Proposition 10

Let $f$ and $g$ be non-zero elements of $A[X]$ of degrees $m$ and $n$ respectively. Let $\alpha_0$ be the leading coefficient of $f$ and $\mu = \sup(n - m + 1, 0)$. There exist $u, v \in A[X]$ such that

If $\alpha_0$ is cancellable in $A$, then $u$ and $v$ are uniquely determined by these properties.

The existence of $u$ and $v$ is clear when $n < m$, because then we can take $u = 0$ and $v = g$. For $n \geq m$ we shall use induction on $n$. Let $\beta$ be the leading coefficient of $g$; if $f = \sum_{k=0}^m \alpha_k X^{m-k}$, we can write $\alpha_0^\mu g = \alpha_0^\mu \beta X^n \cdot f + \alpha_0^{\mu-1} g_1$, where $g_1 \in A[X]$ and $\deg g_1 < n$. By the induction hypothesis there exist $u_1, v \in A[X]$ such that $\alpha_0^{\mu-1} g_1 = u_1 f + v$ and $\deg v < m$. Hence

and it suffices to put $u = \alpha_0^{\mu-1} \beta X^{n-m} + u_1$.

Assuming $\alpha_0$ to be cancellable in $A$, let us now prove the uniqueness of $u$ and $v$. Let $u, v, u_1, v_1 \in A[X]$ be such that

We have $(u - u_1) f = v_1 - v$ and $\deg(v_1 - v) < m$, hence $u - u_1 = 0$ (IV, p. 9, Prop. 7) and therefore $v_1 - v = 0$.