Let $V$ be a vector space (i.e. an $F$-module) and let $\l$ be a linear map on it. Then $V$ can be given a natural $F[\l]$-module structure. For some $v\in V$, I will define its $\l$-span to be submodule $F[\l]v$. Equivalently it is the span of the vectors $\l^iv$. Note that $\ann v$ is an ideal of the PID $F[\l]$ and hence is generated by a polynomial $p(\l)$. Therefore,
$$F[\l]v \cong F[\l]/(p(\l)).$$
In other words, elements of the $\l$-span can be represented as polynomials, and polynomials that are equal mod $p(\l)$ represent the same vector. The action of the map $\l$ on polynomials is simply left multiplication.
If
$$p(\l) = \l^n + a_{n-1}\l^{n-1} + \ldots + a_0,$$
then $F[\l]/(p(\l))$ has a 'cyclic' basis $1,\l,\ldots,\l^{n-1}$, such that the matrix of $\l$ wrt this basis is the companion matrix
$$\begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & & & & -a_1 \\ & 1 & & & -a_2 \\ & & \ddots & & \vdots \\ & & & 1 & -a_{n-1} \\ \end{pmatrix}$$
Thus we can think of $p(\l)$ as encoding the dynamics of $\l$ on the $\l$-span. As a special case, if $p(\l)=\l-\a$ then the $\l$-span is an eigenspace: the vector corresponding to the polynomial 1 is an eigenvector with value $\a$.
It is not too hard to see that the large space $V$ can be decomposed as a direct sum of $\l$-spans,
$$V \cong \bigoplus_i F[\l]/(p_i(\l)).$$
Furthermore, suppose $\l$ satisfies a polynomial relation $p(\l)=0$, and that the $i$th $\l$-span is generated by $v_i$. We have
$$p(\l)\subseteq \ann V\subseteq \ann v = (p_i(\l)),$$
and thus the $p_i(\l)$'s all divide $p(\l)$. (In particular they all divide the minpoly of $\l$.) This is a very useful observation, as can be seen by the following corollaries.
If $\l$ is idempotent, then it satisfies the relation $\l^2-\l=0$. From the preceding remarks, $V$ is a direct sum of $F[\l]$-modules of the form
$$F[\l]/(\l),\quad F[\l]/(\l-1),\quad \text{or}\quad F[\l]/(\l(\l-1)).$$
But by CRT, $F[\l]/(\l(\l-1))\cong F[\l]/(\l)\oplus F[\l]/(\l-1)$, so $V$ is a direct sum of eigenspaces with the eigenvalues 1 or 0.
If $\l$ is nilpotent, then it satisfies a relation $\l^n=0$. Thus we have
$$V \cong \bigoplus_i F[\l]/(\l^{n_i}),\quad\text{each $n_i\le n$}.$$
Within each $\l$-span, the matrix of $\l$ wrt the basis $\\{1,\l,\ldots,\l^{n_i-1}\\}$ is equal to a matrix of the form
$$\begin{pmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & 1 & \\ & & & 0 & 1 \\ & & & & 0 \end{pmatrix},$$
Thus, $\l$ as a whole can be represented by a diagonal block matrix whose blocks have the above form.
Let $V$ be a complex vector space. If the minpoly of $\l$ is separable, then the $p_i(\l)$'s (which divide the minpoly) are also separable and hence split, say $p_i(\l) = \prod_j\,(\l-\a_{ij})$. By CRT we have $\C[\l]/(p_i(\l)) \cong \bigoplus_j \C[\l]/(\l-\a_{ij})$, thus $\l$ is diagonalizable.
Conversely, if $\l$ is diagonalizable then $V$ can be written in the form
$$\bigoplus_i \C[\l]/(\l-\a_i).$$
Each $\l-\a_i$ is an elementary factor. By Smith normal form, we know that the minpoly is the largest invariant factor, which is obtained by multiplying together the highest multiplicities of each elementary factor. But since the highest multiplicities are all 1, the minpoly is separable.