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I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $\sqrt{3+\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3-\sqrt{7}})$ and $\sqrt{3-\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3+\sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.

Any help is appreciated!

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  • $\begingroup$ $\sqrt{3+\sqrt{7}} = a+b\sqrt{3-\sqrt{7}} \implies 3+\sqrt{7} = a^2+2ab\sqrt{3-\sqrt{7}} + b(3-\sqrt{7})$ $ \implies c+d\sqrt{7} = e\sqrt{3-\sqrt{7}} \implies c^2+2cd\sqrt{7}+7d^2 = e^2(3-\sqrt{7}) \implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction. $\endgroup$ – mathworker21 Jan 8 at 15:39
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    $\begingroup$ @mathworker21 What is $a$ and $b$? $\{1,\sqrt{3-\sqrt{7}}\}$ is not a basis for $\mathbb{Q}(\sqrt{3-\sqrt{7}})$ over $\mathbb{Q}$ $\endgroup$ – mouthetics Jan 8 at 15:40
  • $\begingroup$ @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals) $\endgroup$ – mathworker21 Jan 8 at 15:41
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Let $x_{\pm}=\sqrt{3 \pm \sqrt{7}}$.

It is easy to see that we have the following quadratic extensions:

$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_+)$,

$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_-)$.

Assume that $x_+ \in K=\mathbb{Q}(x_-)$. Then $\sqrt{2} = x_+x_- \in K$, thus $L=\mathbb{Q}(\sqrt{2},\sqrt{7}) \subset K$.

Since these fields have the same degree over $\mathbb{Q}$, $K \subset L$, ie $x_+=a+b\sqrt{2}+c\sqrt{7}+d\sqrt{14}$ for rationals $a,b,c,d$.

Taking squares, we get $3+\sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)\sqrt{2} + (2da+2bc)\sqrt{14} + (2ca+4bd)\sqrt{7}$.

Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.

Assume $a=0$: then $d \neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.

Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.

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In general, in this kind of problem, it is better not to mix the two operations + and $\times$. Let me give an illustration here, using only $\times$. Introduce the quadratic field $k=\mathbf Q(\sqrt 7)$. Adopting the notation $x_{\pm}=\sqrt {3 \pm \sqrt 7}$ suggested by @Mindlack, let us write $K_{\pm}=k(x_{\pm})$. These are two extensions of $k$ of degree at most $2$ :

  • if $K_{+}$ or $K_{-} =k$, i.e. $(3 \pm\sqrt 7)\in {k^*}^2$, norming down to $\mathbf Q$ shows that $N(3\pm\sqrt 7)=2$ is a square in $\mathbf Q^*$: impossible

  • if both degrees are 2, $K_{\pm}\subset K_{\mp}$ iff $K_{\pm}= K_{\mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+\sqrt 7)(3-\sqrt 7)\in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $\mathbf Q(\sqrt 2)=\mathbf Q(\sqrt 7)$, iff $2.7$ is a square in $\mathbf Q^*$(again by Kummer): impossible because $\mathbf Z$ is a UFD ./.

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HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $\Bbb{Q}$, and hence that $[\Bbb{Q}(\sqrt{3\pm\sqrt{7}}):\Bbb{Q}]=4$, but that the splitting field of $f$ over $\Bbb{Q}$ has degree greater than $4$.

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  • $\begingroup$ My goal is to justify that the splitting field is $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking? $\endgroup$ – BBC3 Jan 8 at 16:27
  • $\begingroup$ How to prove that the splitting field of $f$ over $\mathbb{Q}$ has degree greater than $4$? Any other method? $\endgroup$ – mouthetics Jan 8 at 16:28
  • $\begingroup$ Do you mean $X^4 - 6X^2 + 2$? $\endgroup$ – Connor Harris Jan 8 at 17:08
  • $\begingroup$ @ConnorHarris Indeed I do, edited. $\endgroup$ – Servaes Jan 8 at 22:46
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    $\begingroup$ @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind. $\endgroup$ – Servaes Jan 8 at 22:48

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