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Abstract Algebra: An Introduction
Found in: Page 139
Abstract Algebra: An Introduction

Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

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Short Answer

Let K be the ring that contains 6 as a subring. Show that px=3x2+16x has no roots in K. Thus, Corollary 5.12 may be false if F is not a field. [Hint: If u were a root, then 0=2·3 , and 3u2+1=0. Derive a contradiction.]

It is proved that px=3x2+16x has no roots in K.

See the step by step solution

Step by Step Solution

Statement of Exercise 3.2.17

Exercise 3.2.17 states that if u is a unit in a ring R with an identity, then u will not be a zero divisor.

Show that px=3x2+1∈ℤ6x has no roots in K

Assume that u is a root of px=3x2+1; therefore, 3u2+1=0. There is 1=3-u2, and as a result, 3 is a unit in K.

Moreover, 3·2=0; therefore, 3 will be a zero divisor in K, and xx2-33.

It is also observed that x2-3 contains a root in 3 (according to Theorem 5.11).

This contradicts the result that units in a ring with unity are not zero divisors.

Hence, it is proved px=3x2+16x has no roots in K.

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