# Roots of $\Phi_{31}(x)$ as roots of unity.

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?

• 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. – Hagen von Eitzen Aug 25 '15 at 21:07
• Thank you for the comment. I edited my question with this understanding. – user7090 Aug 25 '15 at 21:22

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.
• 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 $X^2+X+1$. – Lubin Aug 25 '15 at 21:48
• 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. – user7090 Aug 25 '15 at 21:51
• More precisely, not $\Bbb F_{31}$ but $\Bbb F_{32}$, as you originally specified. – Lubin Aug 26 '15 at 1:44