# Explain why $x^4+1$ is reducible in $\mathbb{F}_p$ for any prime $p$.

We first notice that $x^4+1$ is the eighth cyclotomic polynomial, $\Phi_8(x)$. We know from the theory that $\Phi_8(\alpha)=0$ if and only if $\alpha$ has is a $8$-th primitive root modulo $p$. I've already shown that it is true in the case that $p=2$.

Now consider the case $p \equiv 1 \operatorname{mod} 8$. Then $| ( \mathbb{F}_p^{\times} ) | = p-1$ and so $( \mathbb{F}_p^{\times} )$ has an element of order $8$. Call this element $\alpha$.

From my notes, this element $\alpha$ is a $8$-th primitive root modulo $p$ if $\alpha$ generates $(\mathbb{F}_p)^{\times}$ and $\alpha$ has order $8$.

We know $\alpha$ is a solution of $\Phi_8(x)=0$ if and only if $\alpha$ is the $8$th primitive root mod $p$. We have only shown that $\alpha$ is an element of order $8$ in $( \mathbb{F}_p^{\times} )$, however it does not generate the $( \mathbb{F}_p^{\times} )$ and so we $\textbf{cannot}$ claim (from my understanding) that $\alpha$ is a $8$-th primitive root of $\mathbb{F}_p$ and thus a solution of $\Phi_8(x)=0$.

My notes seem to do this though. What am I misunderstanding??

• If the degree of a polynomial is $4$ or more, it may be reducible in a field and still have no roots in it. Example: $(x^2+1)(x^+2)$ in $\Bbb R$. – ajotatxe Jul 23 '15 at 21:04

Since $x^2 - 1 = (x+1)(x-1)$, if $1 = \alpha^8 = (\alpha^4)^2$ (in any field) we must have either $\alpha^4 = 1$ or $\alpha^4 = -1$. If $\alpha$ has order $8$, $\alpha^4$ can't be $1$ so must be $-1$. That says $\alpha$ is a root of $x^4 + 1$.