proof verification - Prove that $[ mathbb{Q}(sqrt{2},sqrt{3},sqrt{5}):mathbb{Q}]=8.$

I have to solve the following exercise:

Compute $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]$ and $\operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}).$

Here my attempt:

Let $\mathbb{K}=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$ Proved that $[\mathbb{K}:\mathbb{Q}]=8$, compute $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$ is easy, in fact it will be $$\operatorname{Gal}(\mathbb{K}/\mathbb{Q})=\langle \sigma_2,\sigma_3,\sigma_5\rangle,$$ with $\sigma_k$ the automorphism that interchanges $\sqrt{k}$ with $-\sqrt{k}$ and doesn't move the rest of the elements.

I prove that $[\mathbb{K}:\mathbb{Q}]=8$ as follows. I do some observations first:

• If a group $G$ verifies $|G|=4$ then $G$ is isomorphic to $C_4$ (cyclic group of order 4) or to $\mathbb{V}$ (vierergruppe). In particular, $G$ can only have $1$ or $3$ proper subgroups.




• Clearly $\mathbb{K}/\mathbb{Q}$ is a Galois extension since $\mathbb{K}$ is the splitting field of the polynomial $p(x)=(x^2-2)(x^2-3)(x^2-5).$




• We have the four strict chains of extensions, all distinct:

  
  
  
  
  

  1. $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{K}.$

  
  

  2. $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3}) \subset \mathbb{K}.$

  
  

  3. $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{K}.$

  
  

  4. $\mathbb{Q} \subset \mathbb{Q}(\sqrt{6}) \subset \mathbb{K}.$

  
  
  

Let $\mathbb{L}=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Clearly $[\mathbb{L}:\mathbb{Q}]=4$ and $\mathbb{K}=\mathbb{L}(\sqrt{5}).$ Then, $[\mathbb{K}:\mathbb{L}] \in \{1,2\}.$
With this, we have $$[\mathbb{K}:\mathbb{Q}]=[\mathbb{K}:\mathbb{L}][\mathbb{L}:\mathbb{Q}]=4[\mathbb{K},\mathbb{L}].$$ Therefore, $[\mathbb{K}:\mathbb{Q}]\in \{4,8\}.$

If we suppose that $[\mathbb{K}:\mathbb{Q}]=4$, then, since it is a Galois extension, $|\operatorname{Gal}(\mathbb{K}/\mathbb{Q})|=4$ and the Galois correspondence joint with the fact about groups of four elements mentioned above tells us that there is only $1$ or $3$ stricts chains of extensions starting with $\mathbb{Q}$ and ending at $\mathbb{K}.$ But we found $4$ of such chains, and this concludes that must be $[\mathbb{K}:\mathbb{Q}]=8.$

End.

Is correct my solution? Could it be improved? I'm interested in reading other possible solutions and better if it's "faster". How do I prove this result without the using of group theory?

Thanks to everyone!

Edit:

I thought that my solution to the problem has not been posted yet in MSE but I found a question that has my solution as an answer.

Showing field extension $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}$ degree 8 [duplicate]

Other related questions are:

The square roots of different primes are linearly independent over the field of rationals

I'm sorry for duplicating this question.

Answer

Your proof is essentially true, however I feel it can be shortened by proving that $\sqrt{5} \not \in \mathbb{Q}(\sqrt{2},\sqrt{3})$. To see this you can use the fact that $\text{Gal}\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}) = \langle \sigma_2, \sigma_3\rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. Now the quadratic subfields are $\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{6})$ and obviously $\mathbb{Q}(\sqrt{5})$ is not any of them and so $\sqrt{5} \not \in \mathbb{Q}(\sqrt{2},\sqrt{3})$. Thus we must have that $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}] = 8$. From here we conclude that:

$$\text{Gal}\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}) = \langle \sigma_2, \sigma_3,\sigma_5\rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$$

On the other way if you want to avoid group theory you can prove that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}) = \mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{5})$, as it's been done here. Then you can explicitly prove that the minimal polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ over $\mathbb{Q}$ is of degree $8$ and conclude that $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}] = 8$. Once you have this finding the Galois group is an easy task. However I feel this way requires far more work.