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Unformatted text preview: Suggested solutions to selected exercises in sections 26, 27 and 36 1. Section 26 Exercise 20: Let R be a commutative ring with unity of prime characteristic p . Show that the map φ p : R → R given by φ p ( a ) = a p is a homomorphism. Solution: We have to show the following: (1) φ p ( a + b ) = φ p ( a ) + φ p ( b ) for all a, b ∈ R . (2) φ p ( ab ) = φ p ( a ) φ p ( b ) for all a, b ∈ R . Ad (1): φ p ( a + b ) = ( a + b ) p = p summationdisplay i =0 parenleftbigg p i parenrightbigg a p i b i = parenleftbigg p parenrightbigg a p + parenleftbigg p p parenrightbigg b p = a p + b p = φ p ( a )+ φ p ( b ) because ( p i ) ≡ 0 mod p for < i < p . (Check that!) Ad (2): φ p ( ab ) = ( ab ) p = a p b p = φ p ( a ) φ p ( b ) where we used that R is commutative. Exercise 30: An element a of a ring R is called nilpotent if a n = 0 for some n ∈ N . Show that the collection of all nilpotent elements in a commutative ring R is an ideal, called the nilradical of R . Solution: Let N = { a ∈ R  a is nilpotent } . We have to show two things: (1) ( N, +) is a subgroup of ( R, +) . (2) ra ∈ N for all a ∈ N , r ∈ R (Note that since R is commutative we do not have to check that ar ∈ N as well). Ad (1): (i) N is closed under addition: Let a and b be nilpotent elements of R , then there exist n, m ∈ N such that a n = 0 , b m = 0 . Set l = n + m − 1 . Then l ∈ N and ( a + b ) l = l summationdisplay i =0 parenleftbigg l i parenrightbigg a i b l i = n 1 summationdisplay i =0 parenleftbigg l i parenrightbigg a i b l i + l summationdisplay i = n parenleftbigg l i parenrightbigg a i b l i = 0 where the first sum is zero because of l − i ≥ m (so b l i = 0 ) and the second sum is zero because of i ≥ n (so a i = 0 in this case)....
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This note was uploaded on 01/26/2010 for the course MATHM 413 taught by Professor Michaeljolly during the Spring '08 term at Indiana.
 Spring '08
 MichaelJolly

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