Suppose that $a$ and $b$ belong to an integral domain, $b\neq 0$, and $a$ is not a unit. Show that $\langle ab\rangle$ is a proper subset of $\langle b\rangle$.
How can I solve this problem?
Suppose that $a$ and $b$ belong to an integral domain, $b\neq 0$, and $a$ is not a unit. Show that $\langle ab\rangle$ is a proper subset of $\langle b\rangle$.
How can I solve this problem?
Since always $\langle ab\rangle \subseteq \langle b\rangle $ suppose that $\langle b\rangle =\langle ab\rangle$. Then $b=xab$ for some $x$ in your domain. Since $b \neq 0$ you get $xa=1$.
Hint $\rm\ \ (ab)\supset (b)\:\Rightarrow\:ab\:|\:b\:\Rightarrow\: a\:|\:1,\:$ since $\rm\:b\ne 0\:\Rightarrow\:b\,$ cancellable (in a domain).