= Solution
Assume condition (iii), and factor $m=\prod_j f_j^{e_j}$. If some $e_j\ge2$, the <primary decomposition theorem> gives the nonzero summand $V_j=\ker f_j(T)^{e_j}$ on which the <minimal polynomial> is exactly $f_j^{e_j}$. The operator $f_j(T)$ cannot vanish on this whole summand, or its restriction would have <minimal polynomial> dividing $f_j$. Choose $v\in V_j$ with $f_j(T)v\ne0$. Then
$$
v\in\ker f_j(T)^{e_j}\setminus\ker f_j(T),
$$
contradicting (iii) with $p=f_j$ and $n=e_j$. Hence every $e_j=1$, proving \b[(iii) implies (ii)].
Back to article page
