$V$ becoming a $K[x]$-module w.r.t this addition and scalar multiplication

Question

Let $V$ be a vector space over a field $K$ and let $\pi: V \to V$ be a map. With addition defined as usual on $V$ and the scalar multiplication from $K[x]$ as follow $p(x)v=p(\pi)$, that is, $$(p_0 + p_1x + \ldots + p_nx^n)v = p_0v + p_1\pi(v) + \ldots + p_n\pi^n(v).$$ I know the following and it easy to verify that $V$ becomes a $K[x]$-module w.r.t this addition and scalar multiplication.

Now my problems are the following:

1. How to define the $K[x]$-module homomorphism $\phi: V \to V$.

2. How to show that the cyclic $K[x]$-submodules of $V$ is the subspace of $V$ spanned by $v,\pi(v),\pi^2(v),\ldots.$

3. Let $V = K[x]$ be a cyclic $K[x]$-module such that the annihilator of $V$ is $K[x](x-1)^n$.

   How to show that $V$ is $n$ dimensional with basis $(x- a)^{n-1}v,\ldots (x-a)v, v$.

I am not so good in Linear Algebra

Answer 1

1. The definition of module homomorphism remains the same as always - $f:V\to U$ such that $$f(v+u)=f(v)+f(u)\\f(\alpha v)=\alpha f(v)$$ for all $v,u\in V$ and $\alpha\in K[x]$.

2. A cyclic module $M$ is generated by one element, i.e. there exist $v\in V$ such that $$M=\left\{\alpha v:\alpha\in K[x]\right\}=K[x]v$$ Thus a general element in $M$ is $$\left(p_{0}+p_{1}x+\dots+p_{n}x^{n}\right)v\in M$$ for $n\in\mathbb{N}\cup\{0\}$ and $p_{i}\in K, \:\forall\: 0\leq i\leq n$. Can you take it from here?

3. Observe that $$M\cong K[x]/{\rm Ann}_{K[x]}(v)$$ where $${\rm Ann}_{K[x]}(v)=\left\{\alpha\in K[x]:\alpha v=0\right\}$$ by the first isomorphism theorem with $\alpha\in K[x]\mapsto \alpha v\in M$. Thus with your $V=K[x]v$ $$K[x]v\cong K[x]/\left(x-1\right)^{n}K[x]$$ which is indeed an $n$-dimensional vector space with basis ${\rm B}=\left\{\left(x-1\right)^{i}:0\leq i\leq n-1\right\}$.