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Suppose we have 8-th cyclotomic polynomial $\Phi_8(x)=x^4+1$

It is standard fact that any cyclotomic polynomial is irreducible over rational field.

When the coefficient is over $\mathbb{Z}_3$ i.e., $\mod 3$, it can be factorized.

How can we factorize $\Phi_8(x)$?

I know that the number of factor is 2, but am not sure what the factors are.

mallea
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Modulo $3,$ you have $x^4+1= \left(x^2+x+2\right) \left(x^2+2 x+2\right).$

Igor Rivin
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  • @user1952009 I don't understand the question. – Igor Rivin Mar 05 '17 at 04:23
  • @user1952009 how did you find? – mallea Mar 05 '17 at 05:08
  • @IgorRivin Let $f$ such that $\mathbb{F}_3(\zeta_8) = \mathbb{F}_3[x]/(f(x))$. From what you wrote it seems that $f(x) = x^2+x+2$ and hence $[\mathbb{F}_3(\zeta_8):\mathbb{F}_3]=2 $ which is not obvious to me – reuns Mar 05 '17 at 06:56
  • @user1952009 thank you for replying. Did you exhaustively enumerate all possible polynomials, and then found out $f(x)=x^2+x+2$ ? – mallea Mar 05 '17 at 08:13
  • @zorutic Actually, it is not surprising at all. notice that since in the Galois field of order 9 every element satisfies $x^9 - x = 0,$ but $x^9-x = x (x^8 - 1) = x (x^4-1)(x^4+1).$ In other words, the polynomial splits completely in the degree two extenstion. – Igor Rivin Mar 05 '17 at 16:05