MATH 113 - HOMEWORK 6 SOLUTIONS

√ √ √ 18.12. Let Q√[ 2] = {a + b 2 | a, b ∈ Q}. Note that Q[ 2] ⊂ R, and√ the operations of + and · on Q√[ 2] are the usual + and · of real numbers. Not only is Q[ 2] closed under + and ·, but [ 2] is a field (a subfield of ). √Q R √ √ √ • Q[√2] is closed under addition. (a + b 2) + (√c + d 2)√ = (a + c) + (b +√d) 2. √ • Q[ 2] is closed under multiplication. (a + b 2)(c + d 2) = ac + ad 2 + bc 2 + √ 2 √ bd 2 = (ac + 2bd) + (ad + bc) 2. √ • Addition is associative and commutative on Q[ 2], since it is associative and com- mutative in R. √ √ • Identity for addition: 0 = 0 + 0 2 ∈ Q[ 2].√ √ √ √ • Inverses for addition: The inverse of a + b 2 is −(a +√b 2) = −a + −b 2 ∈ Q[ 2]. • Multiplication is associative and commutative on Q[ 2], since it is associative and commutative in R. √ • Distributivity holds in Q[ 2], since it√ holds in√R. • Identity for multiplication: 1 = 1 + 0 2 ∈ Q[ 2]. √ • Inverses for multiplication: This is the interesting one. Given a, b ∈√Q with a√+b 2 6= 0, (either a 6= 0 or b√6= 0), we√ need to find c, d ∈ Q such that√ (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 determinant√ a − 2b , so it’s invertible unless a = 2b . Fortunately, if a2 = 2b2, then a = ± 2b, but a, b ∈ 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 √ a b √ √ so the inverse of a + b 2 is − 2 ∈ [ 2]. a2 − 2b2 a2 − 2b2 Q √ 1 Solution 2: We can invert a + b 2 in R: √ . Rationalizing the denominator, a + b 2 √ √ 1 a − b 2 a − b 2 a b √ √ √ = √ √ = = − 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) = {(a, b, c) | a = ±1, c = ±1, b 6= 0}.

18.38. Suppose R is commutative. Then for all a, b ∈ 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 ∈ 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 ∈ 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 ∈ Z . So no positive integer n kills every element of the ring.

19.8. Z3 × Z3 has characteristic 3. For all (n, m) ∈ 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 | n37 − n if and only if n37 − n ≡ 0 (mod p) if and only if n37 ≡ n (mod p). 37: If 37 | n, then automatically 37 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡ n (mod 37), so by the equivalence noted above, 37 | n37 − n. 19: If 19 | n, then automatically 19 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡ (n19)(n18) ≡ (n)(1) ≡ n (mod 19), as desired. 13: If 13 | n, then automatically 13 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡ (n12)3n ≡ 13n ≡ n (mod 13), as desired. 7: If 7 | n, then automatically 7 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡ (n6)6n ≡ 16n ≡ n (mod 7), as desired. 3: If 3 | n, then automatically 3 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡ (n2)18n ≡ 118n ≡ n (mod 3), as desired. 2: If 2 | n, then automatically 2 | 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)] ∈ 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)] ∈ 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)] ∈ 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)] ∈ 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)] ∈ 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)] ∈ F . [(a, b)] ([(c, d)] + [(e, f)]) = [(a, b)][(cf + de, df)] = [(a(cf + de), b(df))] = [(acf + ade, bdf)]. On the other hand, [(a, b)][(c, d)] + [(a, b)][(e, f)] = [(ac, bd)] + [(ae, bf)] = [((ac)(bf) + (bd)(ae), (bd)(bf))] = [(abcf + abde, b2df)]. Now I claim that [(acf+ade, bdf)] = [(abcf+abde, b2df)]. Indeed, (acf+ade)(b2df) = (bdf)b(acf + ade) = (bdf)(abcf + abde), so (acf + ade, bdf) ∼ (abcf + abde, b2df). We only need to check distributivity on the left, since by (6) multiplication is commutative. (If distributivity on the left holds and · is commutative, then (y+z)x = x(y + z) = xy + xz = yx + zx, which is distributivity on the right.) (8) [(1, 1)] is a multiplicative identity element in F . Let [(a, b)] ∈ F . [(a, b)][(1, 1)] = [(a1, b1)] = [(a, b)]. We only need to check the product in this order, since by (6) multiplication is commutative. (9) If [(a, b)] ∈ F is not the additive identity element, then a 6= 0 in D and [(b, a)] is a multiplicative inverse for [(a, b)]. For the first statement, we’ll prove the contrapositive: If a = 0 in D, then [(a, b)] = [(0, 1)]. Indeed, if a = 0, then a1 = b0 = 0, so (a, b) ∼ (0, 1). Now we know that if [(a, b)] 6= [(0, 1)], then a 6= 0, so [(b, a)] is an element of F . Let’s check that it is [(a, b)]−1. We have [(a, b)][(b, a)] = [(ab, ba)] = [(ab, ab)]. The claim is that [(ab, ab)] = [(1, 1)]. Indeed, ab1 = ab1, so (ab, ab) ∼ (1, 1). We only need to check the product in this order, since by (6) multiplication is commutative.