📂Abstract Algebra

Proof of the Factor Theorem

Theorem¹

Let $f(x) \in F [ x ]$.

$$ f(a) = 0 \iff f(x) = (x-a) q(x) $$

Explanation

This theorem guarantees the existence of the factorization we have done to death since middle school. Note that facts like the division theorem and the factor theorem are meaningful only when the degree of the polynomial is finite.

Proof

$( \implies )$

Division theorem: 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 $m$.

By the division theorem, there uniquely exist $q(x) , r(x) \in F [ x ]$ satisfying $f(x) = (x-a) q(x) + r(x)$. Since the degree of $(x -a )^{1}$ is $1$, the degree of $r(x)$ is $0$, that is, $r(x) = c$ for some constant $c$. Then, substituting $x = a$ into $$ f(x) = (x-a) q(x) + c $$ gives $$ 0 = f(a) = 0 q(a) + c = c $$ Therefore $f(x) = (x-a) q(x)$.

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$( \impliedby )$ Substituting $x = a$ into $$ f(x) = (x-a) q(x) $$ gives $f(a) = 0 q(a) = 0$.

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