# Proving that a polynomial is irreducible over a field

What's the general strategy to show that a particular polynomial is irreducible over a field?

For example, how can I show $x^4 - 10x^2 -19$ is irreducible over $\mathbb Q$?

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$\mathbb Q[x]$ is not a field. We usually say that a polynomial is irreducible over a field $k$ when it is irreducible in the ring $k[x]$. – lhf Sep 30 '11 at 12:46
See for instance en.wikipedia.org/wiki/Rational_root_theorem and en.wikipedia.org/wiki/… – lhf Sep 30 '11 at 12:53

When a polynomial is integer valued, one may appeal to Gauss's lemma which states that if the coefficients of a non-constant polynomial $f$ are relatively prime and $f$ is irreducible in $\mathbb{Z}[X],$ then $f$ is irreducible in $\mathbb{Q}[X].$

This allows you to use the structure of $\mathbb{Z}[X]$ which is richer than that of $\mathbb{Q}[X]$ to prove irreducibility. For example for every prime $p$ there is a reduction map from $\mathbb{Z}[X]$ to $\mathbb{F}_p[X].$ If the reduction of $f$ is irreducible over $\mathbb{F}_p[X]$ then $f$ is irreducible over $\mathbb{Z}[X].$

Note that for your polynomial, $x^4 - 10x^2 - 19 \mod 3$ is irreducible in $\mathbb{F}_3[X].$ Hence, $x^4 - 10x^2 - 19$ is irreducible in $\mathbb{Z}[X].$

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There is no "general strategy" that works for all polynomials and fields, but there are several useful results which can be found here.

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Suppose that $p$ is a fourth degree polynomial which is not irreducible. Then $p$ contains a linear factor or a quadratic factor.

In your case you can easily check using the abc-formula that $p$ does not have a linear factor, i.e., it does not have a root lying in $\mathbf{Q}$.

Thus, now you have to think about why $p$ doesn't have a quadratic factor. You could do this with brute force, i.e., write out $p = (a+bx+x^2)(d+ex+x^2)$ and get to a contradiction. (This used to be $(a+bx+cx^2)(d+ex+fx^2)$, but as lhf points out it's clear that one may assume that $c=f=1$.)

Easier would be to apply one of the methods mentioned by Ragib Zaman.

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You may assume $c=f=1$. – lhf Sep 30 '11 at 13:28
Of course. I'll edit that. – shaye Sep 30 '11 at 13:31