5
$\begingroup$

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 $(I+pA)/pA$.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Let $A$ be the ring. First classify the ideals contains in $2A$: as an $A$-module, $2A$ can be viewed as $(Z/2Z)[x]$-module; it's a PID and submodules are cyclic. Then given such an ideal $I$ you have to classify the ideals $J$ such that $J\cap 2A=I$. For $I=0$ this forces $2J=0$, so $J=0$. For $I=(2f)A\neq 0$ there's a little more work but it sounds easy. But since $A/2A$ is principal too, so it's immediate (without further description) that every ideal is generated by $\le 2$ elements. $\endgroup$
    – YCor
    Commented Apr 28, 2017 at 22:14
  • $\begingroup$ @YCor: Thanks a lot for your answer. It remains to give a criterion when two ideals are equal in terms of the two generators. If $I,J$ are ideals such that $I \cap 2A = J \cap 2A$ and $I + 2A = J + 2A$, does this imply $I=J$? I cannot prove this right now. $\endgroup$
    – Martin Brandenburg
    Commented May 12, 2017 at 8:51
  • $\begingroup$ You can define the Prüfer rank $p(M)$ of a module as the supremum of the minimal number of generators of finitely generated submodules. Then $p$ is sub-additive under extensions. In this case (as in the edit), you have an extension of two modules of Prüfer rank 1 (one isomorphic as $R$-module to $R/P$ for some PID $P$, one isomorphic to a quotient of $R/P$), hence of Prüfer rank $\le 2$. $\endgroup$
    – YCor
    Commented May 14, 2017 at 8:58

1 Answer 1

Reset to default
6
$\begingroup$

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.

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 $(\mathbf{Z}/4\mathbf{Z})^{(\aleph_0)}$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Can you explain why $K_{I_h} \subseteq K$ holds? (The rest is also unclear, but I have to think more about it.) $\endgroup$
    – Martin Brandenburg
    Commented May 13, 2017 at 8:44
  • $\begingroup$ @HeinrichD I'll revert "the generator" which you changed to "a generator". Indeed in polynomials over the field on 2 elements the generator is unique. $\endgroup$
    – YCor
    Commented May 13, 2017 at 9:17
  • $\begingroup$ @MartinBrandenburg Suppose that in polynomials mod 4, we have $A(f+2h)+B(2g)=2P$. Modulo 2 this yields $Af=0$ and since $f$ is nonzero mod 2 and polynomials mod 2 is a domain, it follows that $A=0$ mod 2, say $A=2C$. Hence $2Cf+2Bg=2P$ (mod 4), so, mod 2, $Cf+Bg=P$. Since $g$ divides $f$, it follows that $g$ divides $P$. This precisely shows that $K_{I_k}\subset K$. $\endgroup$
    – YCor
    Commented May 13, 2017 at 9:22
  • $\begingroup$ Thanks a lot. I think this is indeed a complete description. And I think it not only holds for $R/p^2 [X]$ (where $R$ is a PID and $p \in R$ is prime), but also for commutative rings $A$ with an element $p \in A$ such that $\ker(p)=pA$ and $A/pA$ is a PID. $\endgroup$
    – Martin Brandenburg
    Commented May 13, 2017 at 21:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .