Find the splitting field of $x^4+1$ over $\mathbb F_p$

Question

This question was asked in a class exam to find the splitting field of $p(x)= x^4+1$ over $\mathbb F_p$, where $\mathbb F_p$ is a finite field of order $p$, $p$ prime.

My thought is that something is wrong here. It will be different for different primes. The value of $p$ must have been provided. Like,

For $p=2$, $x^4+1=(x^2+1)^2=(x+1)^4$. So from here it seems that $1$ is the only root of $p(x)$ and thus minimal splitting field is $F_2$ itself. But what about $\pm \iota$. They are also roots of $p(x)$. If I go like $x^4=-1=1$, then $x=\sqrt[4]{1}$ and thus $x=\pm 1, \pm \iota$, thus, this way it seems that minimal splitting field is $F_2(\iota)$. Which one is correct here?

Also if I try for $F_3$, $x^4=-1=2 \implies x=\pm \sqrt[4]{2}, \pm \sqrt[4]{2}\iota$ and thus m.s.f would be $F_3(\sqrt[4]{2}, \iota)$.

Similarly we will have to find for each and every $p$ separately and nothing can be said about it in general, right?

Answer 1

The minimal splitting field of $x^4+1$ over $\Bbb{F}_p$ is a finite extension of $\Bbb{F}_p$, and hence it is isomorphic to $\Bbb{F}_{p^k}$ for some $k\geq1$. Suppose $x^4+1$ has a root in $\Bbb{F}_{p^k}$, say $\alpha$. Then $x^4+1$ splits completely over $\Bbb{F}_{p^k}$ because $\alpha^3$, $\alpha^5=-\alpha$ and $\alpha^7=-\alpha^3$ are also roots. So it suffices to find $k$ such that $x^4+1$ has a root in $\Bbb{F}_{p^k}$.

Any root of $x^4+1$ is nonzero and satisfies $x^4=-1$, so it is a unit in $\Bbb{F}_{p^k}$ that satisfies $x^8=1$, i.e. it is an element of order dividing $8$. If $p\neq2$ then $-1\neq1$ and so the roots have order precisely $8$. The unit group $\Bbb{F}_{p^k}^{\times}$ is cyclic of order $p^k-1$, so the question now becomes; what is the minimal $k$ such that the cyclic group of order $p^k-1$ has an element of order $8$?

Answer 2

We have $x^8-1 = (x^4+1)(x^4-1)$ over any field. It follows that each root of $x^4+1$ is an 8-th root of unity in an extension field of ${\Bbb F}_p$. These roots are primitive, since every other root of $x^8-1$ is a root of $x^4-1$ and is a 4th root of unity. So one needs to determine the smallest extension field ${\Bbb F}_{p^k}$ of ${\Bbb F}_p$, $k\geq 1$, which contains the 8th roots of unity. Since the multiplicative group of ${\Bbb F}_{p^k}$ is cyclic with $p^k-1$ elements and the structure of the cyclic subgroups of a cyclic group is fully known, we must have that 8 divides $p^k-1$, where $k$ is minimal.