$X^5 + 1$ into irreducible factors over $\mathbb{Q}$ and $\mathbb{Z}_5$

Question

I need to split this polynomial into irreducible factors. I managed to do this for $\mathbb{C}$ and $\mathbb{R}$, but I can't figure out how to solve this problem for $\mathbb{Q}$ and $\mathbb{Z}_5$.

Answer 1

If you multiply out $(x+1)^5$ by the binomial theorem you get $x^5+1$ modulo 5 so that deals with $\mathbb{Z}_5$.

From your result over the reals, you know that the quadratics are not in $\mathbb{Q}(x)$ and so over $\mathbb{Q}$ we have $x^5+1=(x+1)(x^4-x^3+x^2-x+1). $

Answer 2

This answer only addresses $\mathbb{Z}_5$.

For factoring over (small) finite fields, the easiest choice is to just try all the points in the field.

\begin{align*} &x :& &x^5+1 \pmod{5} \\ &0 :& &1 \pmod{5} \\ &1 :& &2 \pmod{5} \\ &2 :& &3 \pmod{5} \\ &3 :& &4 \pmod{5} \\ &4 :& &0 \pmod{5} \end{align*} So, $x-4 \cong x+1 \pmod{5}$ is a factor and no other linear polynomial is a factor. In $\mathbb{Z}_5$, $$ \frac{x^5+1}{x+1} \cong x^4 + 4x^3 + x^2+4x+1 \pmod{5} $$ and we should check to see if any of the linear factors found in the first table continue to divide this quotient. (This langauge seems a little odd because it also works when the table finds more than one linear factor -- we have to keep checking all the factors we found to see if they divide the subsequent quotients.) \begin{align*} \frac{x^4 + 4x^3 + x^2+4x+1}{x+1} &\cong x^3 + 3x^2+3x+1 \pmod{5} \text{,} \\ \frac{x^3 + 3x^2+3x+1}{x+1} &\cong x^2 + 2x + 1 \pmod{5} \text{, and} \\ \frac{x^2 + 2x + 1}{x+1} &\cong x+1 \pmod{5} \text{.} \end{align*} So we have shown $x^5+1 \cong (x+1)^5 \pmod{5}$.