Is it possible to classify all cyclic $\mathbb{Z}/4$-algebras, i.e. the regular quotients of $\mathbb{Z}/4 [X]$? A typical example is $\mathbb{Z}/4 [X] / \langle X^n , 2 X^k \rangle$. For my purposes it is not necessarily important to decide which quotients are isomorphic (this is already unclear to me for $\mathbb{Z}/2[X]$). So an almost identical question is the following:


Is it possible to classify all ideals of $\mathbb{Z}/4 [X]$? Can we bound the number of generators? Or is this classification not feasible?


It is known that every $\mathbb{Z}/4$-module (also without any finiteness assumptions) is a direct sum of copies of $\mathbb{Z}/2$ and $\mathbb{Z}/4$. This should be useful.


Partial results are also helpful for me. Instead of $\mathbb{Z}/4$ it should be possible to take $R/p^2$ for any principal ideal domain $R$ and a prime element $p \in R$.


*Edit.* Here is how I understand YCor's proof that every ideal of $R/p^2 [X]$ can be generated by two elements; the proof works for every commutative ring $A$ with an element $p \in A$ such that $p^2=0$ and $A/pA$ is a principal ideal ring. Let $I \subseteq A$ be an ideal. Consider the $A$-module $I \cap p A$. Since it is killed by $p$, we may view it as an $A/pA$-module. But $A/pA$ is a principal ideal ring, and $pA$ is a cyclic $A/pA$-module. Hence, $I \cap pA$ is a cyclic $A/pA$-module, i.e. a cyclic $A$-module. On the other hand, $I/(I \cap pA) \cong (I + pA)/pA$ is an $A$-submodule of $A/pA$, and hence cyclic too. Hence, $I$ is generated by a generator of $I \cap pA$ together with any preimage of a generator of $(I+pA)/pA$.



Classification of the quotients of the ring Z/4 [X]
Let $I$ be an ideal in $(\mathbf{Z}/4\mathbf{Z})[X]$. Let $J\_I$ be its projection to $(\mathbf{Z}/2\mathbf{Z})[X]$, and write its intersection with $2(\mathbf{Z}/4\mathbf{Z})[X]$ as $2K\_I$, for some ideal $K\_I$ of $(\mathbf{Z}/2\mathbf{Z})[X]$ (precisely, first define $K'\_I=\{P \in (\mathbf{Z}/4\mathbf{Z})[X]:2P\in I\}$ and then define $K\_I$ as the projection of $K'\_I$ modulo 2).


It's immediate that $J\_I\subset K\_I$. 


Conversely for a given pair of ideals $(J,K)$ of $(\mathbf{Z}/2\mathbf{Z})[X]$ such that $J\subset K$, we can enumerate the $I$ such that $(J\_I,K\_I)=(J,K)$. First denote by $f\mapsto\bar{f}$ the lift $\mathbf{Z}/2\mathbf{Z}\to \mathbf{Z}/4\mathbf{Z}$ mapping $0\mapsto 0$, $1\mapsto 1$, and extend it coefficient-wise to polynomials.


Namely, let $f$ be the (unique) generator of $J$ and $g$ the generator of $K$ (so $g$ divides $f$). If $f=0$ (i.e., $J=0$), clearly $I$ is unique (and equal to the principal ideal $2K'\_I$).


Now suppose $f\neq 0$ (so $g\neq 0$). Then for $h\in (\mathbf{Z}/2\mathbf{Z})[X]$ all $I\_h=\langle \bar{f}+2\bar{h},2\bar{g}\rangle$ have $(J\_{I\_h},K\_{I\_h})=(J,K)$, and $I\_h=I\_{k}$ if and only $h-k\in\langle g\rangle$. Thus these ideals are parameterized by $(\mathbf{Z}/2\mathbf{Z})[X]/\langle g\rangle$, which is finite, and exhaust those ideals $I$ such that $(J\_I,K\_I)=(J,K)$. (In this case, $J\_I=K\_I$ iff $I$ is principal.)


I think this is a full description.


Example: the ideals $\langle X\rangle$ and $\langle X+2\rangle$ both have $(J,K)=(\langle X\rangle,\langle X\rangle)$ and these are the only ones.




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I should add a word on the structure of quotients. When $J\_I\neq 0$, the quotient is finite (i.e., $I$ has finite index). The underlying additive group structure is isomorphic to $(\mathbf{Z}/4\mathbf{Z})^k\times (\mathbf{Z}/2\mathbf{Z})^\ell$ where $k$ is the degree of $f$ and $\ell+k$ is the degree of $g$ (some effort would yield something better, namely the decomposition of the finite length module $(\mathbf{Z}/4\mathbf{Z})[X]/I$ as sum of indecomposable modules). If $J\_I=0$ and $K\_I\neq 0$, the quotient is additely isomorphic to $(\mathbf{Z}/4\mathbf{Z})^k\times (\mathbf{Z}/2\mathbf{Z})^{(\aleph\_0)}$ where $k$ is the degree of $g$, and of course if $I=0$ it is additively isomorphic to $(\mathbf{Z}/4\mathbf{Z})^{(\aleph\_0)}$.


