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?

closed as off-topic by Thomas, user26857, Claude Leibovici, choco_addicted, WatsonMar 10 '16 at 9:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, user26857, Claude Leibovici, choco_addicted, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Jared Mar 9 '16 at 3:58

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.

• – lhf Mar 9 '16 at 1:22
• 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 $\mathbb{Z}_2$. – Taj Mohamed Bandalandabad Mar 9 '16 at 3:19
• @Taj, see my edit answer. – lhf Mar 9 '16 at 9:57

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

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