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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?

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  • $\begingroup$ It' pretty obvious (I think) that $x^8 - 1 = (x^4 - 1)(x^4 + 1) = (x^2 - 1)(x^2 + 1)(x^4 + 1) = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)$...don't know whether or not that helps. $\endgroup$ – Jared Mar 9 '16 at 3:58
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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.

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  • $\begingroup$ See also en.wikipedia.org/wiki/Cyclotomic_polynomial. $\endgroup$ – lhf Mar 9 '16 at 1:22
  • $\begingroup$ Could you complete your argument with $x^2+1$ and $x^4+1$? I am blocked to prove that these polynomials are irreductibles over $\mathbb{Z}_3$ and $\mathbb{Z}_2$. $\endgroup$ – Taj Mohamed Bandalandabad Mar 9 '16 at 3:19
  • $\begingroup$ @Taj, see my edit answer. $\endgroup$ – lhf Mar 9 '16 at 9:57
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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

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

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

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    $\begingroup$ $2x^2+2$, $2x^2+x+1$, $2x^2+2x+1$ are also irreducible. $\endgroup$ – user236182 Mar 9 '16 at 4:41
  • $\begingroup$ You're right. I added the word "monic". $\endgroup$ – Emre Mar 9 '16 at 4:42
  • $\begingroup$ For context: merged from there to reduce the number of questions about the same problem (by the same OP). $\endgroup$ – ccorn Mar 9 '16 at 8:53

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