Is the Cyclic Decomposition the coarsest decomposition among those with terms of cyclic subspaces? Fix a vector space $V$ over a field $F$ and a linear operator $T$ on $V$, we know that we have the Cyclic Decomposition (actually, the Invariant Factor Decomposition) $$V = F[T] \alpha_1 \oplus \dots \oplus F[T] \alpha_m$$ (where $\alpha_1, \dots, \alpha_m$ are non-zero vectors in $V$) which satisfy $p_{\alpha_m} \mid p_{\alpha_{m-1}} \mid\dots \mid p_{\alpha_1},$ where $p_{\alpha_i}$ denotes the annihilator of $\alpha_i$.
Now suppose we have another decomposition $$V = W_1 \oplus \dots \oplus W_n$$ where the $W_i$s are cyclic, which is NOT necessarily the Invariant Factor Decomposition. Is it true that we necessarily have $m \leq n$?
We know that, intuitively, the Invariant Factor Decomposition is the "coarsest" decomposition, the question naturally generates from my effort of trying to make the mentioned intuition precise.
 A: It simplifies notation to do this over a general PID $R$. Let $a_1,\dots,a_m\in R$ be non-zero (in your setting they correspond to the $W_i$ via $W_i\cong F[T]/\langle a_i\rangle$).  We will describe how to obtain the elements $p_1,\dots, p_n\in R$ such that $p_1\mid\dots\mid p_n$, $p_1\notin R^\times$ and $$R/\langle p_1\rangle\oplus\dots\oplus R/\langle p_n\rangle\cong R/\langle a_1\rangle\oplus \dots\oplus R/\langle a_m\rangle$$ From this procedure it will be then clear that $n\leq m$.
Let $q_1,\dots,q_l$ be prime elements in $R$ such that there are integers $r_{ij}\geq0, i=1,\dots,m, j=1,\dots,l$ such that \begin{align*}
a_1 &= q_1^{r_{11}}\cdots q_l^{r_{1l}}\\
a_2 &= q_1^{r_{21}}\cdots q_l^{r_{2l}}\\
&\dots\\
a_m&=q_1^{r_{m1}}\cdots q_l^{r_{ml}}
\end{align*}
Now construct the sequence $b_1,b_2,\dots$ as follows: for each $j=1,\dots,l$ select $i$ so that $r_{ij}$ is minimal, take the corresponding term $q_j^{r_{ij}}$, cross it out the list above and let $b_1$ be the product of these $q_j^{r_{ij}}$ for $j=1,\dots,l$ (the $i$ depends on $j$). Then proceed similarly for $b_2,\dots$ by only considering those powers of the $q_j$ which haven't been crossed out yet. Clearly after constructing $b_1,\dots,b_k$ in each column (in the set of equations above) exactly $k$ elements have been crossed out. Thus this has to stop after $m$ steps and we get by construction a sequence of elements $b_1,\dots,b_m$ such that $b_1\mid \dots\mid b_m$ and $$R/\langle a_1\rangle\oplus \dots\oplus R/\langle a_m\rangle\cong\bigoplus_{i=1}^m\bigoplus_{j=1}^lR/\langle q_j^{r_{ij}}\rangle\cong R/\langle b_1\rangle\oplus \dots\oplus R/\langle b_m\rangle$$
Now some of the factors $R/\langle b_i\rangle$ might be zero because some of the powers $r_{ij}$ might have been $0$, $b_1=\dots b_k=1$ for some $k\geq0$. Then the $p_i$ will be those $b_i$ that are non-units, i.e. $p_i := b_{i+k}$ for $i=1,\dots,m-k=:n$.
