Let $\Phi_{31}(x)$ be the $31$-cyclotomic polynomial. I want to show that $\Phi_{31}(x)$ is the product of six irreducible quintic factors in $\mathbb{F}_2$. I am running into difficulties interpreting roots of unity in the field $\mathbb{F}_{2^5}$.

We know that $\Phi_{31}(x)$ will have an $d$-degree factor if and only if $\alpha^{31}=1$ in $\mathbb{F}_{2^d}^{\times}$. In order for any field extension of $\mathbb{F}_2$ to have an element of order $31$ we must have that $ | \mathbb{F}_{2^d}^{\times} | = 2 ^d -1 \equiv 0 \mod 31$. Notice that $d=5$, gives us such a result. And so $\Phi_{31}(x)$ has a quintic irreducible factor which has all roots in $\mathbb{F}_{2^5}$.

I know that over the rationals any root of $\Phi_{31}(x)$ is a primitive $31$-st root of unity and these primitive roots of unity belong to $\mathbb{C}$.

In this case we have a solution, $\alpha$, such that $\Phi_{31}(\alpha)=0$ but $\alpha \in \mathbb{F}_{2^5}$ and not $\mathbb{C}$.

Do I still interpret $\alpha$ as a primitive $31$-st root of unity??

Furthermore, over the rationals ANY primitive $31$-st root of unity is a solution to $\Phi_{31}(x)$. Again, in $\mathbb{C}$ all $31$-st roots of unity will be primitive, except for $1$.

Does this sort of result hold when we are working over a general field?

  • $\begingroup$ Regarding the last sentence: Just $\zeta^{31}=1$, with equality in $\mathbb F_{32}$. Since $\mathbb F_{32}$ is neither $\mathbb Z/2\mathbb Z$ nor $\mathbb Z/32\mathbb Z$, neither of your modulo suggestions fit. $\endgroup$ – Hagen von Eitzen Aug 25 '15 at 21:07
  • $\begingroup$ Thank you for the comment. I edited my question with this understanding. $\endgroup$ – user7090 Aug 25 '15 at 21:22

It’s hard to see from your question how familiar you are with the general theory of finite fields. Let me make some comments that will probably not answer all your questions, but I’m sure will at least help.

First, $X^{31}-1=(X-1)\Phi_{31}(X)$ is a $\Bbb Z$-factorization, and thus is true over every commutative ring. In particular, a root of $\Phi_{31}$ is a thirty-first root of unity in whatever commutative ring you consider.

Second, every field with $q$ elements has a multiplicative group that is cyclic of order $q-1$. In particular, every nonzero element of $\Bbb F_{32}$ is a thirty-first root of unity.

  • $\begingroup$ It’s actually very easy to write down the six irreducible quintics, when you remember that to be reducible, a quintic must be divisible by a linear or a quadratic, and this means that a reducible must either have an $\Bbb F_2$-root, or be divisible by the one-and-only irreducible quadratic $X^2+X+1$. $\endgroup$ – Lubin Aug 25 '15 at 21:48
  • $\begingroup$ Ah ok. I got it, I realized that in $\mathbb{F}_{31}$ every element besides $1$ and $0$ will have order $31$. And thus all be $31$-st primitive roots of unity and the theory for cyclotomic polynomials holds over ANY field. $\endgroup$ – user7090 Aug 25 '15 at 21:51
  • $\begingroup$ More precisely, not $\Bbb F_{31}$ but $\Bbb F_{32}$, as you originally specified. $\endgroup$ – Lubin Aug 26 '15 at 1:44

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