# Galois Group of $x^4 - 7$ over $\mathbb{F}_5$

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 $$\mathbb{F}_5(\sqrt{2},i)$$ where $$i,\sqrt{2}$$ lie in a fixed algebraic closure of $$\mathbb{F}_5$$ and $$i^2 = -1$$ and $$(\sqrt{2})^4 = 2$$. 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 $$[\mathbb{F}_5(\sqrt{2}): \mathbb{F}_5] = 2$$ or $$4$$ which means the Galois Group is either order $$2$$ or $$4$$. If it were $$2$$, then $$\sqrt{2} = a + b\sqrt{2}$$ 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 $$\sigma: \mathbb{F}_5(\sqrt{2}) \to \mathbb{F}_5(\sqrt{2})$$ given by $$\sigma(\sqrt{2}) = 2\sqrt{2}$$. This is an automorphism of $$\mathbb{F}_5(\sqrt{2})$$ of order $$4$$ and so must be the the galois group is $$\langle \sigma \rangle$$.

• Typo in the question? Aug 1, 2020 at 20:52
• @paulgarrett fixed
– Mike
Aug 1, 2020 at 20:55
• 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. Aug 1, 2020 at 21:01
• 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. Aug 1, 2020 at 21:03
• @JyrkiLahtonen Thank you for the comment
– Mike
Aug 1, 2020 at 21:08

I am stating an alternative path.

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

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 $$\mathbb F_5(\sqrt 2)$$. $$x^4-2=\prod_{a\in \mathbb F_5^*}(x-a\sqrt2)$$

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 $$[\mathbb F_5(\sqrt 2):\mathbb F_5]=4$$.

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

Here’s another way of looking at the problem:

You’re asking for $$\sqrt2$$ 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.)

• Thank you for the comment
– Mike
Aug 2, 2020 at 1:11