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Factor $x^8-x$ in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Here what I get is $x^8-x=x(x^7-1)=x(x-1)(1+x+x^2+\cdots+x^6)$ now what next? Help in both the cases in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Edit: I think $(1+x+x^2+\cdots+x^6)$ is cyclotomic polynomial for $p=7$ so it is irred over $\Bbb Z$. Now the problem remains for $\Bbb Z_2[x]$

Ri-Li
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  • This has been sorta handled many times on our site. This thread covers the most ground. But, also 1, 2,3, 4. – Jyrki Lahtonen Feb 02 '19 at 07:49
  • Difficult to pick a definitive duplicate target. Ideally I would like the target thread to explain the ever useful fact that all irreducible polynomials of degree $d\mid n$ over $\Bbb{F}_p$ appear as factors of $x^{p^n}-x$, AND, the thread to also explain how to quickly find irreducible cubics in $\Bbb{F}_2[x]$. Both of those have been covered in many threads, but finding one that does both proved to be not so easy. – Jyrki Lahtonen Feb 02 '19 at 07:56

2 Answers2

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To answer a question of yours in the comments, here’s how I thought in factoring $(x^7-1)/(x-1)$ over $\Bbb F_2$:

You’re still talking about the six primitive seventh roots of unity here. But what is the smallest field containing $\Bbb F_2$ that also has seventh roots of unity? That is $\Bbb F_8$, the cubic extension of $\Bbb F_2$. So each of those roots must belong to an irreducible cubic polynomial over $\Bbb F_2$. I happened to know that there are only two irreducible $\Bbb F_2$-cubics, namely $x^3+x^2+1$ and $x^3+x+1$. Our sextic had to be the product of these two.

Lubin
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After my edit I finally got the answer it was under my nose the polynomial $1+\cdots +x^6$ does not have zeros at $0$ and $1$ so it can't be factored as a multiple of a one degree polynomial and one other as a whole.

Edit: But as Lubin mentioned in the comment $1+\cdots +x^6=(1+x+x^3)(1+x^2+x^3)$

Ri-Li
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  • It might be divisible by a polynomial with degree greater than one – Lucio Tanzini Feb 01 '19 at 03:20
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    In fact, over the field with two elements, $x^6+x^5+x^4+x^3+x^2+x+1=(x^3+x^2+1)(x^3+x+1)$. – Lubin Feb 01 '19 at 03:20
  • Hi, @Lubin I got that $1+\cdots +x^6=(1+x+x^3)(1+x^2+x^3)$ but what is the general rule to follow in these cases? Are you working with galois conjugates? Even then can you please explain at least for these $\Bbb Z_2$? – Ri-Li Feb 01 '19 at 03:49
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    It's not hard to check that set of roots of $X^{2^n}-X$ is a field containing $\mathbb Z/2$, that is an extension of degree $n$. Using the product formula for intermediate field extensions one can prove that the irreducible polynomials of $X^{2^n}-X$ have degree dividing $n$. The reciprocal is also true – eduard Feb 01 '19 at 13:57
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    Another way of looking at this is to observe that the multiplicative group of the field $\Bbb{F}_8$ has order seven. Therefore all the elements of $\Bbb{F}_8^*$ are zeros of $x^7-1$, and all the elements other then $1$ are zeros of $\Phi_7(x)$. But, $\Bbb{F}_8$ is a degree three extension of the prime field, so the minimal polynomials of its elements are either cubic or linear. And, as eduard explained, all irreducible cubics must be factors of $\Phi_7$. – Jyrki Lahtonen Feb 02 '19 at 07:40