1. We proved thta the ideal < x + 1 > is prime in Z[x] by definition. In this problem, show this ideal is prime, but not maximal, in Z[x] using the FHT as follows: (a) Define the substitution homomorphism μ-1: Z[x] → Z by μ-1 (p(x)) = p(-1) - you can assume this is a homomorphism. Show it is onto (b) Prove keru_1 =< x + 1 > (c) Then apply the FHT to answer the question (d) Now use the FHT again, to explain why is a maximl ideal of F[x], where F is any field

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter6: More On Rings
Section6.4: Maximal Ideals (optional)
Problem 26E: . a. Let, and . Show that and are only ideals of and hence is a maximal ideal. b. Show...
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1. We proved thta the ideal < x + 1 > is prime in Z[x] by definition. In
this problem, show this ideal is prime, but not maximal, in Z[x] using the
FHT as follows:
(a) Define the substitution homomorphism μ-1: Z[x] → Z by μ-1 (p(x)) =
p(-1) - you can assume this is a homomorphism. Show it is onto
(b) Prove keru_1 =< x + 1 >
(c) Then apply the FHT to answer the question
(d) Now use the FHT again, to explain why < x + 1 > is a maximl ideal
of F[x],where F is any field
Transcribed Image Text:1. We proved thta the ideal < x + 1 > is prime in Z[x] by definition. In this problem, show this ideal is prime, but not maximal, in Z[x] using the FHT as follows: (a) Define the substitution homomorphism μ-1: Z[x] → Z by μ-1 (p(x)) = p(-1) - you can assume this is a homomorphism. Show it is onto (b) Prove keru_1 =< x + 1 > (c) Then apply the FHT to answer the question (d) Now use the FHT again, to explain why < x + 1 > is a maximl ideal of F[x],where F is any field
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