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Suppose that the ring $A$ has only a maximal ideal $\mathfrak m$ and that $\mathfrak m$ is principal (denote its generator with $t$). Assume also that $\bigcap_n \mathfrak m^n=0$. I must show that every non-zero ideal $I$ is generated by $t^m$, for a natural number $m$.

I tried to quotient or to construct some map but I couldn't think of anything useful. The only thing that I noticed is that $\bigcap_n I^n=0$ and that exist two positive integers $N>M$ such that $\mathfrak m^N\subseteq I\subseteq\mathfrak m^M$. I know I must use that $\mathfrak m$ is principal, but I don't understand how. Can you only give a hint? Thanks

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Note that since $\mathfrak{m}$ is the only maximal ideal, the complement consists of units. In particular, if $I\neq (1)$ then any element of $I$ can be written as $f=t^Ng$ where $t$ does not divide $g$, and so $g$ is a unit. Take $m$ minimal such that $t^m\in I$ and it follows that $I=(t^m)$.

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  • $\begingroup$ Ok, I have just a question: could we use the same argument if the ideal $\mathfrak m$ was not unique, and $I$ was contained in $\mathfrak m $? I mean, the fact that your $g$ is actually invertible doesn't seem to matter in this context $\endgroup$
    – Dr. Scotti
    Nov 20, 2021 at 16:14
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    $\begingroup$ It matters. We use that since $t^Ng$ is in $I$ and $g$ is invertible we have $t^N\in I$. $\endgroup$
    – Smn
    Nov 20, 2021 at 16:22

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