Eisenstein's Irr. Polynomial Crit. for Poly. over Z that are Irr. over Q

Eisenstein's Irreducible Polynomial Criterion for Polynomials over Z that are Irreducible over Q

Theorem 1 (Eisenstein's Irreducible Polynomial Criterion): Let $f \in \mathbb{Z}[x]$ with $f(x) = a_0 + a_1x + ... + a_nx^n$. If there exists a prime number $p$ such that: 1) $p | a_0, a_1, ..., a_{n-1}$. 2) $p$ does not divide $a_n$. 3) $p^2$ does not divide $a_0$. Then $f$ is irreducible over $\mathbb{Q}$.

For example, for any prime $p$ consider the polynomial:

(1)

\begin{align} \quad f_p(x) = 1 + x + ... + x^{p-1} \end{align}

We claim that $f_p$ is irreducible over $\mathbb{Q}$. To show this, note that:

(2)

\begin{align} \quad f_p(x) = \frac{x^p - 1}{x - 1} \end{align}

Substituting $x$ for $x + 1$ yields:

(3)

\begin{align} \quad f_p(x+1) = \frac{(x+1)^p - 1}{(x + 1)-1} = \frac{(x + 1)^p - 1}{x} = \frac{\left (\binom{p}{0}x^p + \binom{p}{1}x^{p-1} + ... + \binom{p}{p}1 \right ) - 1}{x} = \binom{p}{0} x^{p-1} + \binom{p}{1}x^{p-2} + ... + \binom{p}{p-1} \end{align}

Note that $\displaystyle{p | \binom{p}{p-1}, ..., \binom{p}{1}}$ but $p$ does not divide $\displaystyle{\binom{p}{0} = 1}$, so (1) and (2) holds. Also, $p^2$ does not divide $\binom{p}{p-1} = p$, so (3) holds. Hence, by Eisenstein's criterion we have that $f_p(x + 1)$ is irreducible over $\mathbb{Q}$. Hence $f_p(x)$ is irreducible over $\mathbb{Q}$.