Selected Homework Solutions

SELECTED HOMEWORK SOLUTIONS

Page 75, 1.84: For a given positive integer m, ﬁnd all integers r, with 0 < r < m such that 2r ≡ 0 mod m.

Solution: The positive integers t that satisfy t ≡ 0 mod m are {m, 2m, 3m, . . .}. If 2r is in that set, then 2r = m is the only possible choice, since if 2r = km, for k ≥ 2, then r ≥ m, a contradiction. Thus, if m is even, r = m/2 is the only solution. If m is odd, then 2r 6= m and so there is no solution.

Page 123, 2.34: If n ≥ 3, show that if α ∈ Sn commutes with every β ∈ Sn, then α = (1).

Solution: Let α commute with all β ∈ Sn and assume that α 6= (1). Then there are a and b in {1, 2, . . . , n}, such that a 6= b, and α(a) = b. Thus there is certainly β ∈ Sn such that β(a) = a and β(b) 6= b. Now (αβ)(a) = α(β(a)) = α(a) = b and (βα)(a) = β(α(a)) = β(b). But αβ = βα and so β(b) = b, a contradiction. Thus α = (1).

Page 190, 2.103: Let A and B be groups, let A0 be a normal subgroup fo A and B0 be a normal subgroup of B, and let α : A → B a homomorphism with α(A0) ≤ B0. 0 0 0 (i) Prove that there is a (well-deﬁned) homorphism α∗ : A/A → B/B given by α∗ : aA → α(a)B0.

(ii) Prove that if α is surjective, then α∗ is surjective.

0 0 0 0 0 Solution: (i) α∗ is well-deﬁned: Suppose a1A = a2A . Then a2 = a1a , with a ∈ A . 0 0 0 0 0 0 Hence α(a2) = α(a1)α(a ). But α(A ) ≤ B and therefore α(a ) ∈ B . Thus α∗(a2A ) = 0 0 0 0 0 α(a2)B = α(a1)α(a )B = α(a1)B = α∗(a1A ) and so α∗ is well-deﬁned. 0 0 0 0 0 α∗ is a homomorphism: α∗(a1A · a2A ) = α∗(a1a2A ) = α(a1a2)B = α(a1)α(a2)B = 0 0 0 0 (α(a1)B )(α(a2)B ) = α∗(a1A )α∗(a2A ). Therefore α∗ is a homomorphism. 0 0 (ii) If α is surjective, then α∗ is surjective: Let bB ∈ B/B . Since α is surjective and 0 0 0 b ∈ B, there is a ∈ A such that α(a) = b. Therefore α∗(aA ) = α(a)B = bB . Hence, α∗ is surjective.

Page 191, 2.113: If G is a group and x, y ∈ G, deﬁne their commutator to be xyx−1y−1 and deﬁne the commutator subgroup to be the subgroup generated by all the commutators. (i) Prove that G0 is normal in G. (ii) Prove that G/G0 is abelian. (iii) If φ : G → A is a homomorphism, where A is an abelian group, prove that G0 ≤ ker φ. Conversely, if G0 ≤ ker φ, prove that im φ is abelian. (iv) If G0 ≤ H ≤ G, prove that H is normal in G.

Solution: (i) Let c = xyx−1y−1 and consider gcg−1. Then gcg−1 = g(xyx−1y−1)g−1 = (gxg−1)(gyg−1)(gx−1g−1)(gy−1g−1) = (gxg−1)(gyg−1)(gxg−1)−1(gyg−1)−1) which is again a commutator. Therefore the conjugate of a commutator is a commutator and thus the conjugate of a product of commutators is a product of commutators. Hence G0 is normal in G. (ii) Let gG0, hG0 ∈ G/G0. Then (gG0)(hG0) = (gh)G0 = hg(g−1h−1gh)G0 = (hg)G0 since g−1h−1gh ∈ G0. But (hg)G0 = (hG0)(gG0) and we have (gG0)(hG0) = (hG0)(gG0). Thus G/G0 is abelian. (iii) Since φ : G → A is a homomorphism, it follows from the First Isomorphism Theorem that G/ker φ ' im φ which is abelian, since A is abelian. Hence G/ker φ is abelian. Therefore (g ker φ)(h ker φ)(g ker φ)−1(h ker φ)−1 = ker φ. On the other hand

(g ker φ)(h ker φ)(g ker φ)−1(h ker φ)−1 = (ghg−1h−1)ker φ.

Thus ghg−1h−1 ∈ ker φ for every commutator and therefore, G0 ≤ ker φ. (iv) This was proven in class for the case of N < H < G where G/N is abelian.

Page 234, 3.26: Let k be a ﬁeld, and let R be the subring

R = {n · 1 : n ∈ Z} ⊆ k.

(i) If F is a subfield of k, prove that R ⊆ F. (ii) Prove that a subfield F of k is the prime field of k if and only if it is the smallest subfield of k containing R; that is, there is no subfield of F 0 with R ⊆ F 0 ⊂ F.

Solution: (i) If F is a subfield of k, then 1 ∈ F . Therefore n · 1 is in F for every n ∈ Z. Therefore R ⊆ F. (ii) If F is the prime subfield of k, then it is the intersection of all subfields, and it follows from (i) that each one contains R and hence so does F . If a subfield F 0 exists, then the intersection of all such subfields would not contain all of F , contradiction. Now suppose that there is no such subfield F 0. Then the smallest subfield containing R is a subfield of every subfield of k, each of which contains R. Hence it must be the prime subfield of k. Page 242, 3.37: Assume that (x − a)|f(x) in R[x]. Prove that (x − a)2|f(x) if and only if (x − a)|f 0(x) in R[x].

Solution: Since (x − a)|f(x), it follows that f(x) = (x − a)g(x), for some g(x). Hence f 0(x) = (x − a)g0(x) + g(x). Now assume that (x − a)2|f(x). Then f(x) = (x − a)2r(x), for some r(x). Hence f 0(x) = 2(x − a)r(x) + (x − a)2r0(x), which clearly implies that (x − a)|f 0(x). On the other hand if (x − a)|f 0(x), then it follows from above, that (x − a)|g(x) and so g(x) = (x − a)h(x) for some h(x). Thus f(x) = (x − a)g(x) = (x − a)(x − a)h(x) and (x − a)2|f(x).

Page 273, 3.60: If R is a domain and f(x) ∈ R[x] has degree n, show that f(x) has at most n roots.

Solution: If R is a field, then this result is proven in Theorem 3.50. The proof of that theorem uses only Proposition 3.49 which uses Lemma 3.48 and the Division Algorithm which is proven for domains. Hence the work ”field” could be replaced in all of those theorems and lemmas and the result in Theorem 3.50 would still be true. Another proof is to replace the ring R by the fraction field of R, say S, which is a field in which R is a subring. Then the polynomial f(x) can be thought of as being in S[x] and has at most n roots in S[x]. Hence it has at most n roots in R[x].