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Q11E.

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Abstract Algebra: An Introduction

Found in: Page 139

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 $ℤ_{6}$ as a subring. Show that $p (x) = 3 x^{2} + 1 \in ℤ_{6} [x]$ 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 $0 = 2 \cdot 3$ , and $3 u^{2} + 1 = 0$ . Derive a contradiction.]

It is proved that $p (x) = 3 x^{2} + 1 \in ℤ_{6} [x]$ has no roots in K.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $p (x) = 3 x^{2} + 1$ ; therefore, $3 u^{2} + 1 = 0$ . There is $1 = 3 (- u^{2})$ , and as a result, 3 is a unit in $K$ .

Moreover, $3 \cdot 2 = 0$ ; therefore, 3 will be a zero divisor in $K$ , and $\frac{ℚ [x]}{(x^{2} - 3)} ≅ ℚ (\sqrt{3})$ .

It is also observed that $x^{2} - 3$ contains a root in $ℚ (\sqrt{3})$ (according to Theorem 5.11).

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

Hence, it is proved $p (x) = 3 x^{2} + 1 \in ℤ_{6} [x]$ has no roots in K.

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Show that every polynomial of degree 1, 2, or 4 in $ℤ_{2} [x]$ has a root in $\frac{ℤ_{2} [x]}{(x^{4} + x + 1)}$ .

In Exercises 5-8, each element of the given congruence-class ring can be written in the form [ax + b] (Why?). Determine the rules for addition and multiplication

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