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?


1 Answer 1


The $\mathbf{k}[x]$-module structure of $V$ is defined by: $$\left(\lambda_0 + \lambda_1 x + \cdots + \lambda_n x^n\right)v = \lambda_0 v + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$$

  • $\begingroup$ $\lambda_0 v + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$ or $\lambda_0 + \lambda_1\alpha(v) + \cdots + \lambda_n \alpha^n(v)$ ? $\endgroup$ Oct 23, 2020 at 6:41
  • 1
    $\begingroup$ It is $\lambda v + \cdots$. For the $\mathbf{k}[x]$-submodules, they are just $\mathbf{k}$-vector subspace stable by $\alpha$. $\endgroup$
    – user388615
    Oct 23, 2020 at 9:23

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