# Subfields of $\mathbb{Q}(\sqrt{5})$.

The problem is to find all subfields of $$K=\mathbb{Q}(\sqrt{5})$$.

Big idea approach: Determine the Galois closure of $$K$$, and its Galois group. Then determine the subgroups of this Galois group. Apply the fundamental theorem to the Galois closure, and see which subfields are contained in $$K$$.

The Galois closure will be the splitting field of the minimal polynomial of $$\sqrt{5}$$ (right?), which is $$x^6-5$$. The roots of this polynomial are $$\pm\sqrt{5},\frac{\sqrt{5}\pm i\sqrt{3}\sqrt{5}}{2},$$ and $$\frac{-\sqrt{5}\pm i\sqrt{3}\sqrt{5}}{2}$$ (yes?).

OK, so this means that the splitting field of $$x^6-5$$ is $$E=\mathbb{Q}(\sqrt{5}, i\sqrt{3})$$. Any automorphism of $$E$$ is determined by its action on generators, so we have 4 possibilities: $$1$$, $$\sigma$$, $$\tau$$, and $$\sigma\tau$$, with $$\sigma(\sqrt{5})=\sqrt{5},\;\sigma(i\sqrt{3})=-i\sqrt{3}$$ $$\tau(\sqrt{5})=-\sqrt{5},\;\tau(i\sqrt{3})=i\sqrt{3}.$$

THIS is where my confusion begins. $$E/\mathbb{Q}$$ is a Galois extension, so the number of automorphisms should equal the degree of the extension — which is 16 (?). But there are only 4 automorphisms? What is going wrong here?

EDIT: Going off of Alex Wertheim's comment: Let $$r_1,r_2,\ldots,r_6$$ be the six roots of $$x^6-5$$. Let $$\sigma_j^{+}$$ be the automorphism of $$E$$ which sends $$\sqrt{5}↦ r_j$$, and $$i\sqrt{3}↦ i\sqrt{3}$$, and let $$\sigma_j^{-}$$ be the automorphism of $$E$$ which sends $$\sqrt{5}↦ r_j$$, and $$i\sqrt{3}↦ -i\sqrt{3}$$; for $$j∈\{1,2,\ldots,6\}$$. Then $$\text{Gal}(E/\mathbb{Q})=\{\sigma_1^+,\sigma_2^+,\ldots,\sigma_6^+,\sigma_1^-,\sigma_2^-,\ldots,\sigma_6^-\}$$. From this point is it just a matter of working out the orders of each element//analyzing the relations and figuring out which group it is?

• Because the image of $\sqrt{5}$ has more than $2$ possibilities. Dec 24, 2021 at 5:43
• The degree of the extension is $12$. The problem is that you have not enumerated all possibilities for each field generator. The constraints that you have are that $\sqrt{5}$ must go to a root of $X^{6}-5$, and that $i\sqrt{3}$ must go to either itself or $-i\sqrt{3}$. You can see that there are therefore six possibilities for $\sqrt{5}$ and two possibilities for $i\sqrt{3}$, yielding a maximum of $12$ possible automorphisms defined on these field generators. Since there must be exactly $12$ automorphisms, each of these assignments on generators extends to an automorphism. Dec 24, 2021 at 5:44
• @AlexWertheim Yes that makes a lot of sense, thank you. Dec 24, 2021 at 5:59
• @ondwats no problem, glad to hear it. To respond to your edit, that's more or less exactly what I'd do from here. The only suggestion I'd make is that to make your job easier, you should choose a specific labeling of the roots of $X^{6}-5$ which makes it more apparent what's going on when you are trying to work out the relations. In particular, if $\omega$ denotes a primitive $6$th root of unity, then the roots are $\omega^{i}\sqrt{5}$ for $i = 0, 1, \ldots, 5$. I would set $r_{i} = \omega^{i}\sqrt{5}$. This is just a minor suggestion, though - if it doesn't help, ignore it. Dec 24, 2021 at 7:16
• You may find the subfields via tower law also. The only subfields are of degree $2$ or $3$ over rationals. And with a little more effort one can prove that there exists one field of each degree. Dec 24, 2021 at 13:51

Let $$F$$ be a subfield of $$K$$. Let $$Hom_F(K,\overline{K})$$ be the set of $$F$$-algebra homomorphisms $$K\to \overline{K}$$. You'll have that $$F =\{ a\in K, \forall \sigma\in Hom_F(K,\overline{K}), a=\sigma(a)\}$$ In particular $$F$$ is the subfield fixed by a subset of $$Hom_\Bbb{Q}(K,\overline{K})$$.
Conveniently $$Hom_\Bbb{Q}(K,\overline{K})$$ is very easy to understand (as it is the restriction to $$K$$ of a cyclic group of automorphisms of the Galois closure), as well as the subfield fixed by any set of elements.
You'll find that for each $$d| 6$$ there is a subfield fixed by $$\sigma_d : \sqrt{5}\to e^{2i\pi d/6} \sqrt{5}$$ and no other subfield.