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How do I completely factor $x^5-1$ over $\mathbb{Z_5}$?

I saw one was a root so I divided by $x-1$ and got $(x-1)(x^4+x^3+x^2+x+1)$ which duh is the 5th cyclotomic polynomial and this is the same factorization over the integers. Do I just look for roots from here or is there a better way to go about this? I mean I know I could try to factor it as arbitrary quadratics also, but i'm wondering if there was a more fundamental insight that i'm missing that would make this simpler. Thanks!

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    $\begingroup$ \begin{eqnarray*} (x-1)^5 \equiv x^5-1 \pmod{5}. \end{eqnarray*} $\endgroup$ – Donald Splutterwit Oct 8 '18 at 0:45
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    $\begingroup$ If the comment by @DonaldSplutterwit seems cryptic, try expanding it in $\mathbb{Z}$ and notice something about all of the coefficients besides the two outer ones. $\endgroup$ – Brian Tung Oct 8 '18 at 0:49
  • $\begingroup$ right, binomial expansion, duh, thanks $\endgroup$ – Math is hard Oct 8 '18 at 0:50
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    $\begingroup$ before jumping to the binomial expansion, notice that $x^4 + x^3 + x^2 + x + 1$ has a root $\pmod 5,$ namely $1.$ This means we can factor it as $(x-1) \cdot \mbox{cubic}$ and see how the cubic behaves. I got $x^3 + 2 x^2 + 3x + 4.$ Naturally, $1$ is a root again $\endgroup$ – Will Jagy Oct 8 '18 at 1:11
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Since $1$ is a root, we can divide out $x-1$ by the factor theorem: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.

$1$ is a root of $x^4+x^3+x^2+x+1$, so we can divide it out again: $x^4+x^3+x^2+x+1=(x-1)(x^3+2x^2+3x+4)$

$x^3+2x^2+3x+4=(x-1)(x^2+3x+1)$.

Lastly, $x^2+3x+1=(x-1)(x-1)$. So we can write $x^5-1$ as the product $(x-1)^5$ in $\mathbb{Z}[x]$.

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  • $\begingroup$ hah thanks dawg!!! I could have done this !!!!! but binomial expansion is better imo :P $\endgroup$ – Math is hard Oct 8 '18 at 11:53

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