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36 Rings of Fractions 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) 2 R[x] then p(x) 2 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]. 36.1 Definition. If R is a ring then a subset S ⊆ R is a multiplicative subset if 1) 1 2 S 2) if a; b 2 S then ab 2 S 36.2 Example. Multiplicative subsets of Z: 1) S = Z 137 2) S0 = Z − f0g 3) Sp = fn 2 Z j p - ng where p is a prime number. Note: Sp = Z − hpi. 36.3 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. 36.4. Construction of a ring of fractions. Goal. For a ring R and a multiplicative subset S ⊆ R construct a ring S−1R such that every element of S becomes a unit in S−1R. Consider a relation on the set R × S: 0 0 0 0 (a; s) ∼ (a ; s ) if s0(as − a s) = 0 for some s0 2 S Check: ∼ is an equivalence relation. Elements of S−1R = (equivalence classes of R × S under the relation ∼). Notation. a=s := the equivalence class represented by (a; s) Note. 1) If 0 2 S then (a; s) ∼ (a0; s0) for all (a; s); (a0; s0) 2 R × S, and so S−1R consists of only one element. 2) If R is an integral domain and 0 62 S then (a; s) ∼ (a0; s0) iff as0 = a0s. 138 Multiplication in S−1R: (a=s)(a0=s0) = (aa0)=ss0 Addition in S−1R: a=s + a0=s0 = (as0 + a0s)=ss0 Check: 1) The operations of addition and multiplication in S−1R are well defined. 2) These operations define a commutative ring structure on S−1R. 3) The map −1 'S : R ! S R; 'S(r) = r=1 is a ring homomorphism. −1 Note. If s 2 S then 'S(s) = s=1 is a unit in S R: (s=1)(1=s) = s=s = 1=1 = 1S−1R 36.5 Definition. If S is a multiplicative subset of a ring R then S−1R is called the ring of fractions of R with respect to S. 36.6 Theorem. If S is a multiplicative subset of a ring R and f : R ! R0 is a ring homomorphism such for ever s 2 S the element f(s) 2 R0 is a unit, then there exists a unique homomorphism f¯: S−1R ! R0 such that the following diagram commutes: f R / R0 {= { { { 'S { { f¯ { { { S−1R 139 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 = f'S. 36.7 Examples. −1 ∼ 1) If S ⊆ Z, S = Z then S Z = f0g. −1 ∼ 2) If S0 ⊆ Z, S0 = Z − f0g then S0 Z = Q. −1 3) If S0 ⊆ Z, S0 = Z − hpi then Sp Z is isomorphic to the subring of Q m consisting of all fractions n such that p - n. −1 ∼ 4) If R = Z=6Z, S = f1; 3g ⊆ R then S R = Z=2Z (check!). 36.8 Proposition. If S is a multiplicative subset of a ring R then Ker('S) = fa 2 R j sa = 0 for some s 2 Sg Proof. We have 'S(r) = 0=1 iff s(r · 1 − 0 · 1) = 0 for some s 2 S. 36.9 Corollary. If R is an integral domain and S ⊆ R is a multiplicative subset such that 0 62 S then the homomorphism −1 'S : R ! S R is 1-1. As a consequence in this case we can identify R with a subring of S−1R. 140 36.10 Note. 1) If R is an integral domain and S = R − f0g then the ring S−1R is a field. In this case S−1R 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−1R is called the localization of R at I and it is denoted RI . 36.11 Definition. A ring R is a local ring if R has exactly one maximal ideal. 36.12 Examples. 1) If F is a field then it is a local ring with the maximal ideal I = f0g. 2) Check: if F is a field then F[[x]] is a local ring with the maximal ideal hxi. 3) Check: if F is a field then F[x] is not a local ring since for every a 2 F the ideal hx − ai is a maximal ideal in F[x]. 4) Z is not a local ring. 36.13 Proposition. If R is a local ring then the maximal ideal J C R consists of all non-units of R. Proof. Since J 6= R thus J does not contain any units. Conversely, if a 2 R is a non-unit then 1 62 hai so hai 6= R. Therefore by Theorem 29.1 hai is contained in some maximal ideal of R. Since J is the only maximal ideal we obtain hai ⊆ J, and so a 2 J. 141 36.14 Proposition. If R is a ring and I C R is a prime ideal then the ring RI is a local ring, and the maximal ideal J C RI is given by J := fa=s j a 2 I; s 62 Ig Proof. Exercise. 142 37 Factorization in rings of polynomials Goal: 37.1 Theorem. If R is a UFD then so is R[x]. 37.2 Lemma. If R is an integral domain then p(x) 2 R[x] is a unit in R[x] iff deg p(x) = 0 and p(x) = a0 where a0 is a unit in R. Proof. Exercise. 37.3 Lemma. If R is an integral domain and p(x) = a0 is a polynomial of degree 0 in R[x] then p(x) is irreducible in R[x] iff a0 is irreducible in R. Proof. Exercise. n 37.4 Definition. If R is a UFD and p(x) = a0 + a1x + : : : anx 2 R[x], then p(x) is a primitive polynomial if gcd(a0; : : : ; an) ∼ 1. 37.5 Lemma. If R is a UFD and p(x) 2 R[x] is an irreducible polynomial such that deg p(x) > 0 then p(x) is a primitive polynomial. n Proof. If p(x) = a0 + a1x + : : : anx is not primitive then p(x) = gcd(a0; : : : an)q(x) for some q(x) 2 R[x], deg q(x) > 0. Since both gcd(a0; : : : ; an) and q(x) are non-units in R[x] the polynomial p(x) is not irreducible. 143 37.6 Lemma (Gauss). If R is a UFD and p(x); q(x) 2 R[x] are primitive poly- nomials 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 2 R is an irreducible element, and r~(x) 2 R[x]. Take the canonical epimorphism π : R −! R=hci This defines a homomorphism of rings of polynomials π¯ : R[x] −! (R=hci)[x] given by m n π¯(a0 + a1x + ::: + anx ) := π(a0) + π1(a)x + ::: + π(an)x Note: 1) Since c is irreducible element thus hci is a prime ideal and so R=hci is a domain. As a consequence (R=hci)[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) 2 R[x]. Since p(x) is primitive we get a contradiction. 37.7 Lemma. Let R be a UFD and K be the field of fractions of R. 1) For any non-zero polynomial p(x) 2 K[x] there is an element c(p) 2 K and a primitive polynomial p∗(x) 2 R[x] such that p(x) = c(p)p∗(x) 144 2) If p(x) = c(p)p∗(x) and p(x) =c ~(p)~p∗(x) for some c(p); c~(p) 2 K, and some primitive polynomials p∗(x); p~∗(x) 2 R[x] then there exists a unit u 2 R such that c~(p) = uc(p); p~∗(x) = u−1p∗(x) 3) If p(x); q(x) 2 K[x] are non-zero polynomials then for some unit u 2 R we have c(pq) = uc(p)c(q)(p(x)q(x))∗ = u−1p∗(x)q∗(x) 37.8 Definition. If p(x) 2 K[x] then the element c(p) 2 K is called the content of p(x). Proof of Lemma 37.7. 1) If p(x) 2 K[x] then there is there exists 0 6= c 2 R such that cp(x) 2 R[x].
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