= Solution
If $p=m_{\alpha_W}$ and $q=m_{\alpha_{V/W}}$, then $m_\alpha$ is divisible by both. Also $q(\alpha)V\subseteq W$, so $p(\alpha)q(\alpha)=0$; hence $\operatorname{lcm}(p,q)\mid m_\alpha\mid pq$. A single size-two nilpotent Jordan block with its invariant eigenline makes the lcm too small; a diagonal operator with the same <eigenvalue> on $W$ and $V/W$ makes the product too large. Square-free $m_\alpha$ implies both induced minimal <polynomials> square-free, so diagonalizability descends. The nilpotent-block example disproves the converse.
Back to article page
