# Galois Group

• May 18th 2009, 11:03 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, 04:08 AM
NonCommAlg
$[K:F]$ is 8 not 6. ok, i really need to get some sleep now! haha
• May 19th 2009, 08:16 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, 07:08 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 $2$).