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This is exercise 4.5 from Baker Galois Theory:

Use Kaplansky's theorem to find the Galois group of the splitting field $E$ of the polynomial $x^4 +3 \in \mathbb Q[x]$ over $\mathbb Q$. Determine all the subextensions $F \leq E$ for which $F/\mathbb Q$ is Galois.

The statement of Kaplansky's Theorem is given as follows:

4.28. THEOREM (Kaplansky's Theorem). Let $f(X)=X^4+a X^2+b \in \mathbb{Q}[X]$ be irreducible.
(i) If b is a square in $\mathbb{Q}$ then $\operatorname{Gal}(\mathbb{Q}(f(X)) / \mathbb{Q}) \cong \mathbb{Z} / 2 \times \mathbb{Z} / 2$.
(ii) If $b\left(a^2-4 b\right)$ is a square in $\mathbb{Q}$ then $\operatorname{Gal}(\mathbb{Q}(f(X)) / \mathbb{Q}) \cong \mathbb{Z} / 4$.
(iii) If neither $b$ nor $b\left(a^2-4 b\right)$ is a square in $\mathbb{Q}$ then $\operatorname{Gal}(\mathbb{Q}(f(X)) / \mathbb{Q}) \cong \mathrm{D}_8$.

Attempt: By Eisenstein criterion, $f(x)=x^4+3$ is irreducible. Therefore we can apply Kaplansky's theorem. Note that $f(x)=x^4 + ax^2 +b$ where $a=0$ and $b=3$. Then $b$ is not a square in $\mathbb Q$, and neither is $b(a^2-4b)$, so by Kaplansky's theorem we have $\operatorname{Gal}\left( \mathbb Q(f(x))/ \mathbb Q \right) \cong D_8$, the Dihedral group with $8$ elements.

The second part of the question is where I got a bit confused. I believe we can use the Fundamental Theorem of Galois Theory to establish a one-to-one correspondence between subgroups of $D_8$ and subfields $F$ of $E = \mathbb Q(f(x))$ (splitting field) such that $F/ \mathbb Q$ is Galois. But how do you explicitly determine what the subextensions are using the subgroups of $D_8$?

pyridoxal_trigeminus
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    After a very quick search I can find several theorems named after Kaplansky, and none of them are in Galois theory. Could you please tell us what the theorem says? – Arthur May 01 '23 at 21:48
  • @Arthur J. Rotman, one of Kap's students, on page 106 of his Galois Theory, lists three homework problems attributed to Kap about quartics with rational coefficients. for irreducible $x^4 + ax^2 +b.$ If $b$ is a square, the Galois group is the Klein four group $V$ Next, $b$ not a square but $b(a^2-b)$ is, group $Z/4Z$ Final choice $D_8$ This is problem 113. – Will Jagy May 01 '23 at 22:18
  • @Arthur Sorry! It is stated in the linked PDF on the bottom of Page 60. – pyridoxal_trigeminus May 01 '23 at 22:40
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    The sub-extensions $F\le E$ such that $F/\mathbb Q$ is Galois corresponds to normal subgroups of $D_8$. Such subgroups are classified in https://math.stackexchange.com/questions/484828/normal-subgroups-of-dihedral-groups – Kenta S May 02 '23 at 01:34

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Your attempt is clear and correct. For a subgroup $H\subset D_8$, the corresponding subextension is given by $F=E^H$, the subfield of elements fixed by $H$. For example, the three subgroups of index 2 has the corresponding subextensions $\mathbb{Q}(\sqrt{-1})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{3}).$ I recommend to work out yourself!

Ayaka
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