Is the $F [X]$ module $V$ finitely generated?

Question

Consider a vector space $V$ over $\mathbb F$ as an $\mathbb F [X]$ submodule via $T \in End_{\mathbb F} (V)$. Is $V$ always finitely generated?

It is clear that if $V$ is a finitely dimensional vector space over $\mathbb F$ then it is a finitely generated $\mathbb F [X]$ module via $T$ where the basis for $V$ over $\mathbb F$ can be treated as the finite generating set for $V$ over $\mathbb F [X]$. But what can be said if $V$ is an infinite dimensional vector space? It is quite confusing to me.

Please help me in this regard.

Thank you in advance.

Answer 1

No. In general, this depends on $T$, $V$ and $\mathbb{F}$. Two examples:

1. Let $V = \mathbb{F}[X]$ and $T$ act on $\mathbb{F}[X]$ as multiplication by $X$. Then $V$ is an infinite dimensional vector space but a finitely generated $\mathbb{F}[X]$-module (in fact free, of rank one).
2. Let $V$ be an infinite-dimensional vector space and $T = 0$. Then a generating set of $V$ as an $\mathbb{F}[X]$-module is the same as a generating set of $V$ as an $\mathbb{F}$-vector space so $V$ is not finitely generated.

Answer 2

If you consider $V=\mathbb{R}$ over $F=\mathbb{Q}$, then $V$ can't be finitely generated over $F[X]$ by the cardinality argument. Any finitely generated $F[X]$-module has at most countable elements, while $V$ is an uncountable set.

Answer 3

Every $\mathbb{F}[X]$-module can be described by a pair $(V,T)$ where $V$ is a vector space over $\mathbb{F}$ and $T$ is an endomorphism of $V$.

Indeed, if $M$ is a module over $\mathbb{F}[X]$, then we can set $V=M$, which is a vector space because $\mathbb{F}$ is a subring of $\mathbb{F}[X]$.

Next, defining $T(v)=Xv$, we get an endomorphism of $V$ and we're done. The converse is clear: if $(V,T)$ is a pair as above, we can define an $\mathbb{F}[X]$-module as usual.

Now take a nonfinitely generated module over $\mathbb{F}[X]$ and you're done.

However, if $V$ is finite dimensional, then a finite spanning set as vector space is also a set of generators as $\mathbb{F}[X]$-module.