# Sub-fields of $\mathbb{Q}(a)$ ,$a=\sqrt{4+\sqrt{15}}$ using Galois theory.

Let $$K\subseteq\mathbb{C}$$ be a splitting field of $$f(x)=x^4-8x^2+1$$, $$a=\sqrt{4+\sqrt{15}}$$ , $$b=\sqrt{4-\sqrt{15}}$$ and $$G=Gal(K,\mathbb{Q})$$.Find every subfield of K.

What I've done so far :
I found out that $$f(x)$$ is irreducible using Gauss lemma .Then I showed that $$\mathbb{Q}(a)=K$$ since b is also a solution of $$f(x)$$ and also $$ab=1 \Rightarrow b=a^{-1} \Rightarrow b \in \mathbb{Q}(a)$$ and thus $$\mathbb{Q}(a)$$ contains all four solutions of $$f(x)$$ ,meaning it is a splitting of $$f$$ over $$\mathbb{Q}$$.Of course $$[K:\mathbb{Q}]=4$$ and using the fundmental Galois theorem we conclude that $$[K:\mathbb{Q}]=|G|=4$$.

From here I'm getting a bit unsure on how to continue.

I found that $$G=\{s_1,s_2,s_3,s_4|s_1(a)=a,s_2(a)=-a,s_3(a)=b,s_4(a)=-b\}$$ (I'm not sure about these).I calculated that every s has an order of 2 and thus $$G\simeq\mathbb{Z}_2 \times \mathbb{Z}_2$$.Of course every $$,i\neq1$$ is a subfield of G.From the Galois correspondence we know that $$K$$ has 3 subfields :$$Fix,Fix,Fix$$ .I want to express them as $$\mathbb{Q}(-)$$ but I'm a little bit lost .I expressed $$Fix=\mathbb{Q}(4+\sqrt{15})$$ since $$s_2(a^2)=4+\sqrt{15}$$.I have no idea what to do for the 2 remaining $$Fix-$$ and I think I might have done something wrong.

Any kind of help would be really appreciated!

• Yes, this is a biquadratic extension with Galois group $C_2\times C_2$ - so far, so good. Jun 17, 2022 at 17:37
• Linking this to a summary of Galois groups of biquadratic polynomials. Just to enhance site connectivity, +1 to all. Jun 18, 2022 at 5:05
– GGG
Jun 18, 2022 at 12:50

Everything goes well, so it strongly suggests that $$\sqrt{4+\sqrt{15}}$$ can be broken into a sum of roots, and indeed

$$\sqrt{4\pm\sqrt{15}}=\frac{\sqrt{10}\pm\sqrt{6}}{2}$$

Thus the field contains $$\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}=\sqrt{10}$$ and also $$\sqrt{6}$$.

Once we find those intermediate extensions, we may conclude the Galois group is isomorphic to $$(\mathbb Z/2\mathbb Z)^2$$ instead of $$\mathbb Z/4\mathbb Z$$ without earlier work.

Here is one way to show any element $$\sigma$$ in the Galois group satisfying $$\sigma^2=\text{id}$$. Let $$a=\sqrt{4+\sqrt{15}}$$, then we have $$\sigma(a) = \pm a, \pm\frac{1}{a}$$. If $$\sigma(a)=-a$$, apply $$\sigma$$ on both sides, we get $$\sigma^2(a) = \sigma(-a)=\sigma(-1)\sigma(a)=-\sigma(a)=a$$, hence $$\sigma^2=\text{id}$$. The other cases can be solved in similar fashion. Enjoy.

• Great !Thank you for your help !
– GGG
Jun 17, 2022 at 18:00