# Factor $x^5-1$ into irreducibles in $\mathbb{F}_p[x]$

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!

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If p is 5, then $x^5-1=(x-1)^5$ – Daenerys Naharis Jun 11 '13 at 7:04
Sorry, I forgot that $p \neq 5$ for hypotesis... – Frankenstein Jun 11 '13 at 7:35

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.

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Hints (for you to understand, complete/prove)

Over any field

$$x^5-1=(x-1)(x^4+x^3+x^2+ x +1)$$

Now, the roots of the second factor above are the roots of unity of order $\,5\,$ different from 1 itself. Since these roots (including $\;1\;$) form a cyclic group of order$\; 5\;$, we want to now when $\,5\mid p-1\,$ . In these cases we see the above pol. will split in different linear factors, for example:

$$p=11\;\implies x^5-1=(x-1)(x-4)(x-5)(x-9)(x-3)\in\Bbb F_{11}[x]$$

Also note that $\,a\neq 1\,$, otherwise the above equations have no solution (why?), so the other option for $\,b\,$ still's open. Try now to take it from here.

Note that $\,4=4^1\;,\;5=4^2\;,\;9=4^3\;,\;3=4^4\,$

It though can be the above quartic splits as the product of two irreducible quadratics:

$$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=$$

$$=x^4+(a+c)x^3+(d+ac+b)x^2+(ad+bc)x+bd$$

Comparing corresponding coefficients we get the equations

\begin{align*}a+c&=1\\ac+b+d&=1\\ad+bc&=1\\bd&=1\end{align*}

Thus

$$(1)+(3)\;\;a=1-c\;,\;d=\frac1b\implies (3)\;\;\frac ab+b(1-a)=1\implies$$

$$(1-a)b^2-b+a=0\;:\;\;\Delta=1-4a(1-a)=4a^2-4a+1=(2a-1)^2\implies$$

$$b_{1,2}=\frac{1\pm (2a-1)}{2(1-a)}=\begin{cases}\frac a{1-a}\\{}\\\;\;\;\;1\end{cases}$$

Now

$$b=1\implies d=1\;,\;\;ac=-1\;,\;c=1-a\implies a(1-a)=-1\implies$$

$$a^2-a-1=0\implies a=\frac{1\pm\sqrt 5}2$$

Note that this choice requires $\,\sqrt 5\in\Bbb F_p\iff \binom5p=1\,$ ,but by quadratic reciprocity

$$1=\binom5p=\binom p5\iff p=1,4\pmod 5$$

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