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I am asked to find the Galois Group of the polynomial $x^4 - 7$ over $\mathbb{F}_5$. I am wondering if the following is correct:

The splitting field of $x^4 - 7 = x^4 - 2$ over $\mathbb{F}_5$ is $\mathbb{F}_5(\sqrt[4]{2},i)$ where $i,\sqrt[2]{2}$ lie in a fixed algebraic closure of $\mathbb{F}_5$ and $i^2 = -1$ and $(\sqrt[4]{2})^4 = 2$. Since $2^2 = 4 = -1 \in \mathbb{F}_5$ we see that $i = 2$. Now, $x^4 - 2$ does not have any roots in $\mathbb{F}_5$ and so $[\mathbb{F}_5(\sqrt[4]{2}): \mathbb{F}_5] = 2$ or $4$ which means the Galois Group is either order $2$ or $4$. If it were $2$, then $\sqrt[4]{2} = a + b\sqrt{2}$ where $a,b \in \mathbb{F}_5$. After squaring we must have that $a^2 + b^2 = 0$ and $2ab = 1$. This is an impossibility and so the degree of this extension (and hence the order of the Galois group) is $4$.

Consider $\sigma: \mathbb{F}_5(\sqrt[4]{2}) \to \mathbb{F}_5(\sqrt[4]{2})$ given by $\sigma(\sqrt[4]{2}) = 2\sqrt[4]{2}$. This is an automorphism of $\mathbb{F}_5(\sqrt[4]{2})$ of order $4$ and so must be the the galois group is $\langle \sigma \rangle$.

Mike
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    Typo in the question? – paul garrett Aug 01 '20 at 20:52
  • @paulgarrett fixed – Mike Aug 01 '20 at 20:55
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    I think the first equation you get when squaring should be $a^2+2b^2=0$. Other than that it looks ok to me. You may want to explain why (in the case when the extension has degree $2$), it follows that all the elements have the form $a+b\sqrt2$. That may have been done in class already - cannot tell. – Jyrki Lahtonen Aug 01 '20 at 21:01
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    My go to -technique in questions like this would be to observe that $\root4\of2$ must be a root of unity of order sixteen. A quadratic extension of $\Bbb{F}_5$ has $25$ elements, so its multiplicative group has $25-1=24$. But $16\nmid 24$, so $\root4\of2$ cannot belong to the quadratic extension. – Jyrki Lahtonen Aug 01 '20 at 21:03
  • @JyrkiLahtonen Thank you for the comment – Mike Aug 01 '20 at 21:08
  • It is true in general that the Galois group of $K over L$ (where K is a finite field) is always cyclic of order $[K:L]$. I don't know if you have proved it that's why I stated. It is a fun fact to prove. – Noob mathematician Aug 01 '20 at 21:25

2 Answers2

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I am stating an alternative path.

I will show that $x^4-2$ is irreducible over $\mathbb F_5$ and hence $[\mathbb F_5(\sqrt[4] 2):\mathbb F_5]=4$.

Also $\mathbb F_5$ contains fourth root of unity over $\mathbb F_5$($\because a^{(5-1)}\equiv 1(\mod 5) $ for all $a\in \mathbb F_5^*$ ). So $x^4-2$ splits over $\mathbb F_5(\sqrt[4] 2)$. $$x^4-2=\prod_{a\in \mathbb F_5^*}(x-a\sqrt[4]2)$$

If $g(x)$ is a factor of characteristic $x^4-2$ then $g(-x)$ is also a factor . $x^4-2$ has no root in $\mathbb F_5$.

Then observe that if $x^4-2$ reducible then only possibility is $$x^4-2=(x^2+ax+b)(x^2-ax+b)=x^4+(2b-a^2)x^2+b^2x$$ $\therefore 2b=a^2$ and $b^2=2$ . But this is not possible for any $a,b\in \mathbb F_5$.

So $x^4-2$ is irreducible in $\mathbb F_5$. Then you have $[\mathbb F_5(\sqrt[4] 2):\mathbb F_5]=4$.

After this use the fact that every finite extension over a finite field is cyclic. (you can find a proof here)

Noob mathematician
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Here’s another way of looking at the problem:

You’re asking for $\sqrt[4]2$ and the extension it generates over $\Bbb F_5$.

Now, $2$ is of (multiplicative) order $4$ in $\Bbb F_5^\times$, so its fourth root will be of order $16$. So you’re looking for the smallest power $5^m$ such that $\Bbb F_{5^m}$ has order ($5^m-1$) divisible by $16$. You see that $25$ and $125$ are no good, but surenough, $16\mid624=5^4-1$. So the degree of the splitting field is four. (Note: all extensions of finite fields are normal, abelian, cyclic.)

Lubin
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