Show that any nonzero free $F[x]$-module $V$ is an infinite dimensional vector space over $F$.

Question

The Problem: Show that any nonzero free $F[x]$-module $V$ is an infinite dimensional vector space over $F$.

My Attempt: I understand that $V$ is isomorphic to a direct sum of copies of $F[x]$, but that did not seem to help. Hence I, BWOC, supposed that $V$ IS a finite dimensional vector space over $F$. Then there exists a basis $\mathcal{B}=\{v_1,\dots, v_n\}$ for $V$. Suppose $v\in V$, then $v=r_1v_1+\dots+r_nv_n$, where $r_i\in F$. Screeching Halt.

My Question: Obviously I want to show that $v=f_1(x)v_1+\dots+f_n(x)v_n$ with $f_i(x)\in F[x]$ such that NOT all $f_i(x)=r_i$, then we'd have a contradiction; but I failed. Any help would be greatly appreciated.

Edit: This problem turns out to be a bit more involved than I have anticipated, so I'll add more background here. $F$ is a field, $T: V\to V$ is a linear transformation. Then every $p(x)=a_nx^n+\dots+a_1x+a_0\in F[x]$ acts on $v\in V$ as follows: $p(x)v=a_nT^n(v)+\dots+a_1T(v)+a_0v$. With this action, $V$ is called an $F[x]$-module.

Answer 1

Yes, $V$ is isomorphic to a direct sum of copies of $F[x]$. And each $F[x]$ is infinite-dimensional over $F$. Therefore, $V$ is infinite-dimensional over $F$.

Answer 2

An $F[x]$-module is an abelian group $V$ with a group homomorphism $$\varphi:\ F[x]\ \longrightarrow\ \operatorname{End}_{\Bbb{Z}}V.$$ If such a module is free and nonzero then $\varphi$ is injective, so $\operatorname{End}_{\Bbb{Z}}V$ is infinite dimensional over $F$. Then $V$ itself is also infinite dimensional over $F$.

Answer 3

I believe I finally dug myself out of the hole-I realize that an $F[x]$-module is also an $F$-vector space, and an $F[x]$-module isomorphism is also an $F$-vector space isomorphism. The following is just a formalization of @Another User's answer.

Since $V$ is free, there exists $B=\{v_1, v_2,\dots\}$ such that each $v\in V$ can be written uniquely as a linear combination of elements in $B$ with coefficients in $F[x]$. Thus $V=N_1\oplus N_2\oplus\dots$, where $N_i=\{f(x)v_i|f(x)\in F[x]\}$. Obviously, the $F[x]$-submodule $N_1\subseteq V$ is $F[x]$-isomorphic to $F[x]$ via $\varphi: f(x)v_1\mapsto f(x)$, since $V$ is free implies each submodule is free. Note that $\mathbf{\varphi}$ is also an $\mathbf{F}$-isomorphism when $\mathbf{N_1}$ and $\mathbf{F[x]}$ are viewed as $\mathbf{F}$-vector spaces. Hence the $\mathbf{F}$-vector space $N_1$ is $F$-isomorphic to the $\mathbf{F}$-vector space $F[x]$, and the latter is infinite dimensional over $F$. Since $N_1$ is a $F$-subspace of $V$, it follows that $V$ is also infinite dimensional over $F$.