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This question already has an answer here:

Let $E = \mathbb{Q}(\sqrt{4+\sqrt{7}})$. Prove that $E$ is a normal extension of $\mathbb{Q}$ and find the Galois group $Gal(E/\mathbb{Q})$.

My attempt:

Let $\alpha = \sqrt{4+\sqrt{7}}$. $\alpha$ is a root of the polynomial $(x^2 - 4)^2 - 7$.

1) How can I show $f(x) = (x^2 - 4)^2 - 7 = x^4 - 8x^2 + 9$ is irreducible over $\mathbb{Q}$ because I can't apply Eisenstein condition?

If $f(x)$ is irreducible, $[E:\mathbb{Q}] = 4 = |Gal(E/\mathbb{Q})|$ and I have shown that all roots of $f(x)$, say $\alpha_{1} = \sqrt{4+\sqrt{7}}, \alpha_{2} = -\sqrt{4+\sqrt{7}}, \alpha_{3} = \sqrt{4-\sqrt{7}}, \alpha_{4} = -\sqrt{4-\sqrt{7}}$ are in $E$ and hence $E$ is a normal extension of $\mathbb{Q}$. Let $Gal(E/\mathbb{Q}) = \{ e,\sigma,\tau,\sigma\tau \}$

2) How can I find the order of $\sigma,\tau$ to find $Gal(E/\mathbb{Q})$? Thanks.

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marked as duplicate by Jyrki Lahtonen abstract-algebra May 15 at 5:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There are infinitely many rational numbers, so the tag finite-fields was inappropriate. $\endgroup$ – Jyrki Lahtonen May 15 at 5:01
  • $\begingroup$ See here for an example where a superficially similar extension (= the splitting field of a biquadratic) has a cyclic case, and here for the "full" story. $\endgroup$ – Jyrki Lahtonen May 15 at 5:20
  • $\begingroup$ Arguably the question about a generic biquadratic would make a better duplicate target. $\endgroup$ – Jyrki Lahtonen May 15 at 7:18
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Note that $E$ is a field extension of $\mathbb{Q}(\sqrt{7})$ and hence we can use the tower rule:

$4\geq [E:\mathbb{Q}]=[E:\mathbb{Q}(\sqrt{7})][\mathbb{Q}(\sqrt{7}):\mathbb{Q}]=[E:\mathbb{Q}(\sqrt{7})]\times 2$

The element $\alpha$ does not belong to $\mathbb{Q}(\sqrt{7})$. In order to see this try to write $\alpha=a+b\sqrt{7}$ when $a,b\in\mathbb{Q}$ and you will get that $\sqrt{7}$ is rational which is of course a contradiction. Hence $[E:\mathbb{Q}(\sqrt{7})]$ is at least $2$, so that gives us $4\geq[E:\mathbb{Q}]\geq 4$ which of course implies $[E:\mathbb{Q}]=4$.

Ok, now you know that the extension is Galois of degree $4$. Hence $Gal(E/\mathbb{Q})$ is either $\mathbb{Z_4}$ or $\mathbb{Z_2}\times\mathbb{Z_2}$. However, if we denote $\beta=\sqrt{4-\sqrt{7}}$ then the roots of $f$ are $\alpha,-\alpha,\beta,-\beta$. Now, since the extension $E/\mathbb{Q}$ is simple and generated by $\alpha$ we know that for any $c\in E$ which is a root of the minimal polynomial of $\alpha$ there exists an element in $Gal(E/\mathbb{Q})$ which sends $\alpha\to c$. Hence there is an element in $Gal(E/\mathbb{Q})$ which sends $\alpha\to -\alpha$. Since $\beta=\frac{3}{\alpha}$ it is easy to see that this automorphism sends $\beta\to -\beta$, and hence it defines the permutation $(\alpha,-\alpha)(\beta,-\beta)$ on the roots of $f$. Also, there is an element in $Gal(E/\mathbb{Q})$ which sends $\alpha\to\beta$. Again, it is easy to check that such an automorphism must send $\beta\to\alpha$, and hence it defines the permutaiton $(\alpha,\beta)(-\alpha,-\beta)$. So $Gal(E/\mathbb{Q})$ contains at least two elements of order $2$ and this implies that it must be isomorphic to $\mathbb{Z_2}\times\mathbb{Z_2}$.

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  • $\begingroup$ Thanks, that was super clear. One question: How do we know that for a simple extension, there exists an element in the $Aut$ group which switches between $\alpha$ and $-\alpha$? $\endgroup$ – manifolded May 14 at 21:48
  • $\begingroup$ Sorry, I just realized that the last part of my answer was simply wrong. So I changed the last paragraph. $\endgroup$ – Mark May 14 at 22:55
  • $\begingroup$ Thanks for updating the answer. Should it be $\alpha\rightarrow -\alpha,\beta\rightarrow -\beta$ which leads to the permutation $(\alpha,-\alpha)(\beta,-\beta)$ and not $(\alpha,-\alpha)(\beta,\beta)$? Though both have order 2. $\endgroup$ – manifolded May 14 at 22:59
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    $\begingroup$ Another way of seeing that $4+\sqrt7$ is not a square in its field is to note that $4+\sqrt7=(2-\sqrt7)^2(8+3\sqrt7)$. But $8+3\sqrt7$ is a generator of the unit group of $\Bbb Z[\sqrt7\,]$, so can’t be a square. $\endgroup$ – Lubin May 15 at 4:27
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    $\begingroup$ As a check, I wanted to find the other two quadratic extensions of $\Bbb Q$ in $E$. It turns out that they are $\Bbb Q(\sqrt2\,)$ and $\Bbb Q(\sqrt{14}\,)$. $\endgroup$ – Lubin May 15 at 4:59
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HINTS.

  1. If it were reducible, then it would decompose into two quadratics, a linear and a cubic, or four linear factors. To have a linear factor, it would have to have a root over $\mathbb{Q}$. Does it? This only leaves the quadratic case. Assume it is a product $(x^2+ax+b)(x^2+cx+d)$ and see if you can arrive at a contradiction with the coefficients.

  2. Here recall what I like to think of as 'conjugation for quadratics': $\overline{a+b \sqrt{d}}= a-b\sqrt{d}$. Can you see 'two' of these 'conjugations' (one 'outside' and one 'inside') for the defining element for your field? This should make it easier to see what the order of the elements must be by actually writing down the maps explicitly. The Galois group should then be apparent.

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    $\begingroup$ Useful to know that if it factorises over $\mathbb Q$ it factorises over $\mathbb Z$ (Gauss) $\endgroup$ – Mark Bennet May 14 at 22:00

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