Why is $6\mathbb{Z} + 9\mathbb{Z}$ a principal ideal of $\mathbb{Z}$?

Question

Ler $R$ denote a commutative ring with identity.

For $R = \mathbb{Z}$, why is the ideal $6\mathbb{Z} + 9\mathbb{Z}$ principal?

Any help where to start would be appreciated.

Answer 1

Yes, it is a principle ideal of $\mathbb{Z}$ because it is of the form $aR$. Where a $\in \mathbb{Z}$ and $R = \mathbb{Z}$.

Hence the generator $a = 3$ is found by: $x \in 6\mathbb{Z} + 9\mathbb{Z} \\ x = 6u+9v \\ u,v \in \mathbb{Z} \\ x \in 3(2u + 3v) \subset 3 \mathbb{Z}$