Division Theorem for Polynomial Forms over Field/Proof 1

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Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental over $F$.

Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.

Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$.

Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:

$(1): \quad r = 0_F$


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


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.

Aiming for a contradiction, suppose 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 $F \sqbrk X$.


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

and $f$ would not be a counterexample.

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

So, suppose that:

\(\ds f\) \(=\) \(\ds \sum_{k \mathop = 0}^m {a_k \circ X^k}\)
\(\ds d\) \(=\) \(\ds \sum_{k \mathop = 0}^n {b_k \circ X^k}\)
\(\ds m\) \(\ge\) \(\ds n\)

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 $m$.


$f_1 = q_1 \circ d + r$

for some $q_1, r \in F \sqbrk X$, where either:

$r = 0_F$


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


\(\ds f\) \(=\) \(\ds f_1 + \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d\)
\(\ds \) \(=\) \(\ds \paren {q_1 + a_m \circ b_n^{-1} \circ X^{m - n} } \circ d + r\)

Thus $f$ is not a counterexample.

From this contradiction follows the result.