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My attempt:

I know that $P=\mathbb{Q(\sqrt{3+4i}+\sqrt{3-4i})} \subseteq \mathbb{Q(\sqrt{3+4i},\sqrt{3-4i})}=K$. But I don't know whether $K \subseteq P$. Assuming $K=P$. I can see that K is Galois over $\mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $\sqrt{3+4i}+\sqrt{3-4i}$.

Thanks for any help!

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    $\begingroup$ By $\sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $\sqrt{3-4i}$? $\endgroup$ – Lord Shark the Unknown Jan 11 '19 at 18:36
  • $\begingroup$ @LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $\mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$. $\endgroup$ – 6005 Jan 11 '19 at 19:18
  • $\begingroup$ Hmm, actually maybe it does matter. When adjoining $z = \sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it? $\endgroup$ – 6005 Jan 11 '19 at 19:25
  • $\begingroup$ And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$. $\endgroup$ – 6005 Jan 11 '19 at 19:27
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    $\begingroup$ Cool, $\mathbb{3+4i}$ looks fancy. $\endgroup$ – Dietrich Burde Jan 11 '19 at 19:39
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As the minimal polynomial can be solved, depending on which roots you denote $\;\sqrt{3+4i}$ and $\sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $\mathbf Q$. Hence it can't generate an extension of degree $4$.

On the other hand, it is easy to see that $\;K=\mathbf Q(\sqrt{3+4i})=\mathbf Q(\sqrt{3-4i})$, since $$\sqrt{3+4i}\,\sqrt{3-4i}=\pm 5.$$

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  • $\begingroup$ Oh! yes. (albeit for me, it's a $\mathbf Q$). Thanks for pointing it! $\endgroup$ – Bernard Jan 11 '19 at 19:37
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Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $\mathbb{Q}$.

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