Eisenstein’s criterion

We assume that $P$ is not irreducible over $\mathbb{Q}[X]$, and hence over $\mathbb{Z}[X]$.

There therefore exist non-constant polynomials $A,B\in\mathbb{Z}[X]$ such that

$$P=A\times B.$$

We use the notation

$$A=\sum_{i=0}^{r}a_iX^i \qquad\text{and}\qquad B=\sum_{j=0}^{s}b_jX^j.$$

We have

$$p\mid p_0=a_0b_0.$$

We may therefore assume, without loss of generality, that

$$p\mid a_0\qquad\text{and}\qquad p\nmid b_0,$$

since otherwise $p^2\mid p_0$, which is excluded.

Let us prove by contradiction that $p$ cannot divide all the $a_i$.

Indeed, if $p\mid a_i$ for every $i$, then

$$p\mid \operatorname{dom}(A)\operatorname{dom}(B)=p^n,$$

which is excluded.

We thus denote by $i_0$ the smallest integer such that

$$p\nmid a_{i_0}.$$

We have $i_0<n$, since $B$ is non-constant.

Now,

$$\underbrace{p_{i_0}=a_{i_0}b_0}_{\text{not divisible by }p} + \underbrace{a_{i_0-1}b_1+\cdots+a_0b_{i_0}}_{\text{divisible by }p}.$$

Indeed, by the definition of $i_0$, we have

$$p\mid a_0,\ldots,a_{i_0-1},$$

and hence every term in the second sum is divisible by $p$. On the other hand,

$$p\nmid a_{i_0}b_0,$$

since $p\nmid a_{i_0}$ and $p\nmid b_0$.

Thus,

$$p\nmid p_{i_0},$$

which is a contradiction, since $i_0<n$ and, by hypothesis,

$$p\mid p_0,\ldots,p_{n-1}.$$

Therefore, $P$ is irreducible in $\mathbb{Q}[X]$.