Corollary

— Let $f \in \mathbf{A}[X]$, $a \in \mathbf{A}$ and $p$ an integer $\geq 0$, further, suppose that $p! . 1$ is cancellable in $\mathbf{A}$. Then for $a$ to be a root of order $p$ of $f$, it is necessary and sufficient that $a$ should be a root of $f, Df, ..., D^{p-1}f$ and not a root of $D^pf$.

This follows from Prop. 8 by induction on $p$.