📂Abstract Algebra

Proof of the Division Theorem

Theorem¹

For $a_{n} \ne 0$ and $b_{m} \ne 0$, and $n > m > 0$, let 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 uniquely exist $q(x), r(x) \in F [ x ]$ satisfying $f(x) = g(x) q(x) + r(x)$. The degree of $r$ is less than $m$.

Explanation

This is not a fact that can only be known through a theorem, but it is significant as an algebraically rigorous proof.

Proof

Let $$ S : = \left\{ f(x) - g(x) s(x) \ : \ s(x) \in F [ x ] \right\} $$ Saying that $0 \in S$ means that there exists $s(x)$ satisfying $f(x) - g(x) s(x) = 0$. In that case, one can simply take $q(x) = s(x)$ and $r(x) = 0$, and it must be checked that the theorem holds in the other cases as well.

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Part 1. Existence

Let $r(x)$ be the polynomial of lowest degree in $S$. In other words, $r(x) = f(x) - g(x) s(x)$ for some $q(x) = s(x)$.

For $r(x) := c_{t} x^{t} + \cdots + c_{1} x + c_{0}$, if we assume $t \ge m$, then $$ \begin{align*} & f(x) - q(x) g(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x) \\ =& r(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x) \\ =& r(x) - {{c_{t}} \over {b_{m}}} x^{t-m} \left( b_{m} x^{m} + \cdots + b_{1} x + b_{0} \right) \\ =& r(x) -c_{t} x^{t} - {{c_{t}} \over {b_{m}}} x^{t-m} \left( b_{m-1} x^{m-1} + \cdots + b_{1} x + b_{0} \right) \end{align*} $$ Therefore, the degree of $\displaystyle f(x) - q(x) g(x) - {{c_{t}} \over {b_{m}}} x^{t-m} g(x)$ is less than $t$. However, since $$ f(x) - \left[ q(x) + {{c_{t}} \over {b_{m}}} x^{t-m} \right] g(x) \in S $$ this contradicts the premise that $r(x)$ has the smallest degree among the polynomials in $S$.

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Part 2. Uniqueness

Assume, for $q_{1} \ne q_{2}$ and $r_{1} \ne r_{2}$, that $\begin{cases} f(x) = g(x) q_{1} (x) + r_{1} (x) \\ f(x) = g(x) q_{2} (x) + r_{2} (x) \end{cases}$, and subtracting the two equations from each 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 that $q_{2} - q_{1} = 0$. Then $r_{2}(x) - r_{1}(x) = 0$, which contradicts the assumption.

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