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.
Best Answer
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 $i=1,\dots,m-k=:n$.