Adv Alegbra HW Solutions 91

Adv Alegbra HW Solutions 91 - 91 (iii) If R is a subfield...

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Unformatted text preview: 91 (iii) If R is a subfield of k , prove that R is the prime field of k . Solution. Absent. 3.27 (i) Show that every subfield of C contains Q. Solution. Every subfield of C contains 1, hence contains Z, and hence contains Q, for it must contain the multiplicative inverse of every nonzero integer. (ii) Show that the prime field of R is Q. Solution. This follows from (i). (iii) Show that the prime field of C is Q. Solution. This follows from (i). 3.28 (i) For any field F , prove that (2, F ) ∼ Aff(1, F ), where (2, F ) = denotes the stochastic group. Solution. The only properties of R used in setting up the isomorphism in Example 2.48(iv) is that it is a field. (ii) If F is a finite field with q elements, prove that | (2, F )| = q (q − 1). Solution. Now Aff(1, F ) consists of all functions f : x → ax + b, where a , b ∈ F and a = 0, there are q − 1 choices for a and q choices for b, and so | Aff(1, F )| = q (q − 1). But isomorphic groups have the same order, and so | (2, F )| = q (q − 1). (iii) Prove that (2, F3 ) ∼ S3 . = Solution. By part (ii), | (2, F3 )| = 6. By Proposition 2.135, it is isomorphic to S3 or I6 ; as it is nonabelian, (2, F3 ) ∼ S3 . = 3.29 True or false with reasons. (i) The sequence notation for x 3 − 2x + 5 is (5, −2, 0, 1, 0, · · · ). Solution. True. (ii) If R is a domain, then R [x ] is a domain. Solution. True. (iii) Q [x ] is a field. Solution. False. (iv) If k is a field, then the prime field of k [x ] is k . Solution. True. (v) If R is a domain and f (x ), g (x ) ∈ R [x ] are nonzero, then deg( f g ) = deg( f ) + deg(g ). Solution. True. ...
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