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 $p$ a prime such that:
 * $(1): \quad$ $p \backslash a_i \iff i \ne d$
 * $(2): \quad$ $p^2 \nmid a_0$

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

Then $f$ is irreducible in $\Q[x]$.

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

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

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

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

Therefore, $a_0 = b_0c_0$ and by condition $2$ it follows, without loss of generality, that $p\mid b_0$ and $p\nmid c_0$.

We also have $a_d = b_ec_f$, and by condition $1$ it follows that there exists a smallest positive $i$ such that $p\nmid b_i$.

Note that $if$.

By the minimality of $i$, every term in the sum is divisible by $p$ with the exception of the last term, which is not divisible by $p$ since neither $c_0$ nor $b_i$ is divisible by $p$.

Thus, we conclude that $p \nmid a_i$, which contradicts condition $1$.

Therefore, $f$ is irreducible.