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16

Expert-verifiedFound in: Page 149

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 $\mathrm{\mathbb{Z}}$ , 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 $\mathrm{\mathbb{Z}}$ .**

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

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

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 $\mathrm{\mathbb{Z}}$ .

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

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

We can write $2=\left(-1\right)\xb74+\left(1\right)\xb76$ .

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

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 $\mathrm{\mathbb{Z}}$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 $3=\left(-1\right)\xb76+\left(1\right)\xb79+0\xb715$ .

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

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