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Abstract Algebra: An Introduction

Found in: Page 149

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

(a): Show that $(4, 6) = (2)$ in $ℤ$ , where $(4, 6)$ is the ideal generated by 4 and 6 and $(2)$ is the principal ideal generated by 2.
( b): Show that $(6, 9, 15) = (3)$ in $ℤ$ .

a. It can be proved that $(4, 6) = (2)$ in $ℤ$ .

b. It can be proved $(6, 9, 15) = (3)$ in $ℤ$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Lemma used

Let R be a commutative ring.

Let $(c_{1}, c_{2}, . . . ., c_{n})$ denote the ideal generated by $c_{1}, c_{2}, . . . . . ., c_{n}$ in $R$

If $J$ is an ideal of $R$ and $c_{1}, c_{2}, . . . . ., c_{n} \in J$ , then localid="1648707689871" $(c_{1}, c_{2}, . . . . . c_{n}) \subseteq J$

Proving $(4, 6) \subseteq (2)$

It is clear that $4, 6 \in (2)$ in $ℤ$ .

So, we can conclude $(4, 6) \subseteq (2)$ .

Proving $(2) \subseteq (4, 6)$

We can write $2 = (- 1) \cdot 4 + (1) \cdot 6$ .

This implies $2 \in (4, 6)$ .

Hence, $(2) \subseteq (4, 6)$ .

Conclusion

$(4, 6) \subseteq (2) (2) \subseteq (4, 6)$

Hence, $(2) = (4, 6)$ .

Prove 6,9,15⊆3

Let R be a commutative ring.

Let $(c_{1}, c_{2}, . . ., c_{n})$ denote the ideal generated by $c_{1}, c_{2}, . . . c_{n}$ in $R$

If $J$ is an ideal of $R$ and $c_{1}, c_{2}, . . ., c_{n} \in J$ , then $(c_{1}, c_{2}, . . ., c_{n}) \subseteq J$ .

Proving $(6, 9, 15) \subseteq (3)$

It is clear $4, 6 \in (2)$ in $ℤ$ since 6, 9, 15 are multiples of 3.

So, we can conclude $(6, 9, 15) \subseteq (3)$ .

Proving $(3) \subseteq (6, 9, 15)$

We can write $3 = (- 1) \cdot 6 + (1) \cdot 9 + 0 \cdot 15$ .

This implies $3 \in (6, 9, 15)$ .

Hence, $(3) \subseteq (6, 9, 15)$ .

Conclusion

$(6, 9, 15) \subseteq (3) (3) \subseteq (6, 9, 15)$

Hence, $3 = (6, 9, 15)$ .

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