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I am wondering if $f(x)=x^4+22x^2+1$ is irreducible over $\mathbb{Q}$.

To do this, we use a variant of Gauss's lemma. If $f(x)$ is irreducible over $\mathbb{Z}$, then $f(x)$ is irreducible over $\mathbb{Q}$. Now, I read a comment here https://math.stackexchange.com/questions/68764/proving-that-a-polynomial-is-irreducible-over-a-field#:~:text=For%20example%20for%20every%20prime,Z%5BX%5D that said if there exists a prime $p$ so that the reduction of $f(x)$ is irreducible over $\mathbb{F}_p$, then $f(x)$ is irreducible over $\mathbb{Z}$.

So my question is: how do I choose my prime $p$ and once I choose my prime $p$, how do I show that the reduction of $f$ cannot be broken down into smaller polynomials? Would a proof by contradiction be best? Let me know. Thanks!

Grigor Hakobyan
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    There is, in general, no easy way to pick a prime. You could, of course, just try various small primes. Alternatively, you could note that there are no rational roots so all you need to rule out are quadratic factors. – lulu Dec 05 '22 at 21:18
  • To answer the question in the title: "yes". Is there a reason you want to do this by hand? Any computer algebra system (for instance sage) can quickly do this for you. – HallaSurvivor Dec 05 '22 at 21:27
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    For Galois theory reasons, polynomials of this form are in fact reducible modulo every prime: https://math.stackexchange.com/a/160855/41415 So if you want to show $x^4 + 22x^2 + 1$ is irreducible, you'll have to use a different method. – Daniel Hast Dec 05 '22 at 22:34

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once you see that there are no rational roots, the possible factorizations are, with some integer $a,$ $$ (x^2 + ax + 1)(x^2 - ax +1) = x^4 + (2-a^2) x^2 + 1$$ or $$ (x^2 + ax - 1)(x^2 - ax -1) = x^4 + (-2-a^2) x^2 + 1$$

where the $22$ does not work.

Will Jagy
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