# Galois Group

• May 18th 2009, 10:03 PM
TheEmptySet
Galois Group
I need to find the Galois Group of $\displaystyle (x^2-2)(x^2-3)(x^2-5)$

So to start since the splitting field is a degree 6 extention $\displaystyle [K:F]=6$ so the Galois Group should have 6 isomporphisms.

The splitting field looks like $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

so the automorphisms look like

$\displaystyle 1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$

$\displaystyle \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$ $\displaystyle \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$

$\displaystyle \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ -\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$ $\displaystyle \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$

So I have 8, so I think I am misunderstanding something.

Thanks
• May 19th 2009, 03:08 AM
NonCommAlg
$\displaystyle [K:F]$ is 8 not 6. ok, i really need to get some sleep now! haha
• May 19th 2009, 07:16 AM
TheEmptySet
Quote:

Originally Posted by NonCommAlg
$\displaystyle [K:F]$ is 8 not 6. ok, i really need to get some sleep now! haha

Wow I think I needed some too.

The basis for field extention of $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$
is

$\displaystyle 1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},\sqrt{10},\s qrt{15},\sqrt{30}$

and yes indeeded it does have 8 elements(Crying)

Thanks again
• May 23rd 2009, 06:08 PM
ThePerfectHacker
Quote:

Originally Posted by TheEmptySet
I need to find the Galois Group of $\displaystyle (x^2-2)(x^2-3)(x^2-5)$

So to start since the splitting field is a degree 6 extention $\displaystyle [K:F]=6$ so the Galois Group should have 6 isomporphisms.

The splitting field looks like $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

so the automorphisms look like

$\displaystyle 1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$

$\displaystyle \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$ $\displaystyle \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} \end{cases}$ $\displaystyle \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$

$\displaystyle \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\ -\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$ $\displaystyle \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\ \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} \end{cases}$

So I have 8, so I think I am misunderstanding something.

Thanks

In general, let $\displaystyle p_1,...,p_n$ be distinct prime numbers. Then the Galois group of $\displaystyle (x^2 - p_1)...(x^2 - p_n)$ is $\displaystyle \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ and has order $\displaystyle 2^n$. It happens to be isomorphic to $\displaystyle \mathbb{Z}_2^n$ (since all non-trivial automorphisms have order $\displaystyle 2$).