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HOMEWORK 6 SOLUTIONS 18.12. Let Q[ √ 2] MATH 113 - HOMEWORK 6 SOLUTIONS p p p 18.12. Let Qp[ 2] = fa + b 2 j a; b 2 Qg. Note that Q[ 2] ⊂ R, andp the operations of + and · on Qp[ 2] are the usual + and · of real numbers. Not only is Q[ 2] closed under + and ·, but [ 2] is a field (a subfield of ). pQ R p p p • Q[p2] is closed under addition. (a + b 2) + (pc + d 2)p = (a + c) + (b +pd) 2. p • Q[ 2] is closed under multiplication. (a + b 2)(c + d 2) = ac + ad 2 + bc 2 + p 2 p bd 2 = (ac + 2bd) + (ad + bc) 2. p • Addition is associative and commutative on Q[ 2], since it is associative and com- mutative in R. p p • Identity for addition: 0 = 0 + 0 2 2 Q[ 2].p p p p • Inverses for addition: The inverse of a + b 2 is −(a +pb 2) = −a + −b 2 2 Q[ 2]. • Multiplication is associative and commutative on Q[ 2], since it is associative and commutative in R. p • Distributivity holds in Q[ 2], since itp holds inpR. • Identity for multiplication: 1 = 1 + 0 2 2 Q[ 2]. p • Inverses for multiplication: This is the interesting one. Given a; b 2pQ with ap+b 2 6= 0, (either a 6= 0 or bp6= 0), wep need to find c; d 2 Q such thatp (a + b 2)(c + d 2) = 1. Solution 1: (a+b 2)(c+d 2) = (ac+2bd)+(ad+bc) 2, so we need ac+2bd = 1 and ad + bc = 0. Let's do some linear algebra. a 2b c 1 = b a d 0 2 2 2 2 The matrix on the left has determinantp a − 2b , so it's invertible unless a = 2b . Fortunately, if a2 = 2b2, then a = ± 2b, but a; b 2 Q, so the only way this could happen is if a = b = 0. Multiplying by the inverse matrix 1 a −2b ; a2 − 2b2 −b a we have c 1 a −2b 1 1 a = = ; d a2 − 2b2 −b a 0 a2 − 2b2 −b p a b p p so the inverse of a + b 2 is − 2 2 [ 2]. a2 − 2b2 a2 − 2b2 Q p 1 Solution 2: We can invert a + b 2 in R: p . Rationalizing the denominator, a + b 2 p p 1 a − b 2 a − b 2 a b p p p = p p = = − 2 2 Q[ 2]: a + b 2 (a + b 2)(a − b 2) a2 − 2b2 a2 − 2b2 a2 − 2b2 18.18. (a; b; c) is a unit in the product ring if and only if each of a, b, and c are units in the factor rings. Indeed, if (a; b; c) has an inverse (d; e; f), then (a; b; c)(d; e; f) = (ad; be; cf) = (1; 1; 1), and (d; e; f)(a; b; c) = (da; eb; fc) = (1; 1; 1), so a−1 = d, b−1 = e, and c−1 = f. Conversely, if a, b, and c are units, then (a−1; b−1; c−1) is the inverse of (a; b; c) in the product ring: (a; b; c)(a−1; b−1; c−1) = (aa−1; bb−1; cc−1) = (1; 1; 1) = (a−1a; b−1b; c−1c) = (a−1; b−1; c−1)(a; b; c). So U(Z × Q × Z) = U(Z) × U(Q) × U(Z) = f(a; b; c) j a = ±1; c = ±1; b 6= 0g. 18.38. Suppose R is commutative. Then for all a; b 2 R,(a+b)(a−b) = a(a−b)+b(a−b) = aa − ab + ba − bb = a2 − ab + ab − b2 = a2 − b2. Conversely, suppose that for all a; b 2 R, a2 − b2 = (a + b)(a − b). Then a2 − b2 = (a + b)(a − b) = a2 − ab + ba − b2 by the computation above. Subtracting a2 from both sides and adding b2 to both sides, we have 0 = −ab + ba. Adding ab to both sides, we have ab = ba. So R is commutative. 18.44.a. Let a; b 2 R be idempotent. Then (ab)2 = abab = aabb = a2b2 = ab. So ab is idempotent. b. (a; b) is idempotent in the product ring if and only if each of a and b are idempotent in the factor rings. Indeed, (a; b) is idempotent if and only if (a; b) = (a; b)2 = (a2; b2), if and only if a2 = a and b2 = b. In Z6, by direct inspection, the idempotent elements are 0, 1, and 3. In Z12, the idempotent elements are 0, 1, 4, and 9. So the idempotent elements of Z6 × Z12 are (0; 0), (1; 0), (3; 0), (0; 1), (1; 1), (3; 1), (0; 4), (1; 4), (3; 4), (0; 9), (1; 9), (3; 9), + 19.7. Z3 × 3Z has characteristic 0. Indeed, n · (0; 3) = (0; 3n) 6= (0; 0) for any n 2 Z . So no positive integer n kills every element of the ring. 19.8. Z3 × Z3 has characteristic 3. For all (n; m) 2 Z3 × Z3, 3 · (n; m) = (3n; 3m) = (0; 0). 19.29. Let D be an integral domain, and suppose for contradiction that D has of charac- teristic c, with c a positive composite number. We can write c = nm with n; m < c. Now (n·1) 6= 0, since n < char(D), and similarly (m·1) 6= 0. But (n·1)(m·1) = (1+···+1)(m·1) = (1(m · 1) + ··· + 1(m · 1)) = n · (m · 1) = nm · 1 = c · 1 = 0. So (n · 1) and (m · 1) are a pair of zero divisors in D, contradiction. 20.10. The numbers between 1 and 23 relatively prime to 24 are 1; 5; 7; 11; 13; 17; 19; 23, so φ(24) = 8. Now 7 is relatively prime to 24, so by Euler's theorem, 71000 ≡ (78)125 ≡ 1125 ≡ 1 (mod 24). 20.29. To show that n37 − n is divisible by 383838, we'll show that it is divisible by each of the primes 37, 19, 13, 7, 3, and 2. This suffices, since 383838 = (37)(19)(13)(7)(3)(2). Note that p j n37 − n if and only if n37 − n ≡ 0 (mod p) if and only if n37 ≡ n (mod p). 37: If 37 j n, then automatically 37 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ n (mod 37), so by the equivalence noted above, 37 j n37 − n. 19: If 19 j n, then automatically 19 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ (n19)(n18) ≡ (n)(1) ≡ n (mod 19), as desired. 13: If 13 j n, then automatically 13 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ (n12)3n ≡ 13n ≡ n (mod 13), as desired. 7: If 7 j n, then automatically 7 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ (n6)6n ≡ 16n ≡ n (mod 7), as desired. 3: If 3 j n, then automatically 3 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ (n2)18n ≡ 118n ≡ n (mod 3), as desired. 2: If 2 j n, then automatically 2 j n37 − n. If not, then by Fermat's Little Theorem, n37 ≡ (n1)36n ≡ 136n ≡ n (mod 2), as desired. Alternatively, n is odd, so n37 ≡ 1 ≡ n (mod 2). Fraleigh, Section 21: Prove that the field of fractions of an integral domain really is a field. To do this, refer to \Step 3" of the construction of the field of fractions in Section 21 of Fraleigh, page 193. Parts 1 and 9 of Step 3 are done for you. You should read the proof of Part 1, complete Parts 2-8, and finish by reading Part 9. You can copy down or work out for yourself the proofs of Parts 1 and 9 if you want to have the whole proof in one place, but you don't have to. Solution: (1) Addition in F is commutative. Let [(a; b)]; [(c; d)] 2 F . [(a; b)] + [(c; d)] = [(ad + bc; bd)] = [(cb + da; db)] = [(c; d)] + [(a; b)]: (2) Addition is associative. Let [(a; b)]; [(c; d)]; [(e; f)] 2 F . ([(a; b)] + [(c; d)]) + [(e; f)] = [(ad + bc; bd)] + [(e; f)] = [((ad + bc)f + (bd)e; (bd)f)] = [(adf + bcf + bde; bdf)] = [(a(df) + b(cf + de); b(df))] = [(a; b)] + [(cf + de; df)] = [(a; b)] + ([(c; d)] + [(e; f)]) : (3) [(0; 1)] is an identity element for addition in F . Let [(a; b)] 2 F . [(a; b)] + [(0; 1)] = [(a1 + b0; b1)] = [(a; b)]: We only need to check the sum in this order, since by (1) addition is commutative. (4) [(−a; b)] is an additive inverse for [(a; b)] in F . [(−a; b)] + [(a; b)] = [((−a)b + ba; b2)] = [(−ab + ab; b2)] = [(0; b2)] Now I claim that [(0; b2)] = [(0; 1)]. Indeed, 0 · 1 = b2 · 0 = 0, so (0; b2) ∼ (0; 1). We only need to check the sum in this order, since by (1) addition is commutative. (5) Multiplication in F is associative. Let [(a; b)]; [(c; d)]; [(e; f)] 2 F . ([(a; b)][(c; d)]) [(e; f)] = [(ac; bd)][(e; f)] = [((ac)e; (bd)f)] = [(a(ce); b(df))] = [(a; b)][(ce; df)] = [(a; b)] ([(c; d)][(e; f)]) : (6) Multiplication in F is commutative. Let [(a; b)]; [(c; d)] 2 F . [(a; b)][(c; d)] = [(ac; bd)] = [(ca; db)] = [(c; d)][(a; b)]: (7) The distributive laws hold in F . Let [(a; b)]; [(c; d)]; [(e; f)] 2 F .
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