Cyclic Modules, Characteristic Polynomial and Minimal Polynomial

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.

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.

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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 $T^{n-1}v \neq 0$.

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.