Schönemann-Eisenstein Theorem

Theorem
Let $f \left({x}\right) = a_d x^d + a_{d-1} x^{d-1} + \dotsb + a_0 \in \Z \left[{x}\right]$ be a polynomial over the ring of polynomial forms $\Z \left[{x}\right]$.

Let $p$ be a prime such that:


 * $(1): \quad p \mathrel \backslash a_i \iff i \ne d$
 * $(2): \quad p^2 \nmid a_0$

where $p \mathop \backslash a_i$ signifies that $p$ is a divisor of $a_i$.

Then $f$ is irreducible in $\Q \left[{x}\right]$.

Proof
By Gauss's Lemma, it suffices to show that $f$ is irreducible in $\Z \left[{x}\right]$.

Aiming for a contradiction, suppose that $f = g h$ where $g, h \in \Z \left[{x}\right]$ are both non-constant.

Let:
 * $g \left({x}\right) = b_e x^e + b_{e-1}x^{e-1} + \dotsb + b_0$
 * $h \left({x}\right) = c_f x^f + c_{f-1} x^{f-1} + \dotsb + c_0$

Then we have for each $i$:
 * $\displaystyle a_i = \sum_{j + k \mathop = i} {b_j c_k}$

In particular, it follows that:
 * $a_0 = b_0 c_0$

Possibly after exchanging $g$ and $h$, we may arrange that:
 * $p \nmid c_0$

by condition $(2)$.

From condition $(1)$, it follows that then necessarily:
 * $p \mathrel \backslash b_0$

We also have:
 * $a_d = b_e c_f$

and by condition $(1)$:
 * $p \nmid a_d$

and hence:
 * $p \nmid b_e$

It follows that there exists a smallest positive $i$ such that:
 * $p \nmid b_i$

Naturally, $i \le e$.

By assumption, both $g$ and $h$ are non-constant.

Hence by Degree of Product of Polynomials over Integral Domain:
 * $i < d$

Consider:
 * $a_i = b_0 c_i + b_1 c_{i-1} + \dotsb + b_i c_0$

with the convention that $c_j = 0$ if $j > f$.

By the minimality of $i$, it follows that:
 * $p \mathrel \backslash b_k$

for $0 \le k < i$.

Also, since neither $c_0$ nor $b_i$ is divisible by $p$, the last term $b_i c_0$ is not divisible by $p$.

Thus, we conclude that:
 * $p \nmid a_i$

which contradicts condition $(1)$.

Therefore, $f$ is irreducible.

Also known as
This result is also known as Eisenstein's criterion.