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Let $D_4$ be the dihedral group of order 8. I want to find a field $K/\mathbb Q$ such that $K\supset \mathbb Q(\sqrt 2,\sqrt 3)$ and $Gal(K/\mathbb Q)=D_4$. I know the Galois group of $\mathbb Q(\sqrt 2,\sqrt 3)$ is $C_2^2$ and $D_4$ has a normal subgroup $C_2$ whose quotient is $C_2^2$. So I should be able to find such a field $K$. It should be a quadratic extension of $\mathbb Q(\sqrt 2,\sqrt 3)$. But in practice I cannot find it. I tried to look at splitting field of $\sqrt{\sqrt2+\sqrt3}$, which has Galois group $C_2\times D_4$. But do not know how to find an intermediate field with Galois group $D_4$.

JKDASF
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3 Answers3

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This is a problem well-suited to the L-functions and modular forms database (LMFDB). The LMFDB label of $\newcommand{\Q}{\mathbb{Q}} \Q(\sqrt{2}, \sqrt{3})$ is 4.4.2304.1, and the image of the embedding of $D_4 \hookrightarrow S_8$ has transitive group label 8T4. Searching the LMFDB for number fields with the given Galois group and intermediate field, we find 50 matches:

https://www.lmfdb.org/NumberField/?galois_group=8T4&subfield=4.4.2304.1

The first such example is the field 8.8.3057647616.1, the number field defined by the irreducible polynomial $$ x^8 - 4 x^7 - 4 x^6 + 20 x^5 + 4 x^4 - 20 x^3 - 4 x^2 + 4 x + 1 \, . $$

Viktor Vaughn
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    Cool. An easier one is the splitting field of $x^4 - 6x^2 + 6 $, i.e., $\mathbb Q(\sqrt{3+\sqrt3},\sqrt{3-\sqrt3})$. Thanks! – JKDASF Jan 09 '25 at 06:44
  • @JKDASF In my opinion that example deserves to be fleshed out in the form of an answer. – Jyrki Lahtonen Jan 09 '25 at 06:47
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    @JKDASF I believe that example (whose splitting field is given by the polynomial $x^8 - 60 x^6 + 1104 x^4 - 6120 x^2 + 36$) is isomorphic to the one given in my answer. – Viktor Vaughn Jan 09 '25 at 06:54
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With the help of LMFDB as Viktor Vaughn suggested, here is an explicit example:

$\mathbb Q(\sqrt{3+\sqrt3},\sqrt{3-\sqrt3})\supset\mathbb Q(\sqrt2, \sqrt3)\supset\mathbb Q$

which is the splitting field of $x^4−6x^2+6$, and isomorphic to the field in Viktor Vaughn's answer.

JKDASF
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First consider $$L = ℚ(\sqrt{2}, \sqrt{3}, \sqrt{\sqrt{2}+\sqrt{3}})$$

Over ℚ, the element $\sqrt{2}+\sqrt{3}$ has various Galois conjugates obtained by sending $\sqrt{2} \mapsto \pm\sqrt{2}, \sqrt{3} \mapsto \pm\sqrt{3}$

Once you adjoin $\sqrt{2}$ and $\sqrt{3}$, the polynomial $$x^2-(\sqrt{2}+\sqrt{3})$$ is irreducible over $ℚ(\sqrt{2}, \sqrt{3})$, so adjoining $\sqrt{\sqrt{2}+\sqrt{3}}$ gives a genuine quadratic extension of $ℚ(\sqrt{2}, \sqrt{3})$.

This is not Galois over $ℚ(\sqrt{2}, \sqrt{3})$, so we also adjoin $\sqrt{\sqrt{2}-\sqrt{3}}$, forming $L = ℚ(\sqrt{2}, \sqrt{3}, \sqrt{\sqrt{2}+\sqrt{3}}, \sqrt{\sqrt{2}-\sqrt{3}})$.

A standard calculation then shows [$L:ℚ$] = 16. The full Galois group $Gal(L/ℚ)$ is isomorphic to $C_2 \times D_4$ (a group of order 16).

Since $Gal(L/ℚ)$ $\cong C_2 \times D_4$, it contains a normal subgroup $$N \cong C_2$$ lying entirely in the extra $C_2$-factor. This 'extra' $C_2$ is often generated by an automorphism $\tau$ which does something like $$\tau: \sqrt{\sqrt{2}+\sqrt{3}} \mapsto -\sqrt{\sqrt{2}+\sqrt{3}}, \sqrt{\sqrt{2}-\sqrt{3}} \mapsto -\sqrt{\sqrt{2}-\sqrt{3}}$$ while fixing $\sqrt{2}$ and $\sqrt{3}$. Being a subgroup of index 2, the quotient group $Gal(L/ℚ)/N$ has order 8. One checks (from how all the other generators act) that this quotient of order 8 is the dihedral group $D_4$.

By the Galois correspondence, if $N \subset Gal(L/ℚ)$ is a normal subgroup of index 2, then its fixed field $K = L^N$ is an intermediate field between ℚ and $L$, and $Gal(K/ℚ) \cong Gal(L/ℚ)/N \cong D_4$.

Hence, the field you want is exactly $$K = L^{\langle \tau \rangle} = ℚ(\sqrt{2}, \sqrt{3}, \sqrt{\sqrt{2}+\sqrt{3}}, \sqrt{\sqrt{2}-\sqrt{3}})^{\langle \tau \rangle}$$

Lorenzo
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  • This idea will work, if you have the right $\tau$. I'm just trying to make $K$ more concrete in my head. Is it clear that your automorphism $\tau$ is in the center of $Gal(L/\Bbb{Q})$? – Jyrki Lahtonen Jan 09 '25 at 07:13
  • yes, $\tau$ is in the center because it doesn't touch $\sqrt{2}$ or $\sqrt{3}$. Conjugating $\tau$ by any element that does send $\sqrt{2} \mapsto -\sqrt{2}$ or $\sqrt{3} \mapsto -\sqrt{3}$ flips or unflips the same pair of roots, so $\tau$ commutes with all those actions. Consequently, it is in the center. – Lorenzo Jan 09 '25 at 07:25
  • Does it also commute with the two automorphisms that map $\sqrt{\sqrt2+\sqrt3}\mapsto \sqrt{\sqrt2-\sqrt3}$? The center is isomorphic to Klein four (because $D_4$ also has a non-trivial center), so that doesn't resolve my main doubt which is the need to see your statement this extra C_2 is often generated by... fleshed out a bit. Your mileage may be different, but I haven't seen this ever before, so often is meaningless to me here :-) – Jyrki Lahtonen Jan 09 '25 at 07:31
  • Yes, if you analyze the actions of $\tau$. It only flips the signs of $\sqrt{\sqrt{2} \pm \sqrt{3}}$, it does not alter $\sqrt{2}$ or $\sqrt{3}$. Any automorphism that sends $\sqrt{\sqrt{2} + \sqrt{3}} \mapsto \sqrt{\sqrt{2} - \sqrt{3}}$ necessarily does something nontrivial to $\sqrt{3}$. When you compose those two types of actions (regardless of order), the net effect on each of the four generators ($\sqrt{2}, \sqrt{3}, \sqrt{\sqrt{2} + \sqrt{3}}, \sqrt{\sqrt{2} - \sqrt{3}}$) is the same, so they commute. – Lorenzo Jan 09 '25 at 07:42
  • As for saying the extra $C_2$ is often generated by $\tau$, that's just a standard way of constructing the degree-16 splitting field. – Lorenzo Jan 09 '25 at 07:43
  • I'm afraid I'm still not totally satisfied. I came back to this problem, and it seems to me that with $u=\sqrt{\sqrt2+\sqrt3}$ and $v=\sqrt{\sqrt2-\sqrt3}$ we get $uv=i$ (with suitable choices of branches for the square roots). So if $\tau(u)=-u$ and $\tau(v)=-v$, it follows that $\tau(i)=i$, and $\sqrt2\pm\sqrt3$ are also fixed points of $\tau$. Therefore the fixed field of $\tau$ is $\Bbb{Q}(i,\sqrt2,\sqrt3)$ but this has an elementary abelian Galois group rather than $D_4$. – Jyrki Lahtonen Jan 09 '25 at 21:55
  • I looked at this one more time now that I woke up. I cannot make this work. The non-trivial automorphisms of the center are $\tau_1:u\mapsto -u, i\mapsto i$, $\tau_2:u\mapsto -vi, i\mapsto -i$ and $\tau_3:u\mapsto vi, i\mapsto -i$. Of these only $\tau_1$ fixes $u^2$ and hence $\sqrt2$ and $\sqrt3$. But the fixed field of $\tau_1$ is the elementary 2-abelian extension $\Bbb{Q}(i,\sqrt2,\sqrt3)$. Can you explain this in detail, please? – Jyrki Lahtonen Jan 10 '25 at 05:30
  • So I suspect that your choice of $\tau$ is the non-trivial element in the center of the $D_4$ part. Simply because it is the square of an element of order four. On the other hand, the subgroup $N$ is not generated by a square of an automorphism. Such automorphisms exist, but won't have $\sqrt2\pm\sqrt3$ as fixed points. – Jyrki Lahtonen Jan 10 '25 at 06:01
  • You are seeing a 'wrong' central element because you labeled $uv = i$. If you choose your square root branches differently, the same map $\tau$ will send $i \mapsto -i$ Then it is no longer central in $D_4$, and its fixed field is the nonabelian (dihedral) extension you want. – Lorenzo Jan 11 '25 at 06:12
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    Excuse me! Your definition of $\tau$ maps both $\sqrt{\sqrt2+\sqrt3}$ and $\sqrt{\sqrt2-\sqrt3}$ to their negatives. Yet you claim that their product is not fixed by $\tau$. Can you please make the answer explicit rather than using vague claims. List generating elements of the field. – Jyrki Lahtonen Jan 11 '25 at 06:49
  • Let $x := \sqrt{\sqrt{2}+\sqrt{3}}, y := \sqrt{\sqrt{2}-\sqrt{3}}$. Then L = $ℚ(\sqrt{2}, \sqrt{3}, x, y)$ is the degree-16 extension of ℚ. $x^2 = \sqrt{2}+\sqrt{3}$ and $y^2 = \sqrt{2} - \sqrt{3}$, so $x^2$ and $y^2$ both lie in $ℚ(\sqrt{2}, \sqrt{3})$. Moreover, $(xy)^2 = -1$, so $xy$ is an imaginary unit $i$, but the sign is just a labeling convention. – Lorenzo Jan 11 '25 at 06:58
  • Yes, but irrespective of choice of sign the product $xy$ is a fixed point of $\tau$, so $i$ is in the fixed field regardless. The other elements in the center of the Galois mix $x$ and $y$ (up to a power of $i$), so they won't have $x^2$ in the fixed field. – Jyrki Lahtonen Jan 11 '25 at 07:05
  • What if we define the automorphism as $\tau: \sqrt{2} \mapsto \sqrt{2}, \sqrt{3} \mapsto \sqrt{3}, x \mapsto x, y \mapsto -y$ and label $xy = -y$, so that $xy$ is not fixed by $\tau$? – Lorenzo Jan 11 '25 at 07:12
  • So the automorphism is not central in $D_4$, and its fixed field is nonabelian (dihedral). – Lorenzo Jan 11 '25 at 07:13
  • I listed the automorphisms in the center a few comments up. The fixed field is not Galois unless $\langle\tau\rangle$ is a normal subgroup. Given that $\tau$ has order two, that can be the case only when $\tau$ is in the center of $C_2\times D_4$. Mind you, you only need to specify the images of $x$ and $i$ as the splitting field is $\Bbb{Q}(x,i)$. The possible images of $x$ are $xi^k$ and $yi^k$, $k=0,1,2,3$. Sixteen total (accounting for two choices for the image of $i$), so all combinations occur. – Jyrki Lahtonen Jan 11 '25 at 07:18
  • You may be right -- I don't see a point to extensively discussing semantics. There is no canonically correct sign here, so it really comes down to the way the branches are chosen and labeled. – Lorenzo Jan 11 '25 at 07:26
  • Don't worry about it too much. Welcome to the site! – Jyrki Lahtonen Jan 11 '25 at 10:50