Proposition 8

— Let $f \in \mathbf{A}[X]$ and $a \in \mathbf{A}$, and suppose that $a$ has order $k \geq 1$ relative to $f$: Then $a$ has order $\geq k - 1$ relative to $Df$. If $k . 1$ is cancellable in $\mathbf{A}$, then $a$ has order $k - 1$ relative to $Df$.

By hypothesis there exists $g \in \mathbf{A}[X]$ such that $f = (X - a)^k g$ and $g(a) \neq 0$. Hence $Df = k(X - \alpha)^{k-1}g + (X - \alpha)^k Dg = (X - \alpha)^{k-1}(kg + (X - \alpha)\ Dg)$, which establishes the first part of the proposition. The value of $kg + (X - \alpha)\ Dg$ for $X = \alpha$ is $kg(a)$, and this is non-zero if $k . 1$ is cancellable in $\mathbf{A}$; this proves the second part of the proposition.

Let $k$ be an integer $> 0$ such that $k . 1 = 0$ in $\mathbf{A}$. If $f(X) = X^k$, then $0$ is a root of order $k$ of $f$, and a root of arbitrarily high order of $Df$.