Let $T: \mathbb F^n \longrightarrow \mathbb F^n$ be the shift operator $:$
$$T (x_1,x_2,...,x_n) = (x_2,x_3, \cdots , x_n,0).$$
Let $W_k : = \{x \in \mathbb F^n : x_j = 0, \forall j > k \}$. Show that these are the only $\mathbb F [X]$ submodules of $V$ where $V = \mathbb F^n$.
I know that the $\mathbb F [X]$ submodules of $V$ are precisely the $T$-invariant subspaces of $V$. So I have tried to solve this question by first showing that all the $W_k$'s are indeed $T$-invariant.But I find some trouble in proving the converse part i.e. the only $T$-invariant subspaces of $V$ are $W_k$'s. I found that for $k \geq i$, $T^k (e_i) = 0$ and for $k < i$, $T^k (e_i) = e_{i-k}$.I also proved that if $m < n$ then $(a_m X^m + a_{m-1} X^{m-1} + \cdots + a_0).(e_n) = (a_m T^m + a_{m-1} T^{m-1} + \cdots + a_0 I)(e_n) = (0,0,\cdots,0,a_{m},a_{m-1}, \cdots , a_0)$ and if $m \geq n$ then $(a_m X^m + a_{m-1} X^{m-1} + \cdots + a_0).(e_n) = (a_m T^m + a_{m-1} T^{m-1} + \cdots + a_0 I)(e_n) = (a_{n-1}, a_{n-2}, \cdots , a_0)$.
Now how can I proceed? Please help me.
Thank you in advance.
Given a $T$-invariant subspace $W$, set
$$ j = \min \{ k \, | \, W \subseteq W_k \}. $$
Note that $W_n = V$ so this is well-defined. Then we have $W \subseteq W_j$. Let us show that in fact $W = W_j$.
If $j = 0$, we are done. Otherwise, we have $W \nsubseteq W_{j-1}$ so we can find $w \in W \cap (W_j \setminus W_{j-1})$. Such $w$ has the form
$$ w = a_1 e_1 + \dots + a_{j-1} e_{j-1} + a_j e_j $$
with $a_j \neq 0$. But then $T^j(w) = a_j e_1$ and since $W$ is $T$-invariant, we see that $e_1 \in W$. Replacing $w$ with $w - a_1 e_1$, we see that
$$ T^{j-1}(w - a_1e_1) = T^{j-1}(a_2e_2 + \dots + a_j e_j) = a_j e_2 $$
and so $e_2 \in W$. Repeating the argument, we get that $e_1,\dots,e_{j-1} \in W$ and then
$$ e_j = \frac{w - a_1 e_1 - \dots - a_{j-1} e_{j-1}}{a_j} \in W$$ showing that $W = W_j$.