Proposition 1

— Let $f \in A[X]$ and $\alpha \in A$. The remainder of the division of $f$ by $X - \alpha$ is $f(\alpha)$. For $\alpha$ to be a root of $f$ it is necessary and sufficient that $X - \alpha$ should be a divisor of $f$ in $A[X]$.

For if $u,\ v \in \mathbf{A}[X]$ are such that $f = (X - \alpha)\ u + v,\ \deg v < 1$, then $v$ is a scalar and $f(a) = (a - a)\ u(\alpha) + v = v$. This proves the first assertion and the second follows from it.