By Paul T. Bateman
I first taught an summary algebra path in 1968. utilizing Hcrstein's issues in Algebra. it is tough to enhance on his booklet; the topic could have turn into broader, with purposes to computing and different parts, yet themes comprises the middle of any path. regrettably, the topic hasn't turn into any more uncomplicated, so scholars assembly summary algebra nonetheless fight to profit the hot ideas, in particular because they're most likely nonetheless studying easy methods to write their very own proofs.This "study consultant" is meant to aid scholars who're commencing to know about summary algebra. rather than simply increasing the fabric that's already written down in our textbook, i made a decision to aim to educate through instance, through writing out ideas to difficulties. i have attempted to decide on difficulties that might be instructive, and in various instances i have incorporated reviews to aid the reader see what's quite occurring. in fact, this examine advisor is not an alternative to a superb instructor, or for the opportunity to interact with different scholars on a few tough problems.Finally. i want to gratefully recognize the help of Northern Illinois collage whereas penning this examine consultant. As a part of the popularity as a "Presidential instructing Professor," i used to be given depart in Spring 2000 to paintings on initiatives on the topic of instructing.
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Additional resources for Abstract Algebra: A Study Guide for Beginners
9. Give an example to show that the set of all zero divisors of a ring need not be an ideal of the ring. 10. Let I be the subset of Z[x] consisting of all polynomials with even coefficients. Prove that I is a prime ideal; prove that I is not maximal. 11. Let R be any commutative ring with identity 1. (a) Show that if e is an idempotent element of R, then 1−e is also idempotent. (b) Show that if e is idempotent, then R ∼ = Re ⊕ R(1 − e). 12. Let R be the ring Z2 [x]/ x3 + 1 . (a) Find all ideals of R.
Then we will prove the converse, which states that if gcd(2m − 1, 2n − 1) = 1, then gcd(m, n) = 1, To prove the converse, we will use a proof by contradiction, assuming that gcd(m, n) = 1 and showing that this forces gcd(2m − 1, 2n − 1) = 1. Before beginning the proof, we recall that the following identity holds for all values of x: xk − 1 = (x − 1)(xk−1 + xk−2 + · · · + x + 1). Solution: If gcd(m, n) = 1, then there exist a, b ∈ Z with am + bn = 1. Substituting x = 2m and k = a in the identity given above shows that 2m − 1 CHAPTER 1 SOLUTIONS 41 is a factor of 2am − 1, say 2am − 1 = (2m − 1)(s), for some s ∈ Z.
Are these all different? Suppose that i [−4]j = m [−4]n , for some 0 ≤ i < 4 and 0 ≤ j < 6 and 0 ≤ m < 4 and 0 ≤ n < 6. Then i−m = [−4]n−j , and since the only power of  that is equal to a power of [−4] is  (as shown by our computations), this forces i = m and n = j. 46 CHAPTER 1 SOLUTIONS We conclude that since there are 24 elements of the form i [−4]j , every element in Z35 must be of this form. Finally, (i [−4]j )12 = (4 )3i ([−4]6 )2j = , so no element of Z35 has multiplicative order 24, showing that Z35 is not cyclic.