Question

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

Short answer

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

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

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

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

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

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

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

Conclusion

$$\begin{gathered} \left(4,6\right)\subseteq \left(2\right) \\ \left(2\right)\subseteq \left(4,6\right) \end{gathered}$$

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 $\mathrm{\mathbb{Z}}$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 $3=\left(-1\right)·6+\left(1\right)·9+0·15$.

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

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

Conclusion

$$\begin{gathered} \left(6,9,15\right)\subseteq \left(3\right) \\ \left(3\right)\subseteq \left(6,9,15\right) \end{gathered}$$

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