Cyclic Modules, Characteristic Polynomial and Minimal Polynomial Suppose that $\mathrm{dim}_{F}M<\infty$ for $F$ a field and $M$ an $F$ vector space. Let $T$ be a linear transformation on $M$. Show that $M$ is cyclic (as an $F[x]$ module) if and only if $m(x)$ is the characteristic polynomial of $T$, for $m(x)$ being the minimal polynomial of $T$.

How would one be able to show this? I'm not sure on how to start with either direction. We know that the torsion of $M$ would just be $M$ (since $m(T)=0$) if we consider $M$ as an $F[x]$ module with $x$ being represented as the action of $T$ (i.e. $p(x) \cdot v=p(T)v$). Would the Cayley-Hamilton theorem help in this case?
Thanks for the help.
 A: By the structure theorem of finitely generated modules over principal ideal domains, we can write
$$M \cong F[x]/(e_1) \oplus F[x]/(e_2) \oplus \dotsb \oplus F[x]/(e_s)$$
with $e_1 | e_2 | \dotsc | e_s$.
In particular we have $e_sM=0$, which means the minimal polynomial $m$ is a divisor of $e_s$ and actually one has $m=e_s$. This yields
$$\dim_F M = \deg e_1 + \dotsb + \deg e_s \geq \deg e_s = \deg m.$$
The minimal polynomial and the characteristic polynomial coincide if and only if equality holds, which is the case if and only if $s=1$, which means that $M \cong F[x]/(e_1)$ is cyclic.
A: Let me also give an elementary proof of the hard direction:

If the minimal poylnomial $m \in F[x]$ of $T$ coincides with the
  characteristic polynomial, $M$ is cyclic.

Proof:
Let us first do the case $m=p$ for some irreducible polynomial $p \in F[x]$ of degree $d$. In this case any $v \neq 0$ will generate $M$, since a proper $T$-invariant subspace of $M$ gives rise to a factorization of $m$.
Now let us consider the case $m=p^n$ for some irreducible polynomial $p \in F[x]$ of degree $d$. Let $v$ be any vector with $p^{n-1}(T)v \neq 0$. Then
$$p^j(T)v,p^j(T)Tv, \dotsc, p^j(T)T^{d-1}v, 0 \leq j \leq n-1$$
are $dn$ linear independent vectors, hence they form a basis of $M$, which shows that $v$ generates $M$.
To see the linear independence, apply $p^{n-1}(T)$ to a linear combination of the vectors. Then use the $n=1$-case to see that the remaining coefficients are zero. Then apply $p^{n-2}(T)$ and proceed.
Finally, the general case is a use of the chinese remainder theorem and the decomposition theorem into generalized eigenspaces.
Let $m = p_1^{n_1} \dotsb p_s^{n_s}$. We have the decomposition
$$M = \operatorname{ker}(p_1^{n_1}(T)) \oplus \dotsb \oplus \operatorname{ker}(p_s^{n_s}(T))$$
By the cases already taken care of we obtain that $\operatorname{ker}(p_1^{n_1}(T))$ is cyclic with annihilator $p_1^{n_1}$, i.e. $\operatorname{ker}(p_1^{n_1}(T)) = F[x]/(p_1^{n_1})$. By the chinese remainder theorem we now obtain that
$$M=\operatorname{ker}(p_1^{n_1}(T)) \oplus \dotsb \oplus \operatorname{ker}(p_1^{n_1}(T))=F[x]/(p_1^{n_1}) \oplus \dotsb \oplus F[x]/(p_s^{n_s}) = F[x]/(m)$$
is cyclic.

Note that in the case, where $F$ is algebraically closed, the first two cases collapse into the following very easy statement:
If $T$ is nilpotent and $n$ minimal with $T^n=0$, then $v,Tv, \dotsc, T^{n-1}v$ are linear independent for any $v$ with $T^{n-1}v \neq 0$.
A: Here is one direction (the easy one):
Let $m$ have degree $k$ and let the characteristic polynomial $f$ have degree $n=\dim M$.
We have $m(T)v=0$ for all $v\in M$. If $m\ne f$, then $v, T^2v, \dots T^kv$ cannot form a basis for $M$ because $m(T)v=0$ shows they are linearly dependent already for less than $k+1\le n$ powers. Hence, $M$ cannot be cyclic.
