prime ideals, ring

Question

If $ab \in 6\mathbb{Z}$, it does not follow that $a$ or $b$ is in $6\mathbb{Z}$.

For example, $2 \cdot 3 = 6 \in 6\mathbb{Z}$, but $2$ nor $3$ is in $6\mathbb{Z}$.

Can someone explain why? Why is $2$ or $3$ not contained in the ideal $6\mathbb{Z}$, as it divides both of them?

Answer 1

$$ 6\mathbb{Z}=\{...,6\cdot\pm2,6\cdot\pm1,6\cdot\pm0,6\cdot\pm1,6\cdot\pm2,...\}=\{...,-12,-6,0,6,12,...\} $$

so $2,3\not\in6\mathbb{Z}$.

Answer 2

It's just a definition. $6\mathbb{Z}=\{6n \colon n \in \mathbb{Z}\}$. What integer $n$ gives $6n=2$?