Abstract Algebra: Theory and Applications, Remixed!

Proof.

Suppose that

where the $b_i$'s and the $c_i$'s are integers. We can rewrite $p(x)$ as

where $d_0,\dots ,d_n$ are integers. Let $d$ be the greatest common divisor of $d_0,\dots ,d_n\text{.}$ Then

where $d_i=d{a}_i$ and the $a_i$'s are relatively prime. Reducing $d/(c_0\cdots c_n)$ to its lowest terms, we can write

where $gcd(r,s)=1\text{.}$

Proof.

By Lemma 6.20, we can assume that

where the $a_i$'s are relatively prime and the $b_i$'s are relatively prime. Consequently,

where $c/d$ is the product of $c_1/d_1$ and $c_2/d_2$ expressed in lowest terms. Hence, $dp(x)=c{\alpha }_1(x){\beta }_1(x)\text{.}$

If $d=1\text{,}$ then $c{a}_m{b}_n=1$ since $p(x)$ is a monic polynomial. Hence, either $c=1$ or $c=-1\text{.}$ If $c=1\text{,}$ then either $a_m=b_n=1$ or $a_m=b_n=-1\text{.}$ In the first case $p(x)={\alpha }_1(x){\beta }_1(x)\text{,}$ where ${\alpha }_1(x)$ and ${\beta }_1(x)$ are monic polynomials with $\mathrm{deg}\alpha (x)=\mathrm{deg}{\alpha }_1(x)$ and $\mathrm{deg}\beta (x)=\mathrm{deg}{\beta }_1(x)\text{.}$ In the second case $a(x)=-{\alpha }_1(x)$ and $b(x)=-{\beta }_1(x)$ are the correct monic polynomials since $p(x)=(-{\alpha }_1(x))(-{\beta }_1(x))=a(x)b(x)\text{.}$ The case in which $c=-1$ can be handled similarly.

Now suppose that $d\ne 1\text{.}$ Since $gcd(c,d)=1\text{,}$ there exists a prime $p$ such that $p\mid d$ and $p\nmid c\text{.}$ Also, since the coefficients of ${\alpha }_1(x)$ are relatively prime, there exists a coefficient $a_i$ such that $p\nmid a_i\text{.}$ Similarly, there exists a coefficient $b_j$ of ${\beta }_1(x)$ such that $p\nmid b_j\text{.}$ Let ${\alpha }_1^{\prime }(x)$ and ${\beta }_1^{\prime }(x)$ be the polynomials in ${\mathbb{Z}}_p[x]$ obtained by reducing the coefficients of ${\alpha }_1(x)$ and ${\beta }_1(x)$ modulo $p\text{.}$ Since $p\mid d\text{,}$ ${\alpha }_1^{\prime }(x){\beta }_1^{\prime }(x)=0$ in ${\mathbb{Z}}_p[x]\text{.}$ However, this is impossible since neither ${\alpha }_1^{\prime }(x)$ nor ${\beta }_1^{\prime }(x)$ is the zero polynomial and ${\mathbb{Z}}_p[x]$ is an integral domain. Therefore, $d=1$ and the theorem is proven.

Proof.

Let $p(x)$ have a zero $a\in \mathbb{Q}\text{.}$ Then $p(x)$ must have a linear factor $x-a\text{.}$ By Gauss's Lemma, $p(x)$ has a factorization with a linear factor in $\mathbb{Z}[x]\text{.}$ Hence, for some $\alpha \in \mathbb{Z}$

Thus $a_0/\alpha \in \mathbb{Z}$ and so $\alpha \mid a_0\text{.}$

Proof.

By Gauss's Lemma, we need only show that $f(x)$ does not factor into polynomials of lower degree in $\mathbb{Z}[x]\text{.}$ Let

be a factorization in $\mathbb{Z}[x]\text{,}$ with $b_r$ and $c_s$ not equal to zero and $r,s<n\text{.}$ Since $p^2$ does not divide $a_0=b_0{c}_0\text{,}$ either $b_0$ or $c_0$ is not divisible by $p\text{.}$ Suppose that $p\nmid b_0$ and $p\mid c_0\text{.}$ Since $p\nmid a_n$ and $a_n=b_r{c}_s\text{,}$ neither $b_r$ nor $c_s$ is divisible by $p\text{.}$ Let $m$ be the smallest value of $k$ such that $p\nmid c_k\text{.}$ Then

is not divisible by $p\text{,}$ since each term on the right-hand side of the equation is divisible by $p$ except for $b_0{c}_m\text{.}$ Therefore, $m=n$ since $a_i$ is divisible by $p$ for $m<n\text{.}$ Hence, $f(x)$ cannot be factored into polynomials of lower degree and therefore must be irreducible.