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


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!

I will write down other generalizations when they come to mind.

Published by -

Graduate student

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: