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I'm having some trouble figuring out a solution to this. I understand that $f$ is separable, iff all its roots are distinct, however I'm completely clueless about how to investigate that criterion...so I tried focussing on irreducibility first. That turned out to be quite confusing too.

First, I tried finding the primitive roots of unity, which turned out to be $\pm 1$, so I suspected $f$ might always be irreducible, because 1 is always in $\mathbb{F}_p$, for any prime. But then I plugged them into $f$ over some of the possible fields and that didn't seem to check out.

Next, I figured $\alpha$ is a zero of $f$ over $\mathbb{F}_p$, if $\alpha^4 \equiv p-1 \text{ mod } p \iff \alpha \equiv \sqrt[4]{p-1} \text{ mod } p$. I know $p-1$ must be even, since $p$ is prime and therefore odd, so I can substitute $2k$, $k$ being a suitable element of $\mathbb{Z}$. So I get $\alpha \equiv \sqrt[4]{2k} \text{ mod } p$. $\sqrt[4]{2k}$ is whole (natural, even), precisely if $2k = m^4, m \in \mathbb{F}_p$, so $m = \alpha$ would then be a zero...

I know I'd have to investigate the case that $f$ might be representable as the product of two polynomials of degree less than 4 next, but I'll need to consider that after I crossed this hurdle.

Any ideas?

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    $\begingroup$ This polynomial is always reducible, because it has a root in every field $\mathbb{F_p}$. As for separability, use formal derivatives. $\endgroup$
    – Mark
    Commented Apr 27 at 16:28
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    $\begingroup$ Did $x^4-1$ get edited to $x^4+1$? ... $\endgroup$
    – paul garrett
    Commented Apr 27 at 16:49
  • $\begingroup$ yep that was a typo. idk why Dietrich Burde rolled it back tho.... $\endgroup$
    – Raiden
    Commented Apr 27 at 18:38
  • $\begingroup$ Which polynomial is it: $x^4-1$ or $x^4+1$? Anyway both always factor, and both are duplicates. The first one is obvious, and the second isn't difficult either. As you see, the way it factors depends on the residue class of $p$ mod $8$. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 28 at 4:15
  • $\begingroup$ (cont'd) Mostly, because that residue class determines which of $-1$, $2$, $-2$ have a modular square root. At least one of them always has. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 28 at 4:21

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