Corollary

Suppose further that $\mathbf{A}$ is an integral domain and $F$ a distinguished polynomial of degree $s$. For $F$ to be extremal in $\mathbf{A}[X]$, it is necessary and sufficient that it be extremal in $B = \mathbf{A}[[X]]$.

Suppose that $F$ is not extremal in $\mathbf{A}[X]$, so that $F = f_1 f_2$, where $f_1$ and $f_2$ are non-invertible elements of $\mathbf{A}[X]$; the product of the dominant coefficients of $f_1$ and $f_2$ being equal to 1, these coefficients are invertible in $\mathbf{A}$ and the hypothesis implies that $f_1$ and $f_2$ are of degrees $> 0$ and $< s$; as the reduced polynomials $\bar{f}_1, \bar{f}_2$ satisfy $\bar{f}_1 \bar{f}_2 = X^s$, neither $\bar{f}_1$ nor $\bar{f}_2$ can be invertible in $k[[X]]$, for, if $\bar{f}_1$ were invertible, $\bar{f}_2$ would be of order $s$, which is absurd. A fortiori, neither $f_1$ nor $f_2$ is invertible in $B$ and $F$ is not extremal in $B$.

Conversely, if F is not extremal in $A[[X]]$, then $F = gh$, where neither g nor his invertible in B; their reduced orders are therefore $\geq 1$; then the distinguished polynomials $ug$ and $u^{-1}h$ of Proposition 7 are not constant, which shows that F is not extremal in $A[X]$.