$X^5 + 1$ into irreducible factors over $\mathbb{Q}$ and $\mathbb{Z}_5$ 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 $\mathbb{Z}_5$.
 A: 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 $x^5+1 \cong (x+1)^5 \pmod{5}$.
A: 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). $
