📂Abstract Algebra

Division Theorem Proof

Theorem ¹

Given $a_{n} \ne 0$ and $b_{m} \ne 0$, as well as $n > m > 0$, let the two elements of $F [ x ]$ be $$ f(x) = a_{n} x^{n} + \cdots + a_{1} x + a_{0} \\ g(x) = b_{m} x^{m} + \cdots + b_{1} x + b_{0} $$. Then, there exists a unique $q(x), r(x) \in F [ x ]$ that satisfies $f(x) = g(x) q(x) + r(x)$. The degree of $r$ is less than that of $m$.

Explanation

This fact doesn’t necessarily require a theorem but holds significant value in its algebraically rigorous proof.

Proof

Let’s set $$ S : = \left\{ f(x) - g(x) s(x) \ : \ s(x) \in F [ x ] \right\} $$. Saying $0 \in S$ means that there exists $s(x)$ satisfying $f(x) - g(x) s(x) = 0$. In this case, simply setting $q(x) = s(x)$ and $r(x) = 0$ suffices, and it’s also necessary to check other conditions for consistency.

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Part 1. Existence

Let’s name the polynomial with the lowest degree in $S$ as $r(x)$. Essentially, this means for some $q(x) = s(x)$, it’s $r(x) = f(x) - g(x) s(x)$.

Assuming that $r(x) := c_{t} x^{t} + \cdots + c_{1} x + c_{0}$ equates to $t \ge m$, $$ \begin{align*} & f(x) - q(x) g(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x) \\ =& r(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x) \\ =& r(x) - {{c_{t}} \over {b_{m}}} x^{t-m} \left( b_{m} x^{m} + \cdots + b_{1} x + b_{0} \right) \\ =& r(x) -c_{t} x^{t} - {{c_{t}} \over {b_{m}}} x^{t-m} \left( b_{m-1} x^{m-1} + \cdots + b_{1} x + b_{0} \right) \end{align*} $$ Therefore, the degree of $\displaystyle f(x) - q(x) g(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x)$ is less than that of $t$. However, $$ f(x) - \left[ q(x) + {{c_{t}} \over {b_{m}}} x^{t-m} \right] g(x) \in S $$ implies a contradiction to the assumption that $r(x)$ has the smallest degree among the polynomials in $S$.

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Part 2. Uniqueness

Assuming $q_{1} \ne q_{2}$ and $r_{1} \ne r_{2}$ equate to $\begin{cases} f(x) = g(x) q_{1} (x) + r_{1} (x) \\ f(x) = g(x) q_{2} (x) + r_{2} (x) \end{cases}$ and subtracting one from the other gives $$ g(x) [ q_{2}(x) - q_{1}(x) ] = r_{2}(x) - r_{1}(x) $$ Since the degree of $r_{2}(x) - r_{1}(x)$ is less than that of $g( X)$, it must be $q_{2} - q_{1} = 0$. Therefore, this contradicts the assumption, as it implies $r_{2}(x) - r_{1}(x) = 0$.

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