Proposition 2

— Let $f \in \mathbf{A}[X]$, $a \in \mathbf{A}$, and let $h$ be an integer $\geq 0$. The following conditions are equivalent: (i) $f$ is divisible by $(X - \alpha)^h$ but not by $(X - \alpha)^{h-1}$; (ii) *there exists $g \in \mathbf{A}[X]$ such that $f = (X - \alpha)^h g$ and $g(a) \neq 0$. (i) $\Rightarrow$ (ii) follows at once from Prop. 1. (ii) $\Rightarrow$ (i) : Suppose that $f = (X - \alpha)^h g$, where $g$ does not admit $a$ as a root. Then $f$ is divisible by $(X - \alpha)^h$; if $g_1 \in \mathbf{A}[X]$ existed such that $f = (X - \alpha)^{h+1} g_1$, then since $(X - \alpha)^n$ is not a divisor of zero in $\mathbf{A}[X]$ (IV, p. 9, Prop. 7), we have $g = (X - a)\ g_1$ and so $g(\alpha) = 0$, which is absurd.