ahw9 - Math 3124 Tuesday, December 6 Ninth Homework...

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Unformatted text preview: Math 3124 Tuesday, December 6 Ninth Homework Solutions 1. Let a, b ∈ R. Then θ (a + b) = (a + b) p = (by the Freshman calculus rule) a p + b p = θ (a)θ (b), so θ respects addition. Also θ (ab) = (ab) p = (because R is commutative) a p b p = θ (a)θ (b), consequently θ respects multiplication and it now follows that θ is a homomorphism. To prove that θ is one-to-one, it will be sufficient to show that ker θ = 0. However if θ (a) = 0, then a p = 0. Since R is an integral domain, we see that if a = 0, then ar = 0 implies ar+1 = 0, so by induction on r, we see that ar = 0 for all r, in particular a p = 0. However a ∈ ker θ if and only if a p = 0. We conclude that ker θ = 0 as required. √ 2. Problem 27.23 on page 135. Let R denote the subfield {a + b 2 | a, b ∈ Q} of R, and √ let S denote the subfield {a + b 2 | a, b ∈ Q} of Q. Verify that θ : R → S defined by √ √ θ (a + b 2) = a + b 3 is not a ring isomorphism. √ √ √√ √√ θ ( 2)θ ( 2) = 3 3 = 3 yet θ ( 2 2) = θ (2) = 2. Since 3 = 2, we conclude that θ is not a homomorphism and thus certainly not an isomorphism. 3. Problem 38.20 on page 177. Prove that if F is a field, R is a ring and θ : F → R is a ring homomorphism, then either θ is one-to-one or θ (a) = 0 for all a ∈ F . We know that ker θ = 0 or F because F is a field (Example 38.4 on page 176). If ker θ = 0, then θ is one-to-one. On the other hand if ker θ = F , then θ (a) = 0 for all a ∈ F. 4. Problem 39.6 on page 183. Prove that if I is an ideal of the ring R, then R/I is commutative iff ab − ba ∈ I for all a, b ∈ R. Suppose R/I is commutative. Then (I + a)(I + b) = (I + b)(I + a) for all a, b ∈ R, hence I + ab = I + ba. We deduce that ab − ba ∈ I for all a, b ∈ R. Conversely suppose ab − ba ∈ I for all a, b ∈ R. Then I + ab = I + ba and hence (I + a)(I + b) = (I + b)(I + a) for all a, b ∈ R. We deduce that R/I is commutative and the result is proven. ...
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