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Given the polynomial $f(x)=x^4+1$ in the field $\mathbb{F}_p$, prove that $[F: \mathbb{F}_p]\leq 2$, where $F$ is the splitting field of $f$.

Suppose $a$ is one of the roots of $f$. Then (if $p>2$, case $p=2$ is trivial) all different roots of $f$ are $a,a^3,a^5,a^7$ and therefore $x^4+1=(x-a)(x-a^3)(x-a^5)(x-a^7)$. Clearly $F=\mathbb{F}_p(a)$.

Now I want to prove that $a$ has minimal polynomial of degree not greater than 2. Because minimal polynomial divides other polynomials that have $a$ as a root, I need to prove that one of the following polynomials: $x-a$, $x^2-(a+a^3)x+a^4$, $x^2-(a+a^5)x+a^6$, $x^2-(a+a^7)x+a^8$ is in $\mathbb{F}_p [x]$. Or, equivalently (after simplifications), that at least one of the elements from $\{a,a+a^3,a^2,a-a^3\}$ is in $\mathbb{F}_p$. And here I am stuck. Not even sure this was the right way to start.

So, how can I prove the aforementioned statement?

OAK
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1 Answers1

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My favorite way to prove it is via the following lemma: $f$ splits in $\mathbb F_q$, of characteristic $\neq 2$, if and only if $8 \mid q-1$. Then simply use the fact that $p^2 \equiv 1 \mod 8$ for $p$ an odd prime.

Dustan Levenstein
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  • Suppose $8|q-1$. Then the order of $\mathbb{F}_q^$ is equal to $8s$ for some $s$. Let $\alpha$ be the primitive element of $\mathbb{F}_q^$. Then $\alpha^{4s}$ is the square root of $1$ but not equal to it, therefore $\alpha^{4s}=-1$. So $\alpha^s, \alpha^{3s}, \alpha^{5s}, \alpha^{7s}$ are desired roots of $f$. Like this? I only need one part of the lemma. – OAK Dec 29 '15 at 13:27
  • That's perfect. – Dustan Levenstein Dec 29 '15 at 13:35
  • Thank you! I hope I will do better on exam tomorrow. – OAK Dec 29 '15 at 13:38