Example

For a polynomial $P \in \mathbf{A}[X]$ to be strongly relatively prime to $X - a$ (where $a \in \mathbf{A}$), it is necessary and sufficient that $P(a)$ be invertible in $\mathbf{A}$. For if P and $X - a$ are strongly relatively prime, it follows from Proposition 1 that there exist $c \in \mathbf{A}$ and a polynomial $Q \in \mathbf{A}[X]$ such that $cP + (X - a)Q = 1$, whence $cP(a) = 1$ and $P(a)$ is invertible. Conversely, by Euclidean division

and, if $P(a) = b^{-1}$, where $b \in \mathbf{A}$, we deduce that $1 = bP - b(X - a)R$, which shows that P and $X - a$ are strongly relatively prime.

Let $\mathbf{A}$ and $\mathbf{B}$ be two commutative rings and $f : \mathbf{A} \to \mathbf{B}$ a ring homomorphism. If $P = \sum_{i \geq 0} a_i X^i$ is a formal power series in $\mathbf{A}[[X]]$, let $\bar{f}(P)$ denote the formal power series $\sum_{i \geq 0} f(a_i) X^i$ in $\mathbf{B}[[X]]$. If P is a polynomial, so is $\bar{f}(P)$ and, if further P is monic, then $\bar{f}(P)$ is monic of the same degree as P. Finally, $P \mapsto \bar{f}(P)$ is clearly a homomorphism of $\mathbf{A}[[X]]$ to $\mathbf{B}[[X]]$ which extends $f$ and maps X to X. The notation will be constantly used in this sense in the rest of this paragraph.