Division Theorem for Polynomial Forms over Field

Theorem
Let $\left({F, +, \circ}\right)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental in $F$.

Let $F \left[{X}\right]$ be the ring of polynomial forms in $X$ over $F$.

Let $d$ be an element of $F \left[{X}\right]$ of degree $n \ge 1$.

Then $\forall f \in F \left[{X}\right]: \exists q, r \in F \left[{X}\right]: f = q \circ d + r$ such that either:
 * 1) $r = 0_F$
 * 2) $r \ne 0_F$ and $r$ has degree that is less than $n$.

Proof

 * From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$. So, if there is a counterexample to be found, it will have a degree.


 * Suppose there exists at least one counterexample.

By a version of the well-ordering principle Well-Ordering Principle, we can assign a number $m$ to the lowest degree possessed by any counterexample.

So, let $f$ denote a counterexample which has that minimum degree $m$.


 * If $m < n$, the equation $f = 0_D \circ d + f$ would show that $f$ was not a counterexample, therefore $m \ge n$.


 * If $d \backslash f$ in $F \left[{X}\right]$, there would be $\exists q \in F \left[{X}\right]: f = q \circ d + 0$ and $f$ would not be a counterexample. So $d \nmid f$ in $F \left[{X}\right]$.


 * So, suppose that $\displaystyle f = \sum_{k=0}^m {a_k \circ X^k}, d = \sum_{k=0}^n {b_k \circ X^k}, m \ge n$.

We can create the polynomial $\left({a_m \circ b_n^{-1} \circ X^{m - n}}\right) \circ d$ which has the same degree and leading coefficient as $f$.

Thus $f_1 = f - \left({a_m \circ b_n^{-1} \circ X^{m - n}}\right) \circ d$ is a polynomial of degree less than $m$, and since $d \nmid f$, is a non-zero polynomial.


 * There is no counterexample of degree less than $m$, therefore $f_1 = q_1 \circ d + r$ for some $q_1, r \in F \left[{X}\right]$, where either $r = 0_F$ or $r$ is nonzero with degree $< n$.

Hence:

... thus $f$ is not a counterexample.