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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?

user26857
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blabla
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    I'm guessing $\iota = i$, the imaginary unit? Note that typically $i$ is defined to be one of the square roots of $-1$. In the case of $\Bbb{F}_2$, $-1 = 1$, so $\pm i = \pm 1$. – Theo Bendit Oct 29 '18 at 10:55
  • If $x^4+1$ splits over $\Bbb{F}_2$, then how can any nontrivial extension of $\Bbb{F}_2$ be a minimal splitting field? – Servaes Oct 29 '18 at 11:01
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    @TheoBendit Thanks that clears it that $F_2$ is the m.s.f here. Is my argument for $F_3$ correct? I think it should rather be like, for $F_3$ also $-1=2$, so $i$ must be equal to $\sqrt{2}$ and thus m.s.f will be $F_3(\sqrt[4]{2})$ – blabla Oct 29 '18 at 11:10
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    Endorsing Servaes' answer. The solution is also buried in (at least) this old related thread. Not a dupe, but closely related. – Jyrki Lahtonen Oct 29 '18 at 11:14
  • You are right that it differs between values of $p$! From Servaes answer we see that this effect is rather small since $p$ can influence the answer only through its congruence class mod 8 (which is quite surprising). This leaves only finitely many different classes of $p$'s to check and hence only finitely many different possible answers. So it must be doable to find them all. – Vincent Oct 29 '18 at 11:52

2 Answers2

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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 $8$?

Servaes
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  • why are roots precisely the elements of order 8? – blabla Oct 29 '18 at 11:26
  • So for $p=2$, we have no such $k$ such that a cyclic group of order $2^k-1$ has an element of order $8$ as $8$ does not divide $2^k-1$? – blabla Oct 29 '18 at 11:30
  • @blabla I rushed the answer a bit, the argument does not hold entirely for $p=2$. Let me edit my answer. – Servaes Oct 29 '18 at 11:31
  • So for $F_3$ we get $F_9$ as minimal splitting field. But if I try to find it in the form of $F_3(\alpha_1, \ldots , \alpha_n)$ where $\alpha_1, \ldots , \alpha_n$ are its roots, then I should get a field isomorphic to $F_9$. I somehow got $F_3(\sqrt[4]{2})$, so something must be wrong in my argument as they are not isomorphic. – blabla Oct 29 '18 at 11:49
  • What do you mean by $\sqrt[4]{2}$? (This is a serious question.) – Servaes Oct 29 '18 at 12:57
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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.

Wuestenfux
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