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## Introduction to Abstract Algebra II

by: Melvina Keeling

15

0

2

# Introduction to Abstract Algebra II MATH 567

Marketplace > Colorado State University > Mathematics (M) > MATH 567 > Introduction to Abstract Algebra II
Melvina Keeling
CSU
GPA 3.78

Justin Sawon

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COURSE
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Justin Sawon
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KARMA
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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Melvina Keeling on Monday September 21, 2015. The Class Notes belongs to MATH 567 at Colorado State University taught by Justin Sawon in Fall. Since its upload, it has received 15 views. For similar materials see /class/210074/math-567-colorado-state-university in Mathematics (M) at Colorado State University.

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Date Created: 09/21/15
MATH567 Abstract Algebra II Solutions to Sec 136 exercises 14717 14 Suppose 10171027 710k are the only primes dividing the values P017 71 1727 Since Pz is monic of degree at least one7 there exists a positive integer N such that a PN is non zero Now if PW zm am1m 1 alz 10 then PN 0101102 39 39 3910 N ap1p2pk cgtm am71N ap1p2 pkxm 1 a1N ap1p2 9196 ao N Lm1Nm 1L a1N a0 ap1p2pkp a 0101192 ka96 for some polynomial p E Therefore Q06 a lPUV ap1p2 196 1 Hum p1p06 is an element of ZM and moreover Q01 1 191192 pkpn E 1m0dp1p2 sz for 71 17 27 ln particular7 Q01 is never divisible by 10171027 7 or pk For 71 suf ciently large to ensure that Q01 gt 17 Q01 will have a prime factor different to 10171027 710k Therefore PN ap1p2 mm aQn will also have a prime factor different to p17p277pk This contradiction shows that there must be in nitely many distinct prime divisors of P17 P27 P37 15 Suppose that pfa Then ltIgtma E ltIgtm0 modp7 since ltIgtm has integer coef cients But ltIgtm0 is the constant term of ltIgtmz up to i17 it is the product of the primitive mth roots of unity These all he on the unit circle in C7 and hence so does their product Since ltIgtm0 is an integer7 it must be i1 So if ltIgtma E 0modp7 then p cannot divide a7 and a is relatively prime to p Substituting z a into zm 7 1 H ltIgtdz d m and using ltIgtma E 0rnodp shows that am 7 1 E 0rnodp Therefore the order of a divides m Suppose that ad 7 1 E 0rnodp for some divisor d of m with d lt m Then ad 7 1 H ltIgtd a E 0rnodp d ld so ltIgtd a E 0rnodp for some d S d lt m Therefore z E 1 would have a as a repeated root mod p since it is a root of both the factor ltIgtm and the factor dx But this is impossible recall that DEW E 1 mzmil has only zero as a root so z E 1 is separable over 1917 meaning it has no repeated roots Therefore the order of a in ZpZgtlt is precisely m 16 Suppose that p does not divide m By exercise 15 since ltIgtma E 0rnodp a must be relatively prirne to p and the order of a in ZpZgtlt must be precisely m The order of the group ZpZgtlt is p71 so the order of the element a must divide p71 ie 77111071 Therefore p E 1rnodm 17 Firstly because ltIgtm E ZM is a rnonic polynomial of degree at least one then exercise 14 shows that there are in nitely many prirne divisors of the integers lt1gtm1lt1gtm2lt1gtm3 and hence in nitely many odd prime divisors By exercise 16 each ofthese odd prime divisors must either divide m or be congruent to 1 mod Since m can only have nitely many prirne divisors there must be in nitely many prirnes congruent to 1 mod

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