Primitive polynomials

If all the coefficients of a polynomial with integer coefficients are divisible by an integer $d\not= \pm 1$ then we say that the polynomial is imprimitive. Otherwise, the polynomial is called primitive. For example, $2x^{100}-68x^{47}-2$ is imprimitive but $x+1$ is primitive.

Theorem 2.12.1 (Gauss' lemma)   The product of primitive polynomials is primitive.

proof: Let $p$ be a prime. Let $f(x)=a_0+a_1x+...+a_mx^m$ and $g(x)=b_0+b_1x+...+b_nx^n$ be primitive. Let $a_s$ be the smallest coefficient of $f(x)$ not divisible by $p$ and let $b_r$ be the smallest coefficient of $g(x)$ not divisible by $p$. We have

$$f(x)g(x)=c_0+c_1x+...+c_{r+s}x^{r+s}+...+c_{m+n}x^{m+n},$$

where $c_{r+s}=b_0a_{r+s}+...+b_ra_s+...+b_{r+s}a_0$. By our assumption, $p$ does not divide $c_{r+s}$. Moreover, $c_{r+s}$ is coefficient of $g(x)h(x)$ least degree which is not divisible by $p$. This implies that $f(x)g(x)$ is primitive. $\Box$

This lemma is used to show the following theorem.

Theorem 2.12.2   If $p(x)\in \mathbb{Z}[x]$ and $p(x)$ factors into a product of irreducible polynomials with coefficients in ${\mathbb{Q}}$ then $p(x)$ factors into a product of irreducible polynomials with coefficients in $\mathbb{Z}$.

proof: Suppose $p(x)=f(x)g(x)$, where $f(x),g(x)\in {\mathbb{Q}}[x]$. We can factor out common factors of the denomiators and numerators of the coefficients and put this in the form $p(x)={a\over b}f^*(x)g^*(x)$, where $f^*(x), g^*(x)\in \mathbb{Z}[x]$ are primitive polynomials, and $a,b$ are relatively prime integers. By Gauss' lemma, $f^*(x)g^*(x)$ is also primitive. Since $p(x)$ has integer coefficeints, this implies $b=1$. $\Box$