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### Abstract Algebra: An Introduction

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

# (a): Show that $\left(4,6\right)=\left(2\right)$ in $\mathrm{ℤ}$ , where $\left(4,6\right)$ is the ideal generated by 4 and 6 and $\left(2\right)$ is the principal ideal generated by 2.( b): Show that $\left(6,9,15\right)=\left(3\right)$ in $\mathrm{ℤ}$ .

a. It can be proved that $\left(4,6\right)=\left(2\right)$ in $\mathrm{ℤ}$ .

b. It can be proved $\left(6,9,15\right)=\left(3\right)$ in $\mathrm{ℤ}$ .

See the step by step solution

## Lemma used

Let R be a commutative ring.

Let $\left({c}_{1},{c}_{2},....,{c}_{n}\right)$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" $\left({c}_{1},{c}_{2},.....{c}_{n}\right)\subseteq J$

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

It is clear that $4,6\in \left(2\right)$ in $\mathrm{ℤ}$ .

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

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

We can write $2=\left(-1\right)·4+\left(1\right)·6$ .

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

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

Conclusion

$\left(4,6\right)\subseteq \left(2\right)\phantom{\rule{0ex}{0ex}}\left(2\right)\subseteq \left(4,6\right)$

Hence, $\left(2\right)=\left(4,6\right)$.

## Prove  6,9,15⊆3

Let R be a commutative ring.

Let $\left({c}_{1},{c}_{2},...,{c}_{n}\right)$ 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 $\left({c}_{1},{c}_{2},...,{c}_{n}\right)\subseteq J$.

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

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

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

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

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

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

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

Conclusion

$\left(6,9,15\right)\subseteq \left(3\right)\phantom{\rule{0ex}{0ex}}\left(3\right)\subseteq \left(6,9,15\right)$

Hence, $3=\left(6,9,15\right)$ .