Finding Submodules of $\mathbb{Q}[x]/(x^3)$

Question

I am trying to find all the submodules of $\mathbb{Q}[x]/(x^3)$ over $\mathbb{Q}[x]$, and then using that to determine if it is indecomposable. Is it correct to use the factors of $x^3$: $1,x,x^2$, to create submodules using these elements, or pairs of them such that the sum of all the degrees is less than three, like $(x), (1,x), (x^2,x)$ for example. Is this approach correct or is it not guaranteed that all of them will indeed be submodules? Furthermore, to determine if it is indecomposable, can I express it as the direct sum of $\mathbb{Q}[x]/(x)\bigoplus\mathbb{Q}[x]/(x)\bigoplus\mathbb{Q}[x]/(x)$ or can I not use the same simple sub-module more than once?

Answer 1

Let us denote $A=\mathbb{Q}[x]$, $M=\mathbb{Q}[x]/(x^3)$ and let $N$ be an $A$-submodule. If $N\neq 0$, it has a nonzero element $f$. We denote the least degree term of $f$ by $ax^d\ (d\in \{0,1,2\},\ a\neq 0)$, so we can write $$f=ax^d(1+xg)\ \ \ \ (g \text{ is a polynomial (chosen as a representative)}).$$ Let us pick up $A\ni h=1-xg+(xg)^2$. Since $N$ is closed under $A$-action, we have $$h\cdot f=ax^d(1+x^3g^3)=ax^d\in N.$$ Therefore, $N$ contains either $1$, $x$ or $x^2$. Choosing the least possible such power, it follows that there are exactly four submodules $N$, $$N=0,\ x^2A,\ xA, M.$$ They have different dimensions as $\mathbb{Q}$-vector spaces, so are different.

Besides, the indecomposability can be done independent of the classification of $N$. If $M$ is decomposed as $M\simeq N_1\oplus N_2$, then $\mathrm{End}(M)$ has at least two idempotents $(\mathrm{id}_{N_1},0)$ and $(0,\mathrm{id}_{N_2})$. But, in fact $M$ is a cyclic (namely, generated by one element) module and $\mathrm{End}(M)$ has only one idempotent $\mathrm{id}_M$. So, it is indecomposable.