)) = 0 π( r(x)) = 0 so either π(p(x)) = 0 or π(q(x)) = 0. We can assume that π(p(x)) = 0. Then p(x) = c p(x) for some p(x) R[x]. Since p(x) is primitive we get a contradiction Lemma. Let R be a UFD and K be the field of fractions of R. 1) For any non-zero polynomial p(x) K[x] there is an element c(p) K and a primitive polynomial p (x) R[x] such that p(x) = c(p)p (x) 144

9 2) If p(x) = c(p)p (x) and p(x) = c(p) p (x) for some c(p), c(p) K, and some primitive polynomials p (x), p (x) R[x] then there exists a unit u R such that c(p) = uc(p), p (x) = u 1 p (x) 3) If p(x), q(x) K[x] are non-zero polynomials then for some unit u R we have c(pq) = uc(p)c(q) (p(x)q(x)) = u 1 p (x)q (x) 37.8 Definition. If p(x) K[x] then the element c(p) K is called the content of p(x). Proof of Lemma ) If p(x) K[x] then there is there exists 0 c R such that cp(x) R[x]. Let b R be a greatest common divisor of coefficients of cp(x). Take p (x) = c/b p(x), c(p) = b/c Check that p (x) R[x] and p (x) is a primitive polynomial. 2) We have c(p)p (x) = p(x) = c(p) p (x) where c(p), c(p) K are non-zero elements and p (x), p (x) R[x] are primitive polynomials. Let p (x) = a 0 + a 1 x a n x n. We have p (x) = c(p) c(p) 1 p (x) Since c(p) c(p) 1 K there b, d R such that c(p) c(p) 1 = b/d. We can assume that gcd(b, d) 1. We have d p (x) = bp (x) = ba 0 + ba 1 x ba n x n 145

10 This gives: d ba i for i = 1,..., n. By (35.6) this gives that d a i for all i. Since p (x) is a primitive polynomial it implies that d is a unit in R. By an analogous argument we obtain that b is also a unit in R. As a consequence u = db 1 is a unit in R and we have c(p) = uc(p), p (x) = u 1 p (x) 3) For p(x), q(x) K[x] we have c(p)c(q)p (x)q (x) = p(x)q(x) = c(pq)(p(x)q(x)) We have c(p)c(q), c(pq) K, (p(x)q(x)) R[x] is primitive by construction and p (x)q (x) R[x] is primitive by Lemma Applying part 2) we obtain that there is a unit u R such that c(pq) = uc(p)c(q) (p(x)q(x)) = u 1 p (x)q (x) 37.9 Note. Check: 1) Let p(x) K[x]. Then p(x) R[x] iff c(p) R. 2) p(x) R[x] is primitive iff p(x) = up (x) for some unit u R Proposition. Let R be a UFD and K be the ring of fractions of R. 1) A polynomial p(x) K[x] of non-zero degree is irreducible in K[x] iff p (x) is irreducible in R[x]. 2) A polynomial p(x) R[x] of non-zero degree is irreducible in R[x] iff p(x) is primitive and it is irreducible in K[x]. 146

11 Proof. 1) ( ) If p(x) is not irreducible in K[x] then p(x) = q(x)r(x) for some q(x), r(x) K[x], deg q(x), deg r(x) > 0. By Lemma 37.7 we have p (x) = uq (x)r (x) for some unit u R. Therefore p (x) is not irreducible in R[x]. ( ) If p (x) is not irreducible in R[x] then p (x) = q(x)r(x) where q(x), r(x) are non-units in R[x]. Since p (x) is primitive we must have deg q(x), deg r(x) > 0. Then and so p(x) is not irreducible in K[x]. p(x) = c(p)q(x)r(x) 2) ( ) If p(x) R[x] is irreducible then it must be a primitive polynomial. Therefore p(x) = up (x) for some unit u R. Since p(x) is irreducible in R[x], thus p (x) is also irreducible in R[x], and so by part 1) p(x) is irreducible in K[x]. ( ) Since p(x) is irreducible in K[x] by part 1) we get that p (x) is irreducible in R[x]. Also, since p(x) is primitive we have p(x) = up (x) for some unit u R. It follows that p(x) is irreducible in R[x]. Proof of Theorem By Theorem 31.7 we need to show that: 1) Every non-zero, non-unit p(x) R[x] is a product of irreducible polynomials. 2) Every irreducible polynomial p(x) R[x] is a prime element of R[x]. 147

12 1) Recall that irreducible polynomials of degree 0 in R[x] correspond to irreducible elements of R. Since R is a UFD it follows that if p(x) R[x] has degree 0 then p(x) is a product of irreducibles in R[x]. If deg p(x) > 0 then p(x) is a non-zero non-unit element of K[x]. Since K[x] is a UFD we have p(x) = q 1 (x)q 2 (x)...q k (x) where q 1 (x),..., q k (x) are irreducible polynomials in K[x]. We have p(x) = (c(q 1 )c(q 2 )...c(q k ))q 1(x)q 2(x)...q k(x) By Proposition q i (x) are irreducible polynomials in R[x]. Moreover, by Lemma 37.7 we have c(q 1 )c(q 2 )...c(q k ) = uc(p) for some unit u R, and also since p(x) R[x] we have c(p) R. Therefore c(q 1 )c(q 2 )...c(q k ) R, and since R is a UFD we have c(q 1 )c(q 2 )...c(q k ) = a 1 a 2...a l where a 1,..., a l R are irreducible elements in R[x]. As a consequence we obtain a factorization of p(x): p(x) = a 1 a 2...a l q 1(x)q 2(x)...q k(x) where a 1,..., a l, q1(x),..., qk (x) are irreducible elements in R[x]. 2) Let p(x) R[x] be an irreducible polynomial. We need to show that if for some q(x), r(x) R[x] we have p(x) q(x)r(x) then either p(x) q(x) or p(x) r(x). Exercise: this holds if deg p(x) = 0. If deg p(x) > 0, then since p(x) is irreducible in R[x] by Proposition we obtain that p(x) is primitive and it is irreducible in K[x]. Since K[x] is a UFD we obtain that p(x) is a prime element of K[x], and so p(x) q(x) or p(x) r(x) in K[x]. We can assume that p(x) q(x) in K[x]. Then there exists h(x) K[x] such that p(x)h(x) = q(x) 148

13 We have c(p)c(h) = uc(q) for some unit u R. Also, since q(x) R[x] we have that c(q) R and therefore c(p)c(h) R. Finally, since p(x) is primitive thus c(p) is a unit in R. We obtain: c(h) = c(p) 1 uc(q) R Therefore h(x) R[x], and so p(x) q(x) in R[x] Corollary. Z[x] is a UFD Corollary. If R is a UFD then the ring of polynomials of n variables R[x 1,..., x n ] is also a UFD. Proof. It is enough to notice that (check!). R[x 1,..., x n ] = R[x 1,..., x n 1 ][x n ] 149

38 Irreducibility criteria in rings of polynomials

38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m