Division Theorem for Polynomial Forms over Field

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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 $n$.

Proof 1

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.


Proof 2

Suppose $\map \deg f < \map \deg d$.

Then we take $\map q X = 0$ and $\map r X = \map a X$ and the result holds.

Otherwise, $\map \deg f \ge \map \deg d$.


$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$

We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$:

$\map f X - \map d X \cdot \dfrac {a_m} {b_n} x^{m - n} = \map p X$

where $\map p X$ is some polynomial whose degree is less than that of $f$.

If $\map p X$ still has degree higher than that of $d$, we do the same thing again.

Eventually we reach:

$\map f X - \map d X \cdot \paren {\dfrac {a_m} {b_n} x^{m - n} + \dotsb} = \map r X$

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.


Proof 3

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$

Basis for the Induction

$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.

This is shown trivially to be true by taking $q = r = 0$.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k$

Then we need to show:

$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k + 1$

Induction Step

This is our induction step:

Let $f$ be such that $\map \deg f = n$.


$f = a_0 + a_1 \circ x + a_2 \circ x^2 + \cdots + a_n \circ x^n$ where $a_n \ne 0$


$d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ where $b_j \ne 0$

If $n < l$ then take $q = 0, r = f$.

If $n \ge l$, consider:

$c := f - a_n b_j^{-1} x^{n - j} d$

This has been carefully arranged so that the coefficient of $x^n$ in $c$ is zero.

Thus $\map \deg c < n$.

Therefore, by the induction hypothesis:

$c = d q_0 + r$

where $\map \deg r < \map \deg d$.


\(\ds f\) \(=\) \(\ds d \paren {q_0 + a_n b_j^{-1} x^{n - j} } + r\)
\(\ds \) \(=\) \(\ds d q + r\) where $\map \deg r < \map \deg d$ and $q = q_0 + a_n b_j^{-1} x^{n - j}$

Thus the existence of $q$ and $r$ have been established.

As for uniqueness, assume:

$d q + r = d q' + r'$

with $\map \deg r < \map \deg d, \map \deg {r'} < \map \deg d$


$d \paren {q - q'} = r' - r$

By Degree of Sum of Polynomials:

$\map \deg {r' - r} \le \max \set {\map \deg {r'}, \map \deg r} < \map \deg d$

and by Degree of Product of Polynomials over Integral Domain:

$\map \deg {d \paren {q - q'} } = \map \deg d + \map \deg {q - q'}$

That is:

$\map \deg d < \map \deg d + \map \deg {q - q'}$

and the only way for that to happen is for:

$\map \deg {q - q'} = -\infty$

that is, for $q - q'$ to be the null polynomial.

That is, $q - q' = 0_F$ and by a similar argument $r' - r = 0_F$, demonstrating the uniqueness of $q$ and $r$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
$(1): \quad r = 0_F$
$(2): \quad r \ne 0_F$ and $r$ has degree that is less than $n$.


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