Adv Alegbra HW Solutions 90

Adv Alegbra HW Solutions 90 - 90 3.23(i Show that F ={a bi...

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Unformatted text preview: 90 3.23 (i) Show that F = {a + bi : a , b ∈ Q} is a field. Solution. It is straightforward to check that F is a subring of C, and so it is a commutative ring; it is a field because the inverse of a + bi is r −1 (a − bi ) ∈ F , where r = a 2 + b2 . (ii) Show that every u ∈ F has a factorization u = αβ −1 , where α, β ∈ Z [i ]. (See Exercise 3.50.) Solution. Write a + bi = ( p /q ) + (r /s )i = ( ps + qri )(qs )−1 , where p , q , r, s ∈ Z. 3.24 If R is a commutative ring, define a relation ≡ on R by a ≡ b if there is a unit u ∈ R with b = ua . (i) Prove that ≡ is an equivalence relation. Solution. Absent. (ii) If a ≡ b, prove that (a ) = (b), where (a ) = {ra : r ∈ R }. Conversely, prove that if R is a domain, then (a ) = (b) implies a ≡ b. Solution. Assume that a ≡ b, so there is u , v ∈ R with a = ub and u v = 1. We claim that (a ) = (b). If x ∈ (a ), then x = ra = r ub ∈ (b), and so (a ) ≤ (b). For the reverse inclusion, if y ∈ (b), then y = sb = s v a ∈ (a ). Hence (a ) = (b). Conversely, assume that (a ) = (b). If a = 0, then b = 0 and so a ≡ b. We now assume that a = 0. Since a ∈ (b), there is r ∈ R with a = r b; since b ∈ (a ), there is s ∈ R with b = sa . Therefore, a = r b = r sa , and so (1 − r s )a = 0. Since R is a domain, 1 = r s and r is a unit. Therefore, a ≡ b. 3.25 If R is a domain, prove that there is no subfield K of Frac( R ) such that R⊆K Frac( R ). Solution. Absent. 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 . Solution. Absent. (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 F with R ⊆ F F. Solution. Absent. ...
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