# Thread: just one more question on Galois group

1. ## just one more question on Galois group

So, I have to determine the Galois group of $x^4-5$ over $\mathbb{Q}$, $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$.

$f(x)=x^4-5$
That is irreducible over $\mathbb{Q}$ by Esienstein's criterion.
Resolvent is: $g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
$\alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
$
\mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$

$
[\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$

So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

Is this irreducible or not?

And what to do with
$\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$?

Thank you!

2. Originally Posted by marianne
So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

[COLOR=Red][COLOR=Black]Is this irreducible or not?
Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
This means the Galois group is $D_4$.

3. Originally Posted by ThePerfectHacker
Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
This means the Galois group is $D_4$.
Oh, I missed that, thank you!!!

Could you possibly help me out with , ?

4. Originally Posted by marianne
So, I have to determine the Galois group of $x^4-5$ over $\mathbb{Q}$, $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$.

$f(x)=x^4-5$
That is irreducible over $\mathbb{Q}$ by Esienstein's criterion.
Resolvent is: $g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
$\alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
$
\mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$

$
[\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$

So $G$ is isomorphic to $D_4$ or $\mathbb{Z}_4$.
To determine which, we check is $f$ irreducible over $\mathbb{Q}[\sqrt{-5}]$
$
f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

Is this irreducible or not?

And what to do with
$\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[\sqrt{-5}]$?
Let $K$ be the splitting field for the resolvent, $x(x^2+20)$. As you have shown $K = \mathbb{Q}(\sqrt{-5})$ and so $[K:\mathbb{Q}]=2$. Therefore, the Galois group is either $\mathbb{Z}_4$ or $D_4$. It all depends whether or not $f$ is irreducible over $K$. First $f$ has NO roots in $K$ because the roots of $f$ are: $\pm\sqrt[4]{5}, \pm i\sqrt[4]{5}$. These numbers have degree four over $\mathbb{Q}$ while $K$ is only a degree two extension. Thus, you need to show that $(x^2 + ax + b)(x^2 + cx + d) = x^4 - 5$ is impossible for $a,b,c,d\in K$. To simplify things we will appeal to a result from number theory, that the ring of integers in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$ if $d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Q}(\sqrt{-5})$. By Gauss' Lemma (the general one) to prove irreduciblity of $x^4 - 5$ over $K$ it sufficies to prove irreducibility of $x^4 - 5$ over $\mathbb{Z}[\sqrt{-5}]$. Since we have $(x^2 + ax + b)(x^2 + bx + d) = x^4 - 5$ where $a,b,c,d,\in \mathbb{Z}[\sqrt{-5}]$ if we expand we get $x^4+(a+c)x^3+(b+ac + d)x^2 +(ad+bc)x+bd = x^4 - 5$. I think (I did not check this!) that it is only possible to get $bd = -5$ if $b=d=\pm \sqrt{-5}$, now show that both those cases lead to impossibilities in the coefficiensts $a,b,c,d$. Can you be the mathematician of the gaps and fill them in?

5. Originally Posted by ThePerfectHacker
To simplify things we will appeal to a result from number theory, that the ring of integers in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$ if $d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Q}(\sqrt{-5})$.
I have something similar, but it's $x^4-7$ over $\mathbb{Q}[\sqrt{-7}]$. How do I check is $f$ irreducible? I can't use the number theory result you provided.

6. Originally Posted by georgel
I have something similar, but it's $x^4-7$ over $\mathbb{Q}[\sqrt{-7}]$. How do I check is $f$ irreducible? I can't use the number theory result you provided.
If $d\equiv 1(\bmod 4)$ the ring of integers of $\mathbb{Q}(\sqrt{d})$ is given by $\{ \tfrac{a+b\sqrt{d}}{2} | a\equiv b(\bmod 2) \}$.