Division Theorem for Polynomial Forms over Field/Proof 2

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

Let $X$ be transcendental over $F$.

Let $F \left[{X}\right]$ be the ring of polynomials 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:
 * $(1): \quad r = 0_F$

or:
 * $(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.

Proof
Suppose $\deg \left({f}\right) < \deg \left({d}\right)$.

Then we take $q \left({X}\right) = 0$ and $r \left({X}\right) = a \left({X}\right)$ and the result holds.

Otherwise, $\deg \left({f}\right) \ge \deg \left({d}\right)$.

Let:
 * $f \left({X}\right) = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$


 * $d \left({X}\right) = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$

We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$:
 * $f \left({X}\right) - d \left({X}\right) \cdot \dfrac {a_m} {b_n}x^{m-n} = p \left({X}\right)$

where $p \left({X}\right)$ is some polynomial whose degree is less than that of $f$.

If $p \left({X}\right)$ still has degree higher than that of $d$, we do the same thing again.

Eventually we reach:
 * $f \left({X}\right) - d \left({X}\right) \cdot \left({\dfrac {a_m} {b_n}x^{m-n} + \cdots}\right) = r \left({X}\right)$

where either $r = 0_F$ or $r$ has degree that is less than $n$.

This approach can be formalised using the Principle of Complete Induction.