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
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 $I=(t^m)$.