Proposition 2

Let $\mathbf{A}$ and $\mathbf{B}$ be two commutative rings, $f$ a homomorphism from $\mathbf{A}$ to $\mathbf{B}$ and $P, P'$ two polynomials in $\mathbf{A}[X]$. If $P$ and $P'$ are strongly relatively prime in $\mathbf{A}[X]$, then $\bar{f}(P)$ and $\bar{f}(P')$ are strongly relatively prime in $\mathbf{B}[X]$. The converse is true iff is surjective, if its kernel is contained in the Jacobson radical of $\mathbf{A}$ and if $P$ is monic.

Suppose that P and P' are strongly relatively prime; then there exist polynomials Q, Q' in $\mathbf{A}[X]$ such that $PQ + P'Q' = 1$; we deduce that

whence the first assertion. To show the second, let a denote the kernel of $f$;

let $E = A[X]$ and $F$ the ideal of $E$ generated by $P$ and $P'$; as $f$ is surjective and $\bar{f}(P)$ monic, Proposition 1 shows that for every polynomial $T \in A[X]$ there exist two polynomials $Q, Q'$ in $A[X]$ such that

whence the relation $E = F + aE$. Now, $E/F$ is a finitely generated $A$-module, for every polynomial is congruent mod. $P$ to a polynomial of degree $< \deg(P)$, $P$ being monic. As $E/F = a(E/F)$ and $a$ is contained in the Jacobson radical of $A$, Nakayama's Lemma shows that $E/F = 0$ (Algebra, Chapter VIII, § 6, no. 3, Corollary 2 to Proposition 6 ), which means that $P$ and $P'$ are strongly relatively prime.