find all the maximal ideal of $\mathbb{Z} /p^n\mathbb{Z}?$ find all the maximal ideal of $\mathbb{Z} /p^n\mathbb{Z}?$
I know that  $p\mathbb{Z}$ is a maximal ideal in $\mathbb{Z}$  whenever  $p$ is prime
Here  $\mathbb{Z} /p^n\mathbb{Z}\cong \mathbb{Z}_{p^n}$
Edit: From Arthur comment, I take  $\mathbb{Z}_{2^3}=\mathbb{Z}_8$  here  divisor of  $8$ are $1,2$ and $4$  by analysis  $(2)$ is the only maximal ideal. Similarly in $\mathbb{Z_{2^2}} $, $(2)$ is the only maximal ideal.
My question : How to find all the maximal ideal of $\mathbb{Z} /p^n\mathbb{Z}?$
 A: Question: "My question : How to find all the maximal ideal of $\mathbb{Z}/p^n\mathbb{Z}$?"
Answer: If $I:=(p)$ with $p$ a non-zero prime, it follows $I^l=(p^l)$. If $I^l \subseteq J$ with $J$ a prime ideal, there is a maximal ideal $J \subseteq I'$, hence there is an inclusion $I^l \subseteq I'$ of maximal ideals. It follows, since $I'$ is a prime ideal that $p\in I'$, hence $I'=I=(p)$. Hence the only maximal (and prime) ideal in $\mathbb{Z}/p^n\mathbb{Z}$  is $(p)$.
Note: This question points to a general fact: If $\mathfrak{m} \subseteq A$ is a maximal ideal and if $\mathfrak{m}^l \subseteq \mathfrak{p}$ is a prime ideal containing $\mathfrak{m}^l$, it follows $\mathfrak{p}=\mathfrak{m}$: If $x\in \mathfrak{m}$, it follows $x^l \in \mathfrak{m}^l \subseteq \mathfrak{p}$ and since $\mathfrak{p}$ is a prime ideal it follows by induction that $x\in \mathfrak{p}$ hence $\mathfrak{p}=\mathfrak{m}$.
Hence the ring $A/\mathfrak{m}^l$ is an Artinian ring with maximal ideal $\mathfrak{m}/\mathfrak{m}^l$.
