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:
 * $(1): \quad r = 0_F$
 * $(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.

Proof
As a field is also an integral domain, we can directly use the result Division Theorem for Polynomial Forms over an Integral Domain.