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I want to factor $x^8-1$ into a product of irreducibles over $\mathbb{Z}$ and $\mathbb{Z_2}$ and then EXPLAIN how i know that polynomials I obtain are irreducible.

So over $\mathbb{Z}$, $x^8-1=(x^4+1)(x^4-1)=(x^4+1)(x^2+1)(x^2-1)=(x^4+1)(x^2+1)(x+1)(x-1)$.

And then it looks like it would factor the same way over $\mathbb{Z_2}$, with $-1=1$ of course. Is this correct? Perhaps there is a way to factor the quartic term but i'm not sure. Anyway, my lack of sureness surely reflects a lack of some insight that i'm missing. Can anyone help me out here? I'd appreciate it!

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    $\begingroup$ Note that $x^2+1$ has two roots over $\Bbb{F}_2$. $\endgroup$ – Dietrich Burde Sep 24 '18 at 19:28
  • $\begingroup$ $x^4+1=\Phi_8(x)$ is never irreducible over a finite field. $\endgroup$ – Jack D'Aurizio Sep 24 '18 at 19:38
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Hint: $a^2+b^2=(a+b)^2$ over $\mathbb{Z}_2$.

Solution: $ x^8-1=x^8+1=(x+1)^8 $

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Since $-1=1$ in $\mathbb{Z_2}$, $x^4+1=x^4-1$ and $x^2+1=x^2-1$.

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