I have to factor the polynomial $f(x)=x^5-1$ in $\mathbb{F}_p[x]$, where $p \neq 5$ is a generic prime number.
I showhed that, if $5 \mid p-1$, then $f(x)$ splits into linear irreducible.
Now I believe that, if $5 \nmid p-1$ but $5 \mid p+1$, then $f(x)$ splits into three irreducible polynomial, one of degree $1$ and two of degree $2$.
Oterwise, if $5 \nmid p-1$ and $5 \nmid p+1$ (so $p \mid p^2+1$), then $f(x)$ splits into two irreducible factors, one of degree $1$ and one of degree $4$.
How can I prove this statements (if I'm right, well…)? Thank you!
Best Answer
For $p\neq 5$, it is always true that $$p^4\equiv 1 \textrm{ mod }5$$ On the other hand, $$p^4-1=(p-1)(p+1)(p^2+1)$$ If $5|p-1$, then roots of $x^5-1=0$ are included in $\mathbb{F}_p^{*}\simeq\mathbb{Z}/(p-1)\mathbb{Z}$. So, you don't need an extension field to include roots. Thus, $x^5-1$ factors into linear polynomials.
If $5\nmid p-1$, but $5|p+1$, then $5|p^2-1$, and all roots of $x^5-1=0$ are in $\mathbb{F}_{p^2}^{*}\simeq\mathbb{Z}/(p^2-1)\mathbb{Z}$, but not in $\mathbb{F}_p^{*}$. So, you need degree 2 extension of $\mathbb{F}_p$ to include roots. Thus, $x^5-1$ factors into $(x-1)$ and two quadratic irreducible polynomials.
Finally, if $5\nmid p^2-1$, and $5\mid p^2+1$, then all roots of $x^5-1=0$ are in $\mathbb{F}_{p^4}^{*}\simeq\mathbb{Z}/(p^4-1)\mathbb{Z}$, but not in $\mathbb{F}_{p^2}^{*}$. So, you need degree 4 extension. Thus, $x^5-1$ factors into $(x-1)$ and degree 4 irreducible polynomial.