Eisenstein's Irreducibility Criterion

Eisenstein's irreducibility criterion is a method for proving that a polynomial with integer coefficients is irreducible (that is, cannot be written as a product of two polynomials of smaller degree with integer coefficients). Due to its specific requirements, it is not generally applicable to most polynomials, but it is useful for exhibiting examples of carefully chosen polynomials which are irreducible.

Eisenstein's Irreducibility Criterion

Let $f(x) = a_nx^n + a_{n-1} x^{n-1} + \cdots + a_0$ be a polynomial with integer coefficients. Suppose that there exists a prime $p$, such that

1. $p \not \mid a_n$
2. $p \mid a_{n-1} , a_{n-2} , \ldots, a_1, a_0$
3. $p^2 \not \mid a_0$.

Then $f(x)$ is irreducible over the integers. $_\square$

It is a corollary of Gauss's lemma that $f(x)$ is also irreducible over the rational numbers.

Suppose not, then

$$f(x) = \left(x^r + b_{r-1} x^{r-1} + \cdots + b_0\right)\left(x^s + c_{s-1} x^{s-1} + \cdots + c_0\right)$$

with $0 <r$ and $s < n$. Take $i$ least such that $p \not \mid b_i$ (where we allow the case $i = r$ and $b_r = 1$), and $j$ least such that $p \not \mid c_j$ (where we allow the case $j = s$ and $c_s = 1$). Now the $(i+j)^\text{th}$ coefficient $a_{i+j}$ of the product is

$$a_{i+j} = b_i c_j + (b_{i-1} c_{j+1} + \cdots) + (b_{i+1} c_{j-1} + \cdots)$$

and so is a sum of $b_i c_j$ and terms divisible by $p$. As $p \not \mid b_i c_j$, $p \not \mid a_{i+j}$. But $p^2 \not \mid a_0 = b_0 c_0,$ so $p$ cannot divide both $b_0$ and $c_0$. Thus, either $i=0$ or $j=0,$ so $i+j < n$. So we contradict $p \mid a_{i+j}$.

Thus, by contradiction, Eisenstein's irreducibility criterion holds true. $_\square$

Show that there is an irreducible polynomial with integer coefficients of degree $n,$ for any positive integer $n.$

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The polynomial $x^n-2$ is irreducible by Eisenstein's criterion (with $p=2).$ $_\square$

Let $p$ be a prime number. Show that $\Phi_p(x) = \frac{x^p-1}{x-1}$ is irreducible.

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$\Phi_p(x) = x^{p-1} + x^{p-2} + \cdots + 1$ does not satisfy the conditions of Eisenstein's criterion. But $$\Phi_p(x+1) = \frac{(x+1)^p-1}{x} = x^{p-1} + \binom{p}{1} x^{p-2} + \binom{p}{2} x^{p-3} + \cdots + \binom{p}{p-2} x + \binom{p}{p-1}$$ does, because $\binom{p}{k}$ is divisible by $p$ for $1\le k \le p-1$, and $\binom{p}{p-1} = p$ is not divisible by $p^2$. So $\Phi_p(x+1)$ is irreducible. But if $\Phi_p(x)$ factored as $f(x)g(x)$, then so would $\Phi_p(x+1) = f(x+1)g(x+1)$. So $\Phi_p(x)$ is irreducible as well. $_\square$