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103b_06w_2s - F ⊆ E(b Probably the simplest basis to ²nd...

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Math 103B Second Hour Exam Solutions 8 March 2006 1. 1 - 2 x since (1 + 2 x )(1 - 2 x ) = 1 - 4 x 2 = 1 in Z 4 [ x ]. 2. Five, since by looking at 0 1 0 0 0 1 0 1 1 1 1 1 0 0 1 1 we see 5 differences. Calling the desired word x , we have 5 = d ( u,v ) d ( u,x ) + d ( x,v ) 2 + 2 = 4, a contradiction. Hence there is no such word. 3. The three zeros of x 3 - 2 are a = 2 1 / 3 , b = and c = 2 where ω = e 2 πi/ 3 . There are many possibilities. Here are three. Adjoin any two of them to Q . Adjoin one of them and ω to Q . Adjoin a + ω to Q . The first two are obviously splitting fields. The third is not so clear—but you weren’t asked to prove the result. 4. (a) Since ( 2 + 5 ) 2 = 7 + 2 10, it follows that 10 E and so
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Unformatted text preview: F ⊆ E . (b) Probably the simplest basis to ²nd is 1, √ 2 + √ 5; however, there are others such as 1, √ 2. (c) One possibility is 1, √ 2, √ 5, √ 10. 5. (a) | F | must be a power o± p and all powers p k with k a positive integer are possible. (b) I± [ K : F ] = n , then K is a vector space over F o± dimension n and so | K | = | F | n . (c) Suppose | F | = p k and | K | = | F | n = p kn . By the uniqueness o± ²nite ²elds (Theorem 22.1), we have F = GF( p k ) and K = GF( p kn ). By the sub²eld theorem (Theorem 22.3), F is a sub²eld o± K ....
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