Proposition 1

Let $\mathbf{A}$ be a commutative ring and $P$ and $P'$ two strongly relatively prime polynomials in $\mathbf{A}[X]$. Suppose that $P$ is monic and of degree $s$. Then every polynomial $T$ in $\mathbf{A}[X]$ may be written uniquely in the form

where $Q \in \mathbf{A}[X]$, $Q' \in \mathbf{A}[X]$ and $\deg(Q') < s$. If further $\deg(T) \leq t$ and $\deg(P') \leq t - s$, then $\deg(Q) \leq t - s$.

As $P$ is monic, $PR \neq 0$ for every polynomial $R \neq 0$ of $\mathbf{A}[X]$ and in this case $\deg(PR) = s + \deg(R)$.

Let $T$ be any polynomial in $\mathbf{A}[X]$. As the ideal generated by $P$ and $P'$ is the whole of $\mathbf{A}[X]$, there exist polynomials $Q_1$ and $Q'$ such that

as $P$ is monic of degree $s$, Euclidean division (Algebra, Chapter IV, § 1, no. 5) shows that there exist two polynomials $Q', Q''$ such that $Q'_1 = PQ'' + Q'$ where $\deg(Q') < s$; then we deduce that where $Q = Q_1 + P'Q''$. To show the uniqueness of formula (1), it is sufficient to prove that the relations imply $Q = Q' = 0$. Now, if (2) holds, $P$ divides $-PQ = P'Q'$ and, as $P$ and

P' are strongly relatively prime, P divides Q' by Lemma 1; if Q' ≠ 0, there would exist a polynomial S ≠ 0 such that Q' = PS, whence

which is a contradiction. We conclude that Q' = 0, whence PQ = 0 and finally Q = 0 by the remark at the beginning.

Finally, suppose that $\deg(T) \leq t$ and $\deg(P') \leq t - s$; with the polynomial T in the form (1),

and therefore

whence $\deg(Q) \leq t - s$.