Galois Group

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May 18th 2009, 10:03 PM
TheEmptySet

Galois Group

I need to find the Galois Group of $(x^2-2)(x^2-3)(x^2-5)$

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

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

so the automorphisms look like

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

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

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

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

Thanks
May 19th 2009, 03:08 AM
NonCommAlg

$[K:F]$ 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

$[K:F]$ 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 $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$
is

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

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 $(x^2-2)(x^2-3)(x^2-5)$

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

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

so the automorphisms look like

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

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

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

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

Thanks

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