Schönemann-Eisenstein Theorem

Theorem
Let $f(x) = a_dx^d + a_{d-1}x^{d-1} + \ldots + a_0 \in \Z \left[{x}\right]$ be a polynomial over the Ring of Polynomial Forms $\Z \left[{x}\right]$.

Let $p$ be a prime such that:


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

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 $\Z \left[{x}\right]$.

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

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

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

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 $p \mathop \backslash b_0$.

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, $i \le e$.

Since both $g$ and $h$ are assumed to be non-constant, it follows that $i < d$ by Degree of Product of Polynomials over Integral Domain.

Consider $a_i = b_0 c_i + b_1 c_{i-1} + \ldots + b_i c_0$, with the convention that $c_j = 0$ if $j > f$.

By the minimality of $i$, it follows that $p \mathop \backslash b_k$ 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.