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Determine the Galois group of $f=X^4+10$ and the lattice of intermediary fields.

I know that $f$ is irreducible by Eisenstein for $p=5$. I calculated the roots to be $\sqrt[4]{10}\zeta_8^i$ for $i\in\{\pm 1,\pm 3\}$. I know that $\sqrt[4]{10}\zeta_8^{-1}/\sqrt[4]{10}\zeta_8=i$, thus we can say that the splitting field is $\mathbb{Q}(\sqrt[4]{10}\zeta_8,i)$. Now I have to calculate the degree of this splitting field over $\mathbb{Q}$ to calculate the order of the Galois group. The trivial bound is given by $$[\mathbb{Q}(\sqrt[4]{10}\zeta_8,i):\mathbb{Q}]\leqslant[\mathbb{Q}(\sqrt[4]{10}\zeta_8):\mathbb{Q}]\cdot[\mathbb{Q}(i):\mathbb{Q}]=4\cdot 2=8.$$ Pari affirms that the Galois group is $D_4$, so this bound must be sharp.

I tried several things. For example, we have the tower of extensions $\mathbb{Q}\subset \mathbb{Q}(i)\subset\mathbb{Q}(\sqrt[4]{10}\zeta_8,i)$. Now my question boils down to proving that $f$ is also the minimal polynomial of $\sqrt[4]{10}\zeta_8$ over $\mathbb{Q}(i)$. Assume it splits in quadratic factors $(X^2+aX+b)(X^2-aX+b)$ for $a,b\in\mathbb{Q}(i)$, then the constant term $b^2=20$ for $b\in\mathbb{Q}(i)$. However, is equivalent to the (much more difficult question of) finding $x,y\in\mathbb{Q}$ for which $x^2-y^2=20$. I also tried to exploit the tower of extensions $\mathbb{Q}\subset \mathbb{Q}(\sqrt[4]{10}\zeta_8)\subset\mathbb{Q}(\sqrt[4]{10}\zeta_8,i)$. This boils down to the question $i\stackrel{?}{=}\mathbb{Q}(\sqrt[4]{10}\zeta_8)$.

Could someone give any hint? (This is not a duplicate, I am aware of this question, but it does not answer my question.)

Dr. Heinz Doofenshmirtz
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    You obviously know the zeros of $f(x)$, call them $\alpha_j,j=1,2,3,4$. Those putative quadratic factors would have to be of the form $g_{jk}(x)=(x-\alpha_j)(x-\alpha_k),j\neq k$, so one thing you can do is simply to expand those polynomials $g_{jk}(x)$ and check whether any of the have coefficients in $\Bbb{Q}(i)$. Mind you, this is not very elegant, but there are only three cases to check because you can concentrate on the quadratic factor that has $\alpha_1$ as a zero. In other words, just expand $g_{12},g_{13}$ and $g_{14}$. – Jyrki Lahtonen Apr 20 '19 at 07:55
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    If you know that $\Bbb{Z}[i]$ is a Euclidean domain, then you can also apply Eisenstein's criterion to prove that $x^4+10$ is irreducible over $\Bbb{Z}[i]$, hence over $\Bbb{Q}(i)$ by Gauss' Lemma. $10$ is associate to $(1+i)^2(2+i)(2-i)$ so it has prime factors with multiplicity one. – Jyrki Lahtonen Apr 20 '19 at 07:57
  • Related; https://math.stackexchange.com/questions/3180910/degree-of-splitting-field-of-x42x22-over-mathbfq – Servaes Apr 20 '19 at 08:00
  • Another thread dedicated irreducible quartic binomials of this type. Note that $x^4+2^\ell$ is different because $\sqrt2\in\Bbb{Q}(\zeta_8)$. – Jyrki Lahtonen Apr 20 '19 at 08:04
  • @JyrkiLahtonen Both your comments seem like fine answers to me. – Servaes Apr 20 '19 at 08:05
  • @Servaes. In the sense that I could flesh them out to an answer, sure. I'm trying to induce Heinz to do the work himself. Also, IIRC I have already used both techniques in other answers, and I seek to avoid such duplication. I'm fine with another user (a relatively new one in particular) using those (standard) techniques, but I try and avoid attempts at "getting paid twice for the same work". I am not able to always follow that maxim myself, but I try. Others feel differently. – Jyrki Lahtonen Apr 20 '19 at 08:09
  • ... and I am still looking for better duplicates to save everybody the time. – Jyrki Lahtonen Apr 20 '19 at 08:11
  • This may be useful as a duplicate target. Still a different constant, but not sure it makes a difference. Yet another. One more. – Jyrki Lahtonen Apr 20 '19 at 08:12
  • @Servaes To be clear. I'm definitely not suggesting that you should follow a similar policy. I do feel that users with six digit rep figures should show some restraint :-) – Jyrki Lahtonen Apr 20 '19 at 08:17
  • @JyrkiLahtonen if we take $(X-\sqrt[4]{10}\zeta_8)(X-\sqrt[4]{10}\zeta_8^{-1}$, this has constant coefficient $\sqrt[4]{10}$. How would you prove that this is not in $\mathbb{Q}(i)$? – Dr. Heinz Doofenshmirtz Apr 20 '19 at 18:05
  • Well, it’s hardly difficult to find integers with $x^2-y^2=20$, since $(n+2)^2-n^2=4n+4$, yielding $n=4$, $n+2=6$. Clearly, $(n+4)^2-n^2=20$ has no integer solutions, so $4$ and $6$ are the only pair of integers. Once you have that one point $(6,4)$ on the hyperbola $X^2-Y^2=20$, familiar methods allow you to find all rational pairs, same as you can find all the rational points on the unit circle. – Lubin Apr 20 '19 at 18:07
  • Heinz, $\root4\of{10}$ is real, so it is in $\Bbb{Q}(i)$ only if it is in $\Bbb{Q}$. But, isn't that constant coefficient equal to $\sqrt{10}$. No matter, that ain't in $\Bbb{Q}(i)$ either. – Jyrki Lahtonen Apr 20 '19 at 19:05

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