One to one correspondence between all $F[x]$-module $V$ and all linear transformations $T\colon V\to V$, $V$ being a vector space over $F$.

Question

I am fairly a beginner in module theory. While discussing the $F[x]$-modules, my text book (Dummit & Foote) describes:

$\left\{ V \text{ an } F[x] \text{-module} \right\}\longleftrightarrow \left\{ \begin{aligned} V \text{ is a vector space }\\ \text{ and }\\ T\colon V\to V \text{ a linear transformation }\end{aligned} \right\}\tag*{}$

given by

$\text{ the element } x \text{ acts on } V \text{ as the linear transformation } T\tag*{}$

This is saying that we can't find any $F[x]$-module without specifying the linear transformation $T$ .

But how can we exclude the possibility of having some other $F[x]$-module which can be obtained without the help of the linear transformation $T\colon V\to V$?

Is it the case that if $x$ acts on the vectors $v\in V$, then it must be a linear transformation? If it is the case, how can I prove it?

Answer 1

If $S$ is a subring of a ring $R$ and $M$ is a (left) $R$-module, then $M$ becomes in a natural way a (left) $S$-module by “restriction of scalars”.

I'm not saying that $M$ (keeping the same addition) can be endowed with the structure of $S$-module in only this way. But this structure of $S$-module can be defined.

Suppose $V$ is an $F[x]$-module. Then it can be considered an $F$-module (that is, vector space over $F$): the action of $F$ on $V$ is “multiplication by constant polynomials”.

Now consider the map $T\colon V\to V$ defined by $T(v)=xv$. This is obviously $F$-linear when $V$ is considered an $F$-vector space. Note that with this linear map we are able to reconstruct the action of $F[x]$ on $V$; namely, $$ (a_0+a_1x+\dots+a_nx^n)v=a_0v+a_1T(v)+\dots+a_nT^n(v) $$ Just compare the two sides of the stated equality.

Conversely, given an $F$-vector space $V$ and a linear map $T\colon V\to V$, we can define an $F[x]$-module structure on $V$, by defining $$ xv=T(v) $$ and using linearity to complete the definition. Note here that the map induced on $V$ by multiplication by $x$ is exactly the linear map $T$.

Thus we have the required bijection.

There may certainly be different ways to define an $F[x]$-module structure on an abelian group $V$. However, as shown above, any $F[x]$-module structure can be defined via a (unique) linear map. That's what the theorem is about, nothing else.

Answer 2

If $V$ is a $k[x]$-module, then multiplication with $x$ defines a linear transformation $T: V \to V, v \mapsto x v$.

Conversely, if $V$ is a $k$-vector space and $T:V \to V$ is a linear transformation, then you can define a $k[x]$-module structure by letting a polynomial $p \in k[x]$ act on $V$ by $$ p \cdot v = p(T)(v) $$

The point is that no matter what you start with (a $k[x]$-module structure or a linear transformation) you always get the other thing for free.

Answer 3

Let us take any $F[x]$-module V. We shall prove this module is obtained from a Linear Transformation $T\colon V \rightarrow V$. We consider the action of $x$ on $V$. Let $v,w \in V$ then from the definition of module we have, $$x.(v+w)=x.v+x.w,.......(1)$$ Again let $p \in F \subset F[x]$ then from the definition of modules, we have, $$x.(p.v)=(xp).v=(px).v=p.(x.v),.......(2)$$ $(1)$ and $(2)$ show the action of $x$ is linear, so this induce a linear transformation.

Hence we show that any action of $F[x]$ on V is obtained from a linear transformation.

I think from this it will be clear to you that why we exclude the possibility of having some other $F[x]$-module which can be obtained without the help of the linear transformation $T\colon V\to V$.