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Let $p(x):=x^4-2x^2-2 \in F_5[x]$. I am trying to find $\mathrm{Gal}(L/F_5)$, where $L$ is the splitting field of $p$ over $F_5$.

My approach: So the first thing would be to prove that, $p$ is irreducible over $F_5$. But as far as I know because the degree of $p$ is $4,$ the simple method of testing by absence of zeros does not work (as far as I know).

Well, here begins the problem. I know that if I can prove irreducibility and determine the splitting field $L$, I should somehow get something like $[L:F_5]=n$ which means that $L$ has to be cyclic.

Note: I did try this exercise for some time, but couldn't really figure something out. All by approaches seem to either not work or have errors in them. I would be very thankful if someone could explain the method to me.

aqualubix
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wanymose
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  • Do you need a hint or a complete solution? – Andres2003 May 24 '23 at 04:33
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    Hint: $p(x)$ is quartic, but it can be written as $q(x^2)$ for a quadratic polynomial $q$. So, you can solve for the roots of $p$ using the quadratic formula! This should help you determine irreducibility – diracdeltafunk May 24 '23 at 04:36
  • @Andres2003 I don't think that a hint would help me. As mentioned, I did try this myself. I think there is something in this topic that I didn't understand correctly or can't connect the dots. I would prefer a solution, maybe then I could see which point I am missing. – wanymose May 24 '23 at 04:39
  • @diracdeltafunk ok, then I get the polynomial $q(x)=x^2-2x-2$, now by trying elements of $F_5$, and confirming that there is no zero, I can argue that $q(x)$ is irreducible. But does this then mean that $p$ is irreducible, too? – wanymose May 24 '23 at 04:42
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    @wanymose If you know that your polynomial $p$ is not a product of quadratic factors, you know that $p$ is irreducible (I gather that you already checked $p$ has no linear factor and that exhausts all the possibility of $p$ being reducible). – daruma May 24 '23 at 04:45
  • @daruma thank you. – wanymose May 24 '23 at 04:48
  • @daruma Now that I know that $p$ is irreducible, I calculate the roots: $\sqrt{1-\sqrt{3}},-\sqrt{1-\sqrt{3}},\sqrt{1+\sqrt{3}},-\sqrt{1-\sqrt{3}}$. I think the splitting field then is, $L=F_5(\sqrt{1+\sqrt{3}},i\sqrt{-1+\sqrt{3}})$, is that correct? – wanymose May 24 '23 at 04:49
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    Once know that $p$ is irreducible, as it is of degree 4, then you know that the field $F_5[X]/P$ is $F_{5^4}$, and $F_{5^4}$ is a Galois extension of $F_5$ with Galois group cyclic of order 4 generated by the Frobenius. – Thomas May 24 '23 at 05:06
  • @Thomas Ok, the Frobenius Map is $\phi(x)=x^5$, does this mean that I just take some root $a$ of $x^4-2x^2-2$, and then calculate $(\phi(a))^0=(a^5)^0$,..., $(\phi(a))^3=(a^5)^3$. Is this correct? – wanymose May 24 '23 at 05:55
  • No This is a general result $Gal(F_{p^n}:F_p) $is cyclic generated by the Frobenius. This follows from the fact that $F_{p^n}^*$ is a cyclic group. – Thomas May 24 '23 at 18:15
  • A simple proof is : let. $\phi$ acts on $F_{p^n}$. let us prove that $\Phi $ is of order $n$. As every element in $F_{p^n}$ satisfies $x^{p^n}=x$ , one sees that $\Phi ^n=id$ so the order of $\Phi $ divides $n$, but the set of fixed points of $\Phi ^k$ is exactly $F_{p^k}$, as this is the set of $x^{p^k}-x=0$.. – Thomas May 24 '23 at 18:23

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Let's call $f: = x^4-2x^2-2$. You could use the fundamental theorem of Galois theory. A field extension $\mathbf{F}_5 \subset L$ of finite fields of degree $n$ will be cyclic of order $n$. Therefore, you should determine the degree of this extension.

Actually, you know that $f$ is irreducible over $\mathbf{F}_5$ (you already know that it doesn't have a linear factor), because a product of two quadratic factors let's say $f = (x^2+ax+b)(x^2+cd+d)$ where $a,b,c,d \in \mathbf{F}_5$ gives that $b = d$ (use that $\mathbf{F}_5$ is a field therefore a domain). Now, the last term $b^2 = d^2$ should be equal to $-2$, but $-2$ is not a square in $\mathbf{F}_5$. Therefore, $f$ is irreducible over $\mathbf{F}_5$.

Use the abc-formula which still holds in characteristic $5$ to show that $x^4-2x^2-2 = (x^2 -1 -2\sqrt{2})(x^2-1+2\sqrt{2})$. Notice that $\mathbf{F}_5$ doesn't contain a square root of $2$. Then you can see that $L = \mathbf{F}_5\left(\sqrt{1+2\sqrt{2}}, \sqrt{1 - 2\sqrt{2}}\right)$. Because $f \in \mathbf{F}_5[x]$ is separable (look at the roots), you can see that the Galois group $G:= \text{Gal}(L/\mathbf{F}_5)$ can be imbedded into $S_4$. Note that $f \in \mathbf{F}_5[x]$ is irreducible, therefore the order of $G$ is divisible by $4$ (and $G$ is a transitive subgroup of $S_4$). You can conclude that $G$ is a cyclic group of order $4$.

Let's call $\alpha := \sqrt{1+2\sqrt{2}}$ and $\beta := \sqrt{1-2\sqrt{2}}$, then because of the irreducibility of $f$ over $\mathbf{F}_5$ you know that $[\mathbf{F}_5(\alpha) : \mathbf{F}_5] = 4$. We know (because of Galois theory) that $ \# G = [\mathbf{F}_5(\alpha,\beta) : \mathbf{F}_5] = 4$, so you conclude that $\mathbf{F}_5(\alpha) = \mathbf{F}_5(\alpha,\beta)$. Using this you can explicitly determine the Galois group. You should find an expression of $\beta$ in terms of $\alpha$ which can be `legally' obtained ( I disagree with Aphelli's comment).

Anton Odina
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  • Let $u=\sqrt{1+2\sqrt{2}}$, then $\sqrt{1-2\sqrt{2}}u=\sqrt{-7}=2\sqrt{2}=u^2-1$, so $L=\mathbb{F}_5(u)$. – Aphelli May 24 '23 at 09:00
  • @Aphelli I'm not sure if you are allowed to multiply the roots like that. – Anton Odina May 24 '23 at 09:11
  • As long as you’re careful about the exact meaning of what you write, there’s no problem. In my case, what I mean is the following: let $u$ be a square root of $2$ (in $\mathbb{F}{25}$), and $v$ a square root of $1+2u$ (in $\mathbb{F}{625}$), then $\frac{2u}{v}$ is a square root of $1-2u$. – Aphelli May 24 '23 at 11:35
  • Ah, when you put it like that it's better. – Anton Odina May 24 '23 at 12:14