General principles in factoring

A general principle in factoring polynomials is to determine whether or not the polynomial has multiple factors is the following one. First, for any polynomial $f(x)=a_0+a_1x+a_2x^2+...+a_nx^n\in F[x]$, let

$$f'(x)=a_1+2a_2x+...+na_nx^{n-1}$$

denote the derivative polynomial.

Example 2.4.5   If the field is not characteristic zero then funny things can happen when you differentiate! For example, if $f(x)=x^{25}+x^7\in {\mathbb{F}}_5[x]$ then $f'(x)=25x^{24}+7x^6=2x^6$, so $deg(f(x))=25$ while $deg(f'(x))=6$. Even stranger things than this can happen: if $f(x)=x^5+1\in {\mathbb{F}}_5[x]$ then $f'(x)=5x^4=0$. In other words, some non-constant polynomials have derivative equal to 0!

Lemma 2.4.6   If $F$ is a ring, $f(x)$ is a polynomial in $F[x]$, and $f'(x)\not=0$ (i.e., $f'(x)$ is not the zero polynomial) and then $f(x)$ has a multiple factor if and only if $gcd(f(x),f'(x))\not= 1$.

proof: The proof uses the fact that if $f(x)=g(x)^kh(x)$ then we have the ``product rule'' $f'(x)=kg(x)^{k-1}g'(x)h(x)+g(x)^kh'(x)$. This implies $g(x)^{k-1}\vert gcd(f(x),f'(x))$, which proves the lemma. $\Box$

Example 2.4.7   Let

$$f(x)=(x+2)^3(x^2+1)^5,$$

so

$$\begin{array}{c} f'(x)= 3\,\left (x+2\right )^{2}\left ({x}^... ...left (x+2\right )^{2}\left ({x}^{2} +1\right )^{4}, \end{array}$$

so $gcd(f(x),f'(x))=(x+2)^2(x^2+1)^4$.

Lemma 2.4.8   If $F$ is a field of characteristic zero and $f(x)$ is an irreducible polynomial in $F[x]$, $f(x)$ has cannot have a multiple root in any field extension $E/F$.

proof: Since $f(x)$ is an irreducible polynomial in $F[x]$, by the previous lemma and the Euclidean algorithm, there are polynomials $p(x)$ and $q(x)$ such that $f(x)p(x)+f'(x)q(x)=1$. This also holds over any field extension.

Suppose, to get a contradition, $f(x)$ had a multiple root in $E$. By the previous lemma, it would have a root in common with $f'(x)$, say $c\in E$. Plugging this into the equation $f(x)p(x)+f'(x)q(x)=1$ would yield $0=1$, which is impossible. $\Box$