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Unformatted text preview: Homework #3 Solutions pp 268-271: 12, 14, 18, 42, 44 pp 268-271: 6, 30, 32, 36, 52, 56 p 268, #6 We will find the maximal ideals in the general case of Z n only. The ideals of Z n are, first of all, additive subgroups of Z n . These we know to all have the form h d i where d divides n . But, as we know, the set h d i is the ideal generated by d . So we have just proven that The ideals in Z n are precisely the sets of the form h d i where d divides n. Since we are interested in maximal ideals, and this concept is defined in terms of containment of ideals in one another, we now need to determine when we can have h d 1 i ⊂ h d 2 i . This is the case if and only if d 1 ∈ h d 2 i , which is true if and only if there is an element a ∈ Z so that ad 2 = d 1 , i.e. if and only if d 2 divides d 1 . We are now ready to prove the main result: an ideal I in Z n is maximal if and only if I = h p i where p is a prime dividing n . If I has this form and J is another ideal in Z n with I ⊂ J then J = h d i for some d dividing n . By our comments above this means that d divides p , i.e. d = 1 or d = p , which means that J = Z n or J = I , proving that I is maximal. For the converse, suppose I = h d i ( d dividing n ) is maximal but d is not prime. Then d = kl with d > k, l > 1. But then I $ h k i $ Z n . The first inequality follows from the fact that k < d implies k 6∈ I . The second follows from the fact that k is a divisor of n but is not 1, therefore is not a unit in Z n and so 1 6∈ h k i . But this string of inequalities implies that I is not maximal, a contradiction. Therefore d must be prime, and we are finished. p 269, #12 As usual, we use the two step ideal test. It is clear that AB is nonempty since ∈ A, B so that 0 = 0 · ∈ AB . Let x, y ∈ AB . Then x = a 1 b 1 + a 2 b 2 + ··· a n b n and y = c 1 d 1 + c 2 d 2 + ··· + c m d m for some a i , c i ∈ A , b i , d i ∈ B and m, n ∈ Z + . Then x- y = a 1 b 1 + a 2...
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- Spring '07
- Prime number, Divisor, Integral domain, Commutative ring, maximal ideal