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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 $\langle b\rangle$.

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First observe that if $b=0$, then $\langle b\rangle=\langle ab\rangle=\langle 0\rangle$. Now, suppose $a$ is a unit. Then $a^{-1}(ab)=b\in\langle ab\rangle$, while $ab\in\langle b\rangle$, so $\langle ab\rangle=\langle b\rangle$. Does that help? –  Tabes Bridges Oct 27 '12 at 5:22

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Hint $\rm\ \ (ab)\supset (b)\:\Rightarrow\:ab\:|\:b\:\Rightarrow\: a\:|\:1,\:$ since $\rm\:b\ne 0\:\Rightarrow\:b\,$ cancellable (in a domain).

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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 $xa=1$.

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