$Gal_{\mathbb{Q}}[\mathbb{Q}[\sqrt{3}, \sqrt{5}, \sqrt{7}]$ Okay so I'm at a loss. Could someone just give me a general direction as to how to solve this? Much appreciated.
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The group is clearly a subgroup of $(\Bbb{Z}/2\Bbb{Z})^3$ (why?). Then it comes down to calculating the degree of the extension. There is a general theorem stating that if $p_1, p_2, ... p_n$ are distinct primes, then $[\Bbb{Q}[\sqrt{p_1}, ..., \sqrt{p_n}]: \Bbb{Q}] = 2^n$. If you don't want to use this you may need to do some fiddling with equations to find the degree. – Vik78 Apr 03 '17 at 02:19
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How come and how does that help me construct the galois group? Thanks and sorry – Kaity Scarlett Parsons Apr 03 '17 at 02:21
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What are the possible images of $\sqrt{3}, \sqrt{5}, \sqrt{7}$ under an element of the Galois group? How many possible images are there for each one? An element of the Galois group is completely determined by its action on these three elements. – Vik78 Apr 03 '17 at 02:22
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1Your question is very similar to this. If you fancy a generalization to more primes I sketched one way here. That argument relies on our ability to identify the fixed fields of maximal subgroups, so I recommend that you study the other highly upvoted answers in the latter thread also. Good stuff there! Following up on the links there gives more. – Jyrki Lahtonen Apr 03 '17 at 05:25
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You have three different plain-to-see automorphisms of $K=\mathbb Q[\sqrt3,\sqrt5,\sqrt7\,]$:
$\rho:\sqrt3\leftrightarrow-\sqrt3$, $\sigma:\sqrt5\leftrightarrow-\sqrt5$, $\tau:\sqrt7\leftrightarrow-\sqrt7$. Does that help? You do need to show that messing up $\sqrt3$ does not mess $\sqrt5$ up, etc.
Lubin
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