How could it possible to factorise $x^8-1$ in product of irreducibles in the rings $(\mathbb{Z}/2\mathbb{Z})[x]$ and $(\mathbb{Z}/3\mathbb{Z})[x]$? [closed]

How could it possible to factorize $x^8-1$ in product of irreducibles in the rings $(\mathbb{Z}/2\mathbb{Z})[x]$ and $(\mathbb{Z}/3\mathbb{Z})[x]$?

I'm having a hard time starting the problem. Could anyone help me at this point?

Start with the irreducible factorization over $\mathbb Z$: $x^8-1=(x-1) (x+1) (x^2+1) (x^4+1)$.

Now factor $x^2+1$ and $x^4+1$ over the given rings.

Over $\mathbb{Z}/2\mathbb{Z}$, we have $x^2+1=(x+1)^2$ and $x^4+1=(x+1)^4$. Since $-1=1$, we get $x^8-1=(x+1)^8$.

Over $\mathbb{Z}/3\mathbb{Z}$, we have that $x^2+1$ has no roots and so is irreducible. On the other hand, $x^4+1$ does not have roots but can be factoried into quadratics: $x^4+1=(x^2+x-1) (x^2-x-1)$. This is the only step that needs some work.

Over $\mathbf Z/2\mathbf Z$, this is trivial, since squaring is a homomorphism: $$x^8-1=(x-1)^8.$$

Over $\mathbf Z/3\mathbf Z$, you can check $x^2+1$ is irreducible since it has no root, and try to factor $x^4+1$ as $(x^2+ax+b)(x^2+a'x+b')$, you obtain the system of equations $$a+a'=0,\quad aa'+b+b'=0,\quad ab'+ba',\quad bb'=1.$$ The last equation tells us $b=b'=\pm1$, and the first equation that $a'=-a$, whence $a^2=b+b'=\pm 1$. However, only $1$ is a square mod. $3$, so we arrive at

$$x^4+1=(x^2-x-1)(x^2+x-1).$$

You can't use that problem as $\mathbb{Z}/3\mathbb{Z}$ is a field and has no proper ideals.

$$x^8-1=(x^4+1)(x^2+1)(x+1)(x-1)$$

First of all, $x^2+1$ is irreducible as it has no roots in $\mathbb{Z}/3\mathbb{Z}$. Similarly, $x^4+1$ also has no roots in $\mathbb{Z}/3\mathbb{Z}$. So, it has no linear factors. So, it is enough to check if monic irreducible quadratic polynomials divide $x^4+1$. Luckily, we only have three of them $x^2+1,x^2-2x+2,x^2+2x+2$. Plugging them in we have: $$x^4+1=(x^2-2x+2)(x^2+2x+2)$$