Division Theorem for Polynomial Forms over Field/Proof 1

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.

there exists at least one counterexample.

By a version of the 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 $f = 0_F \circ d + f$ would show that $f$ was not a counterexample.

Therefore $m \ge n$.

Suppose $d \divides f$ in $F \sqbrk X$.

Then:
 * $\exists q \in F \sqbrk X: f = q \circ d + 0_F$

and $f$ would not be a counterexample.

So $d \nmid f$ in $F \sqbrk X$.

So, suppose that:

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

Thus $f_1 = f - \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ is a polynomial of degree less than $m$.

Since $d \nmid f$, $f_1$ is a non-zero polynomial.

There is no counterexample of degree less than $m$.

Therefore:
 * $f_1 = q_1 \circ d + r$

for some $q_1, r \in F \sqbrk X$, where either:
 * $r = 0_F$

or:
 * $r$ is non-zero with degree strictly less than $n$.

Hence:

Thus $f$ is not a counterexample.

From this contradiction follows the result.