Division Theorem for Polynomial Forms over a 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

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.

• Suppose there exists at least one counterexample.

By a version of the well-ordering principle 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 $m \ge n$.

• If $d \backslash f$ in $F \left[{X}\right]$, there would be $\exists q \in F \left[{X}\right]: f = q \circ d + 0$ and $f$ would not be a counterexample. So $d \nmid f$ in $F \left[{X}\right]$.

• So, suppose that $\displaystyle f = \sum_{k=0}^m {a_k \circ X^k}, d = \sum_{k=0}^n {b_k \circ X^k}, m \ge n$.

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

Thus $f_1 = f - \left({a_m \circ b_n^{-1} \circ X^{m - n}}\right) \circ d$ is a polynomial of degree less than $m$, and since $d \nmid f$, is a non-zero polynomial.

• There is no counterexample of degree less than $m$, therefore $f_1 = q_1 \circ d + r$ for some $q_1, r \in F \left[{X}\right]$, where either $r = 0_F$ or $r$ is nonzero with degree $< n$.

Hence:

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

... thus $f$ is not a counterexample.

$\blacksquare$

Proof 2

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.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.27$: Theorem $52$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 65.1$