Irreducibility criteria of polynomials

In this article, I will prove both the Gauss primitive theorem and Eisenstein’s criterion using one proof method, and then try to generalize irreducibility criteria in $\Bbb{Q}[x]$ . I think this article does a pretty neat job of unifying two criteria that most textbooks present as arbitrary and unrelated. As always, I claim this work to be completely original. I have not been influenced by any texts or other sources that I might have come across.

Gauss’s lemma states that

$(a_0+a_1x+\dots +a_mx^m)(b_0+b_1x+\dots+b_nx^n)\in\Bbb{Z}[x]$

is primitive if the individual polynomials are primitive (they don’t have any non-unit constant factors dividing all the coefficients). Proof: The expanded polynomial is

$a_0b_0+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_0)x^2+\dots +a_mb_n x^{m+n}$ .

Let us assume that $p$ divides all coefficients, where $p$ is a prime element in $\Bbb{Z}$ . There are $m+n+1$ coefficients. If $p|a_0b_0$ , then it must divide at least one of $a_0$ and $b_0$ . Let us assume it divides $a_0$ . Moving on to the coefficient of $x$ , we see that for $p|(a_0b_1+a_1b_0)$ to be true, $p$ must divide at least one of $a_1$ and $b_0$ .

Continuing like this for $m+n+1$ coefficients, regardless of whether they are $0$ or not, we see that at every step, we get a new $a_i$ or $b_j$ that is proved to be divisible by $p$ . By the $(m+n+1)$ th step, even in the worst case scenario, we have one polynomial such that all of its coefficients are divisible by $p$ . This contradicts the assumption that both polynomials were primitive.

Now we move on to Eisenstein’s criterion: It states that for a polynomial

$a_0+a_1x+a_2x^2+\dots+a_mx^m\in\Bbb{Z}[x]$

if $p|\{a_0,a_1,\dots,a_{m-1}\}$ , $p\not| a_m$ and $p^2\not|a_0$ , then the polynomial is irreducible. Let us assume that the polynomial is reducible-

$a_0+a_1x+a_2x^2+\dots+a_mx^m=(c_0+c_1x+\dots c_hx^h)(d_0+d_1x+\dots+d_kx^k)$ .

As $p|c_0d_0=a_0$ , but $p^2\not|c_0d_0$ , we know that $p$ divides only one of $c_0$ and $d_0$ . Let us assume that $p|c_0$ . Now it is also true that $p|(c_0d_1+c_1d_0)=a_1$ . This implies that $p|c_1$ . Going on like this, we see that $p$ divides every coefficient of $c_0+c_1x+c_2x^2+\dots c_hx^h$ , which is a contradiction as it should not divide $c_hd_k=a_m$ . You may assume that $p|d_0$ , and come upon an analogous contradiction.

Now how can we generalize Eisenstein’s theorem? For one, we may conclude that if $p|\{a_1,a_2,\dots,a_m\},p\not|a_0$ and $p^2\not|a_m$ , then the polynomial is irreducible too!