How to check whether a module is Artinian? Support $V$ is a vector space over $k$ with a countable basis $\\{e_1,e_2,e_3,\cdots \\}$. Let $T$ is a linear transformation on V, satisfying $T(e_1)=0$ and $T(e_{i+1})=e_{i}$ for $i\ge 1$. Let $R=k[x]$ with indeterminate x, then we make V an $R$-module by $$x\cdot v=T(v),\forall v\in V.$$
Show that V is Artinian as an $R$-module.
I wanted to find an Artinian submodule $M$ of $V$ such that $V/M$ is Artinian. Let $M$ is an $R$-submodule generated by $e_2$. I lost my way after I realized that $V$ and $V/M$ are similar. How to prove it?
 A: With a little work, you can show that the nontrivial submodules are of the form $N_j=Span(\{e_i\mid i\leq j\})$.
Let $N$ be a nonzero submodule of $M$. Suppose that $x$ is a nonzero element of $M$, and that $j$ is maximal such that $e_j$ is the basis element has a nonzero coefficient in the expression for $x$.  Then $T^{j-1}(x)=\lambda e_1\in N$, and thus so is $e_1$.  Furthermore $T^{j-2}(x)=\alpha e_2+\beta e_1\in N$, and so $\alpha e_2\in N$ (and $e_2$ itself too) because $\beta e_1\in N$.  Going on this way, we can see that every $e_i$ with $i\leq j$ is in $N$.
Now one of two things is going to happen:

*

*There will be a maximal $j$ such that $e_j$ is in $N$

*There is no maximal $j$ such that $e_j$ is in $N$.

In the latter case, clearly $N=M$. In the former case, $N=N_j$.
So as you can see, all of the proper submodules form a chain, indexed from the bottom up (you may call $\{0\}=N_0$ if you like) and since $\mathbb N$ is well-ordered, the submodules clearly satisfy the minimum condition on submodules.
At the same time, you can see it does not satisfy the maximum condition on submodules.
