Let $K=Q(\sqrt3,\sqrt5)$. Show that the extension is $K/Q$ is simple and also Galois extension. Determine its Galois group.
I showed that extension is simple because $K=Q(\sqrt{3}+\sqrt{5})$ But I can not find its Galois group?
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2It's not $Q(\sqrt{3+5})$, but $Q(\sqrt{3}+ \sqrt{5})$ – Crostul May 05 '15 at 09:58
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@Crostul they are same – corcia candy May 05 '15 at 09:59
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4Absolutely not: $\sqrt{3+5}=\sqrt{8}=2\sqrt{2}$, so $Q(\sqrt{3+5})=Q(\sqrt{2})$. – Crostul May 05 '15 at 10:04
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1@corciacandy I've corrected the typo Crostul points out; if this really is not what was intended, please revert it (ideally with some explanation, because as Crostul points out the statement was not correct as written). – Travis Willse May 05 '15 at 10:07
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sorry you are right it is my carelessness – corcia candy May 05 '15 at 10:14
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Let $\sigma$ be a Galois automorphism of $K$ over $\mathbb Q$. What can $\sigma(\sqrt{3})$ be ? (Hint: $x = \sqrt(3)$ satisfies $x^2-3 = 0$, and $\sigma$ is a field automorphism stabilizing $\mathbb Q$, so that $y = \sigma(x)$ must satisfy the same equation). Same question for $\sqrt{5}$.
Now you have a maximum set of a few candidates for Galois automorphisms. You only need to prove that all of them actually are Galois automorphisms. This should not be the hardest part.
Circonflexe
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1You still need to prove that these four automorphisms do exist, but yes, this is the idea. – Circonflexe May 05 '15 at 10:07
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one more question, why did you choose $\sqrt3$ and $\sqrt5$.. can I choose $\sqrt3 + \sqrt5$ – corcia candy May 05 '15 at 10:17
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1Yes, you can. But you need to write down its minimal polynomial over $\mathbb Q$. – Circonflexe May 05 '15 at 10:20
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Hint: Notice that any $\sigma \in \mathrm {Aut} L$ is determined by the images of $\sigma (\sqrt{3})$ and $\sigma (\sqrt 5)$ then the possibilities are $\sigma (\sqrt 3) \in \{-\sqrt 3, \sqrt 3 \}$ and $\sigma (\sqrt 5) \in \{-\sqrt 5, \sqrt 5\}$.
Can you see how this is related to $\mathbb Z_2 \times \mathbb Z_2$?
Aaron Maroja
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Actually I think that this Galois group isomorphic to Klein 4 group – corcia candy May 07 '15 at 13:26
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1There are only two groups of order $4$, apart from isomorphisms: $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. Now if you check the order of each element then you'll notice that they all have order $2$, what does that say? – Aaron Maroja May 07 '15 at 13:33
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Notice that the Klein group is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$, because all of its elements have order $2$. – Aaron Maroja May 07 '15 at 13:46