# Describe all the submodules of $V$ considered as $\mathbf{k}[x]$-modul via $\alpha$.

This problem taken from Hartley Hawkes's book (Rings, Modules, and Linear Algebra) Exercises for Chapter 5 pp. 82 number 2.

Let $$V$$ be a vector space over a field $$\mathbf{k}$$ with basis $$\{v_1,v_2\}$$, and let $$\alpha : V\to V$$ be the map defined by $$\alpha(\lambda_1v_1+\lambda_2v_2)=\lambda_2 v_1+\lambda_1 v_2$$ for all $$\lambda_1,\lambda_2\in\mathbf{k}$$. Show $$\alpha\in \operatorname{End}_{\mathbf{k}}V$$, and describe (by finding bases) all the submodules of $$V$$ considered as $$\mathbf{k}[x]$$-modul via $$\alpha$$. Contrast this with the case where $$V$$ considered as $$\mathbf{k}$$-module. (Caution: $$\mathbf{k}$$ may have characteristic $$2$$.)

I have proved $$\alpha\in \operatorname{End}_{\mathbf{k}}V$$. Now I go to second question in this problem. Describe (by finding bases) all the submodules of $$V$$ considered as $$\mathbf{k}[x]$$-modul via $$\alpha$$.

I know $$\mathbf{k}[x]$$ is polynomial field, i.e. $$\mathbf{k}[x]=\{\lambda_0+\lambda_1 x+\lambda_2 x^2+\ldots+\lambda_n x^n\mid \lambda_i\in \mathbf{k}, i=0,1,2,\ldots,n\}.$$

Is that true if we define $$\begin{array}{l@{\;}c@{\;}c@{\;}c} \alpha\colon & V &\to& V\\ & (\lambda_0+\lambda_1 x+\lambda_2 x^2+\ldots+\lambda_n x^n) v_1+(\lambda_0'+\lambda_1' x+\lambda_2' x^2+\ldots+\lambda_n' x^n)v_2 &\mapsto &(\lambda_0'+\lambda_1' x+\lambda_2' x^2+\ldots+\lambda_n' x^n)v_1+(\lambda_0+\lambda_1 x+\lambda_2 x^2+\ldots+\lambda_n x^n) v_2 \end{array}$$ ?

I don't know how to finding bases to describe all the submodules of $$V$$ considered as $$\mathbf{k}[x]$$-modul via $$\alpha$$. Any hint to solve this problem?

The $$\mathbf{k}[x]$$-module structure of $$V$$ is defined by: $$\left(\lambda_0 + \lambda_1 x + \cdots + \lambda_n x^n\right)v = \lambda_0 v + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$$
• $\lambda_0 v + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$ or $\lambda_0 + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$ ? Oct 23, 2020 at 6:41
• It is $\lambda v + \cdots$. For the $\mathbf{k}[x]$-submodules, they are just $\mathbf{k}$-vector subspace stable by $\alpha$. Oct 23, 2020 at 9:23