Is $x^4-1$ reducible over every $\mathbb{F}_p$ for every prime p?

Question

Is $x^4-1$ reducible over every $\mathbb{F}_p$ for every prime p? I think I need to split into the cases of $p=2$ and $p$ being odd, but not idea how to proceed from there...

This is trivial, as Igor's factorization shows. Did you perhaps mean $x^4 + 1$? In that case, see http://math.stackexchange.com/questions/427439/why-is-x41-reducible-over-mathbb-f-p-with-p-geq-3-prime — Travis Willse, Feb 20 '16 at 20:34

score 4 · Answer 1 · answered Feb 20 '16 at 20:32

4

Yes, because $x^4 - 1 = (x-1)(x^3 + x^2 + x + 1).$

answered Feb 20 '16 at 20:32

Igor Rivin

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So how does show that it is reducible over any $\mathbb{F}_p$? – dtgo Feb 20 '16 at 20:38
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Because I factored it over every $\mathbb{F}_p?$ – Igor Rivin Feb 20 '16 at 20:39
Oh I see... this seems to be a trivial question then... Thanks – dtgo Feb 20 '16 at 20:41

Is $x^4-1$ reducible over every $\mathbb{F}_p$ for every prime p?

1 Answers1