Corollary (« Euclidean division of polynomials »)

— Let $f$ be a non-zero element of $A[X]$ whose leading coefficient is invertible and $m = \deg f$. (i) For every $g \in A[X]$ there exist $u,\ v \in A[X]$ such that

Moreover, these conditions determine $u$ and $v$ uniquely. (ii) The sub-A-modules $A + AX + \ldots + AX^{m-1}$ and $fA[X]$ of $A[X]$ are supplementary in $A[X]$. (iii) Assume $f$ non-constant and consider $A[X]$ as an $A[T]$-*module* by means of the homomorphism $u(T) \mapsto u(f(X))$ of $A[T]$ into $A[X]$. Then $A[X]$ is a free $A[T]$-*module* with basis $(1, X, \ldots, X^{m-1})$. Assertions (i) and (ii) are immediate consequences of Prop. 10. Let us prove (iii). Let $\psi$ be the homomorphism $v \mapsto v(f(X), X)$ of $A[T, X]$ into $A[X]$. Consider $A[T, X]$ first as polynomial ring in $T$ with coefficients in $A[X]$; Cor. 1 of IV, p. 5 shows that the kernel $a$ of $\psi$ is the ideal $(T - f(X))$ of $A[T, X]$. Consider now $A[T, X]$ as polynomial ring in $X$ with coefficients in $A[T]$; then $\psi$ is an $A[T]$-linear mapping of $A[T][X]$ into $A[X]$. Assertion (ii) above (applied to the polynomial $f(X) - T$ in $X$ with coefficients in $A[T]$) shows that $(1, X, \ldots, X^{m-1})$ is a basis of an $A[T]$-submodule of $A[T, X]$ supplementary to $a$. Since $\psi(X^i) = X^i$ for every integer $i \geq 0$, (iii) follows at once.

With the notations of (i) we shall say that $u$ is the quotient and $v$ the remainder in the Euclidean division of $g$ by $f$; for the remainder to vanish it is necessary and sufficient that $f$ should divide $g$.