Reducing $x^8-1$ over $\mathbb{Z}$ and $\mathbb{Z_2}$

Question

I think this will factor completely over $\mathbb{Z_2}$ and will factor to maybe something like $(x^4-1)(x^4+1)$=$(x^2-1)(x^2+1)(x+1)^4$=$(x-1)(x+1)(x^2+1)(x^4+1)$.

I don't know how to explain with words why $(x^4+1)$ won't factor over $\mathbb{Z}$... to show that it won't factor do I just show it has no roots and then write it as a product of arbitrary quadratics and look for a contradiction?

Also, what are the various different ways to argue that $(x^8-1)$ factors completely over $\mathbb{Z_2}$?

Thanks yall!

Answer 1

To show $x^2+1$ does not factor over $\Bbb Z$ you can appeal to the rational root theorem because any factor must be linear and must give a rational root. That also shows $x^4+1$ does not split off a linear factor but does not show it will not factor as a product of quadratics. Since the only way to get $1$ as a product is $(\pm1)^2$you know it must be $x^2+ax\pm 1$ and $x^2+bx\pm 1$ with the same choice of sign. Just multiply them and show that it fails.

Over $\Bbb {Z_2}$ you can just multiply to show $x^2+1=(x+1)^2, x^4+1=(x+1)^4$ and in fact $x^8-1=(x+1)^8$

Answer 2

Here’s another way to show that $f(X)=X^4+1$ is $\Bbb Z$-irreducible: surely its reducibility is the same as that of $f(X+1)$, and this is $X^4+4X^3+6X^2+4X+2$, which is irreducible by Eisenstein.