Not artinian right module

Let's consider the ring $R = \begin{bmatrix}\Bbb{Q} & 0\\\Bbb{Q(x)} & \Bbb{Q(x)}\end{bmatrix}$, where $\Bbb{Q(x)}$ is the field of fractions of the polynomial ring $\Bbb{Q[x]}$, and the right $R$-module $M = \begin{bmatrix}0 & 0\\\Bbb{Q(x)} & \Bbb{Q(x)}\end{bmatrix}$.

I want to prove that $M$ is not artinian, that is, there exists a descending chain of submodules of $M$ that is not stationary. How can I define this chain?

• Can you describe at least some submodules? Doing so seems like a good first step: you need to exhibit infinitely many submodules in a special arrangement... but it is certainly a good idea to find many submodules! – Mariano Suárez-Álvarez Jan 7 '18 at 19:13
• $N = \begin{bmatrix}0 & 0\\\Bbb{Q(x)} & 0\end{bmatrix}$ is a submodule of M, but I'm not able to find submodules of N... – effezeta Jan 8 '18 at 1:37
• What other submodules are there? – Mariano Suárez-Álvarez Jan 8 '18 at 3:14
• I don't know...who are the submodules of $\Bbb{Q(x)}$? – effezeta Jan 8 '18 at 13:44

$\def\QQ{\mathbb{Q}}$We compute that $$\begin{pmatrix}0&0\\a&b\end{pmatrix}\cdot\begin{pmatrix}\lambda&0\\f&g\end{pmatrix}=\begin{pmatrix}0&0\\\lambda a+fb&bg\end{pmatrix}$$ for all $a$, $b$, $f$, $g\in\QQ(x)$ and all $\lambda\in\QQ$. This describes the action of the ring on the module.
As you noticed, the subset of all elements of $M$ of the form $\begin{pmatrix}0&0\\a&0\end{pmatrix}$ with $a\in\QQ(x)$ ius a submodule $N$ of $M$. The formula above implies —and you should have no difficulty showing this— that if $V$ is any $\QQ$-subspace of $\QQ(x)$ then the set $$S_V=\left\{\begin{pmatrix}0&0\\a&0\end{pmatrix}:a\in V\right\}$$ is a submodule of $N$. This gives you many, many submodules.
A good excercise you can do is showing that in fact all proper submodules of $M$ are of the form $S_V$ for some $\QQ$-subspace of $\QQ(x)$.
• For example, I can define a descending chain of submodules setting $A_n = \begin{bmatrix}0 & 0\\\Bbb{Q[x]}x^n & 0\end{bmatrix}$...is it correct? – effezeta Jan 9 '18 at 2:18