Let $V$ be an $n$-dimensional vector space where $n\in \mathbb{N}, n>0$.
If the sub-$\mathbb{F}[x]$-module $\langle{p_v}\rangle=\{g\in \mathbb{F}[x] \mid g(A)v=0\}$ satisfies $p_v=m_A$ for all $v \in V$ that is non-zero ($m_A$ is the minimal polynomial of $A$), am I right to conclude that $A$'s rational canonical form consists of one block (the companion matrix for $m_A$)?
My thought process was that one can show there's an $F[x]$-module isomorphism: $$\langle v \rangle \cong \mathbb{F}[x]/\langle p_v \rangle$$
which means that for all $v \in V$:
$$\langle v \rangle \cong \mathbb{F}[x]/\langle m_A \rangle$$
but even if my statement is true, how does one pass from the submodules $\langle v \rangle $ to the entire vector space $V$ to complete the claim?
If my claim is not true, is there anything we can conclude about $A$'s rational canonical form?
In general, the different values of $p_v$ for $v\in V$ are exactly the factors of $m_A$. In one direction, $p_v$ must always divide $m_A$ since $m_A(A)v=0$. Conversely, if $fg=m_A$ and you start with $v$ such that $p_v=m_A$ (the existence of such a $v$ is a consequence of the Chinese remainder theorem, or alternatively follows by just taking the last block of the rational canonical form), then $p_{g(A)v}=f$.
So, $p_v=m_A$ for all nonzero $v$ iff $m_A$ has no factors besides $1$ and $m_A$, i.e. iff $m_A$ is irreducible (or $m_A=1$, but that means $\dim V=0$). In this case $V$ can be considered as a vector space over the field $K=\mathbb{F}[x]/(m_A)$ so it is a direct sum of copies of $K$. This direct sum decomposition is the rational canonical form of $A$, so every block of the rational canonical form is the companion matrix of $m_A$. However, there can be any number of blocks, since $V$ could have any finite dimension over $K$. For a very simple example, $V$ could have any dimension and $A$ could be $0$, so $m_A=x$ and there are $\dim V$ different blocks which are all just $0$.