Division Theorem for Polynomial Forms over Field/Proof 3

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Theorem

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:

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

or:

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


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\forall f \in F \left[{X}\right]: \exists q, r \in F \left[{X}\right]: f = q \circ d + r$ provided that $\deg \left({f}\right) < n$


$P \left({0}\right)$ is the statement that $q$ and $r$ exist when $f = 0$.

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


Basis for the Induction

$P \left({0}\right)$ 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 $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$\forall f \in F \left[{X}\right]: \exists q, r \in F \left[{X}\right]: f = q \circ d + r$ provided that $\deg \left({f}\right) < k$


Then we need to show:

$\forall f \in F \left[{X}\right]: \exists q, r \in F \left[{X}\right]: f = q \circ d + r$ provided that $\deg \left({f}\right) < k + 1$


Induction Step

This is our induction step:

Let $f$ be such that $\deg \left({f}\right) = n$.

Let:

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

Let:

$\displaystyle d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ 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 $\deg \left({c}\right) < n$.

Therefore, by the induction hypothesis:

$c = d q_0 + r$

where $\deg \left({r}\right) < \deg \left({d}\right)$.

Therefore:

$f = d \left({q_0 + a_n b_j^{-1} x^{n-j}}\right) + r$
\(\displaystyle f\) \(=\) \(\displaystyle d \left({q_0 + a_n b_j^{-1} x^{n-j} }\right) + r\)
\(\displaystyle \) \(=\) \(\displaystyle d q + r\) where $\deg \left({r}\right) < \deg \left({d}\right)$ 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 $\deg \left({r}\right) < \deg \left({d}\right), \deg \left({r'}\right) < \deg \left({d}\right)$

Then:

$d \left({q - q'}\right) = r' - r$

By Degree of Sum of Polynomials:

$\deg \left({r' - r}\right) \le \max \left\{{\deg \left({r'}\right), \deg \left({r}\right)}\right\} < \deg \left({d}\right)$

and by Degree of Product of Polynomials over Integral Domain:

$\deg \left({d \left({q - q'}\right)}\right) = \deg \left({d}\right) \deg \left({d}\right) + \deg \left({q - q'}\right)$

That is:

$\deg \left({d}\right) < \deg \left({d}\right) + \deg \left({q - q'}\right)$

and the only way for that to happen is for:

$\deg \left({q - q'}\right) = - \infty$

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 $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

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

$\blacksquare$


Sources