1.2-1.5 - Math 431 - Assignment 1 Solutions Questions on...

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Unformatted text preview: Math 431 - Assignment 1 Solutions Questions on fields 1. Let a be an element in F , a 6 = 0. Then there exist an element b ∈ F such that a · b = 1. We want to prove that b is unique. Indeed, assume that b is in F such that a · b = 1. Hence a · b = a · b . From the cancellation law, which applies because a 6 = 0, it follows that b = b . Thus the multiplicative inverse of a is unique. 2. Let a,b ∈ F such that a · b = 0. If a = 0, we are done, so assume that a 6 = 0 and let a- 1 denote the (unique) multiplicative inverse of a . Then we have a- 1 · ( a · b ) = a- 1 · , ( a- 1 · a ) · b = 0 , 1 · b = 0 , b = 0 . That is, b = 0, and our solution is complete. 5. We will show that the characteristic of the field Z p = { , 1 ,..., p- 1 } is p . Indeed, since p · 1 = 0 the set { n ∈ Z | n ≥ 1 and n · 1 = } is non-empty, so the characteristic of Z p is definitely greater than zero and must be given by the minimum of the above set. But { n ∈ Z | n ≥ 1 and n · 1 = } = { n ∈ Z | n ≥ 1 and n ≡ modp } = { kp | k = 1 , 2 , 3 ,... } , so its minimum is clearly p , and our proof is complete. 6. Let char ( F ) = p 6 = 0 and note that since 1 F 6 = 0 F , we must have p > 1. Moreover, observe that since the characteristic is non-zero, it must equal to the minimum positive integer k such that k · 1 F = 0 F . I.e. if 1 ≤ k < p , then k · 1 F 6 = 0 F . We want to prove that p is prime. Indeed, assume to the contrary that p is a composite number. Then there exist integers n,m > 1 such that p = mn ....
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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1.2-1.5 - Math 431 - Assignment 1 Solutions Questions on...

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