# Finding the Galois group of $\mathbb Q (\sqrt 5 +\sqrt 7) \big/ \mathbb Q$

I know that this extension has degree $4$. Thus, the Galois group is embedded in $S_4$. I know that the groups of order $4$ are $\mathbb Z_4$ and $V_4$, but both can be embedded in $S_4$. So, since I know that one is cyclic meanwhile the other is not, I've tried to determine if the Galois group is cyclic but I couldn't make it. Is there any other way?

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Have you shown that $\Bbb{Q}(\sqrt{5}+\sqrt{7})=\Bbb{Q}(\sqrt{5},\sqrt{7})$? –  Chris Eagle Jan 26 '13 at 18:16
yes, I've already proved it. thanks –  Kits89 Jan 26 '13 at 18:45
Have you proved the fundamental theorem of Galois theory? –  JSchlather Jan 26 '13 at 21:41
For this problem I assume it as known –  Kits89 Jan 27 '13 at 8:18

Indeed this is an extension of degree $4$.

Show $$\mathbb{Q}(\sqrt{5}+\sqrt{7})=\mathbb{Q}(\sqrt{5},\sqrt{7})$$ and use that fact that any element must go to a conjugate of his under automorphism to find all the elements of $Gal(\mathbb{Q}(\sqrt{5}+\sqrt{7})/\mathbb{Q})$.

Can you find there an element of order $4$ ?

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but, first, shouldn't I need the irreducible polynomial $Irr(\sqrt{5}+\sqrt{7},\mathbb Q)$ for the mapping of its roots? –  Kits89 Jan 26 '13 at 18:47
@Kits89 - No, this is sine with the first equality you have two generators for the extension and you know what is the minimal polynomial of both of them –  Belgi Jan 26 '13 at 19:29
You mean the polynomials $x^2-5$ and $x^2-7$. I still don't see how can I find the automorphisms using these. I mean, $\sqrt{5}$ can only be mapped to $\sqrt{5}$ or $-\sqrt{5}$, right? An element of order $4$ may be $\sqrt{5}+\sqrt{7}$ ? –  Kits89 Jan 27 '13 at 8:22
@Kits89 - The elements of $Gal(\mathbb{Q}(\sqrt{5}+\sqrt{7})/\mathbb{Q})$ are functions, $\sqrt{5}+\sqrt{7}$ is not a function. You have the polynomials right, so you have two options to map $\sqrt{5}$ and two options to map $\sqrt{7}$ so can you think of the four functions you know exist since the Galois group is of order $4$ ? –  Belgi Jan 27 '13 at 11:45
Oh, I understand, then is isomorphic to $V_4$ because $\mathbb Z_4$ is cyclic but $Gal(\mathbb Q(\sqrt{5}+\sqrt{7})$ does not have any element of order 4 (maximum order 2). –  Kits89 Jan 27 '13 at 11:53

You should first prove that $\mathbf{Q}(\sqrt{5}+\sqrt{7})/\mathbf{Q}$ is a Galois extension. For this it may be useful to verify that $\mathbf{Q}(\sqrt{5}+\sqrt{7}) = \mathbf{Q}(\sqrt{5},\sqrt{7})$. Then you might consider the Galois groups of $\mathbf{Q}(\sqrt{5})/\mathbf{Q}$ and $\mathbf{Q}(\sqrt{7})/\mathbf{Q}$.

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