How can one find all ideals of $ (\mathbb Z/6\mathbb Z)[x]/(x^3)$?

Question

I am working on Homework Problem which asks me to find all nontrivial ideals of the ring $ R = S[x]/J$ where`$ S = \mathbb Z/6\mathbb Z$ and $ J = (x^3)$. I could construct a homomorphism from $ R = S[x]/J$ to $ [(\mathbb Z/2\mathbb Z)[x]/J]\times [(\mathbb Z/3\mathbb Z)[x]/J] $. However, I cannot go further...What should I do?

Answer 1

Hint (a general procedure to find all the ideals in a finite ring): First, find all the distinct principal ideals. Then for each $n$, find all distinct ideals generated by at most $n$ elements (by looking at the sums $I+J$, where $I$ is principal and $J$ is generated by at most $n-1$ elements). Since your ring has $6^3$ elements, after at most $6^3$ steps you will not get any new ideals (likely, much sooner).

Answer 2

$(\mathbb Z/p\mathbb Z)[x]$ is a prinicipal ideal domain when $p$ is prime, so the ideals of the quotient by $(x^3)$ will correspond to divisors of $x^3$.

There are four (classes of) divisors: $\{1, x, x^2, x^3\}$, so there are four ideals in each half of your product.

Combining them in pairs, you get the ideals of the product, which you already related to your original ring. Writing out all 16 possibilities probably isn't hard, but I leave it to you to translate this back to your original ring as an exercise.

Answer 3

The ideals of $R$ are into 1-1 correspondence with ideals of $S[x]$ containing $x^3$.

Now, as you noticed, $S\simeq \mathbb{Z}/2\mathbb{Z}[x]\times \mathbb{Z}/3\mathbb{Z}[x]$ is a product of two principal rings, so it is a principal ring (since the ideals of a direct product ring is a direct product of ideals).

A principal ideal $(f)$ of $\mathbb{Z}/6\mathbb{Z}[x]$ contains $x^3$ if and only if $f\mid x^3$. Since you work up to units, you may assume that $f$ is monic. You get $f=1,x,x^2,x^3$. Finally you will get four ideals: $R, (x), (x^2), (x^3)=(0).$