AtiyahSolutions

AtiyahSolutions - Adam Allan Solutions to Atiyah Macdonald...

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Unformatted text preview: Adam Allan Solutions to Atiyah Macdonald 1 Chapter 1 : Rings and Ideals 1.1. Show that the sum of a nilpotent element and a unit is a unit. If x is nilpotent, then 1- x is a unit with inverse ∑ ∞ i =0 x i . So if u is a unit and x is nilpotent, then v = 1- (- u- 1 x ) is a unit since- u- 1 x is nilpotent. Hence, u + x = uv is a unit as well. 1.2. Let A be a ring with f = a + a 1 x + ··· + a n x n in A [ x ] . a. Show that f is a unit iff a is a unit and a 1 ,...,a n are nilpotent. If a 1 ,...,a n are nilpotent in A , then a 1 x,...,a n x n are nilpotent in A [ x ]. Since the sum of nilpotent elements is nilpotent, a 1 x + ··· + a n x n is nilpotent. So f = a + ( a 1 x + ··· + a n x n ) is a unit when a is a unit by exercise 1.1. Now suppose that f is a unit in A [ x ] and let g = b + b 1 x + ··· + b m x m satisfy fg = 1. Then a b = 1, and so a is a unit in A [ x ]. Notice that a n b m = 0, and suppose that 0 ≤ r ≤ m- 1 satisfies a r +1 n b m- r = a r n b m- r- 1 = ··· = a n b m = 0 Notice that 0 = fg = m + n X i =0 i X j =0 a j b i- j x i = m + n X i =0 c i x i where we define a j = 0 for j > n and b j = 0 for j > m . This means that each c i = 0, and so 0 = a r +1 n c m + n- r- 1 = n X j =0 a j a r +1 n b m + n- r- 1- j = a r +2 n b m- r- 1 since m + n- r- 1- j ≥ m- r for j ≤ n- 1. So by induction a m +1 n b = 0. Since b is a unit, we conclude that a n is nilpotent. This means that f- a n x n is a unit since a n x n is nilpotent and f is a unit. By induction, a 1 ,...,a n are all nilpotent. b. Show that f is nilpotent iff a ,...,a n are nilpotent. Clearly f = a + a 1 x + ... + a n x n is nilpotent if a ,...,a n are nilpotent. Assume f is nilpotent and that f m = 0 for m ∈ N . Then in particular ( a n x n ) m = 0, and so a n x n is nilpotent. Thus, f- a n x n is nilpotent. By induction, a k x k is nilpotent for all k . This means that a ,...,a n are nilpotent. c. Show that f is zero-divisor iff bf = 0 for some b 6 = 0 . If there is b 6 = 0 for which bf = 0, then f is clearly a zero-divisor. So suppose f is a zero-divisor and choose a nonzero g = b + b 1 x + ··· + b m x m of minimal degree for which fg = 0. Then in particular, a n b m = 0. Since a n g · f = 0 and a n g = a n b + ··· + a n b m- 1 x m- 1 , we conclude that a n g = 0 by minimality. Hence, a n b k = 0 for all k . Suppose that a n- r b k = a n- r +1 b k = ··· = a n b k = 0 for all k Then as in part a we obtain the equation 0 = m + n- r- 1 X j =0 a m + n- r- 1- j b j = a n- r- 1 b m 2 Again we conclude that a n- r- 1 g = 0. Hence, by induction a j b k = 0 for all j,k . Choose k so that b = b k 6 = 0. Then bf = 0 with b 6 = 0. d. Prove that f,g are primitive iff fg is primitive....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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AtiyahSolutions - Adam Allan Solutions to Atiyah Macdonald...

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