Proposition 5

Suppose that $A$ is an integral domain. Let $f$ be a non-zero element of $A[X]$, and $\alpha_1, \ldots, \alpha_p$ pairwise distinct roots of $f$ in $A$, of orders $k_1, \ldots, k_r$. We have

where $g \in A[X]$ and $\alpha_1, \ldots, \alpha_p$ are not roots of $g$.

We proceed by induction on $p$, the proposition being evident for $p = 1$, by Def. 1. Suppose then that $f(X) = g_1(X)g_2(X)$, where

Since $A$ is an integral domain and $\alpha_p$ is distinct from $\alpha_1, \ldots, \alpha_{p-1}$ it follows that $\alpha_p$ is not a root of $g_1(X)$, hence $\alpha_p$ is a root of order $k_p$ of $g_2(X)$ (Prop. 4, (ii)). It follows that $g_2(X)$ is divisible by $(X - \alpha_p)^{k_p}$, and so

where $g(X) \in A[X]$. Clearly $\alpha_1, \ldots, \alpha_p$ are not roots of $g$.