= Monic polynomial quotient is finite free
{title2=$A[T]/(F)\cong A^{\deg F}$}
For any commutative <ring> $A$ and a <monic polynomial> $F$ of positive degree $D$, <monic polynomial division over a ring> gives unique representatives of degree less than $D$. Thus the classes of $1,T,\ldots,T^{D-1}$ form a <basis of a module> over $A$. This is a finite free algebra, hence flat, and the map $A\to A[T]/(F)$ is injective. The monic hypothesis makes leading-degree arguments valid even when $A$ has zero divisors.
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