Generator of $\text{Ann}(V)$ in $F[x]$ I'm reading the text by Dummit and Foote, and I'm on section 12.2 where they discuss how the classification theorems for finitely generated modules over PIDs can be used to give the rational canonical form of a matrix/linear transformation.
In the beginning of this section, I came across the following two lines:

We have seen that $V$ considered as a module over $F[x]$ via the linear transformation $T$ is a torsion $F[x]$-module.  Let $m(x) \in F[x]$ be the unique monic polynomial generating the annihilator of $V$ in $F[x]$.  

The second sentence is the one I am perplexed about.  I have two questions:


*

*How do we even know that $\text{Ann}(V)$ is larger than $\{0_{F[x]}\}$.  I understand that $V = \text{Tor}(V)$, but this only means that for every $v \in V$ there is some $f(x) \in F[x]$ that annihilates $v$.  But we seem to need one $f(x) \in F[x]$ that annihilates every $v \in V$...

*Assuming that there is a nonzero $f(x) \in \text{Ann}(V)$, why can't there be two monic polynomials that both generate $\text{Ann}(V)$?
Thanks!
EDIT:  I believe we are assuming $V$ is a finite dimensional vector space over $F$ in this entire section.
 A: As you said, a polynomial $p(x) \in F[x]$ is in  ${\rm Ann}(V)$ iff $p(T)$ is the zero matrix. There is clearly at least one non-zero polynomial in ${\rm Ann}(V)$: for example, the characteristic polynomial $c_{T}(x)$ obeys $c_{T}(T) = 0$ by the Cayley-Hamilton theorem, so $c_{T}(x) \in {\rm Ann}(V)$.
Next, observe that ${\rm Ann}(V)$ is an ideal. (This should be easy to check.) Moreover, $F[x]$ is a principal ideal domain, so ${\rm Ann}(V)$ is a principal ideal, i.e. there exists a polynomial that generates the whole of ${\rm Ann}(V)$ over $F[x]$.
Furthermore, if we are given two different polynomials that both generate ${\rm Ann}(V)$, then they must be equal up to multiplication by a unit in $F[x]$, i.e. they must be scalar multiples of each other.
(By convention, the minimal polynomial of $T$ is defined to be the generator of ${\rm Ann}(V)$ whose leading coefficient is $1$. Thus all other generators of ${\rm Ann}(V)$ are scalar multiples of the minimal polynomial.)
A: The evaluation map $F[X] \to L(V)$ given by $p \mapsto p(T)$ is a linear transformation.
If this map were injective, then there'd be a copy of $F[X]$ inside $L(V)$, which cannot happen because $L(V)$ has finite dimension but $F[X]$ doesn't.
So the kernel of the evaluation map is not zero. That kernel is ${\rm Ann}(V)$.
