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Let $R$ is a commutative ring have identity element, and $R$ have a unique maximal ideal. Let $a,b \in R$ such that $\langle a \rangle = \langle b \rangle$. Show that $$a=bu$$ for some $u\in R^*$.

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What've you done so far? This kind of rings are called local, and they're characaterised by the maximal ideal being the set of all non-units, so...? – DonAntonio Nov 19 '12 at 2:54

If $a=0$, then $b=0$ and there's nothing to prove, so assume $a,b\neq 0$. Since $a\in(b)$, you can write $a=br$ and similarly $b=as$ for $r,s\in R$. Thus $a=br=ars$, so $(rs-1)a=0$. If $r$ is not a unit, then neither is $rs$. So $rs$ lies in the unique maximal ideal of $R$. This means that $rs-1$ is not in the unique maximal ideal, because then we would have that $1=rs-(rs-1)$ is in it. So $rs-1$ is a unit, which implies $a=0$, contrary to assumption.

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I feel confused. Are you assume $r\notin R^*$?, then $rs\ne 1$ – Muniain Nov 19 '12 at 5:52
Can you tell me why condition $R$ have a unique maximal ideal can't loss?. Thank you – Muniain Nov 19 '12 at 5:58
or give me a counterexample if we loss condition "$R$ have a unique maximal ideal" – Muniain Nov 19 '12 at 6:00
Dear @Muniain, Yes, I am assuming $r$ is not a unit to derive a contradiction. Regarding the condition that $R$ has a unique maximal ideal, it can be weakened as in Bill Dubuque's answer below. Presumably there are counterexamples in the paper referenced in that answer. – Keenan Kidwell Nov 19 '12 at 11:27
I just checked, and indeed there are counterexamples on the first two pages of the paper referenced by Bill. – Keenan Kidwell Nov 19 '12 at 11:50

More generally, let $R$ be a ring where zero-divisors lie in every max ideal. It's trivial if $a = 0$. Else $b = sa,\ a = rb = rsa,\:$ so $\,a(1-rs) = 0.\:$ So $1\!-\!rs$ is $\,0\,$ or a zero-divisor, so is in every max ideal, hence $rs$ is a unit, lying in no max ideal $M$ (else both $\:1\!-\!rs,\ rs \in M\:$ so their sum $\:1\in M)$.

Remark $ $ For further generalizations see D. Anderson et al. When are Associates Unit Multiples?

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