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

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.