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 $[\mathbb{K}:\mathbb{Q}]=8.$


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!


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.

  • $\begingroup$ This is a nice solution, if you ask me. And it also generalizes. See my answer to this question. Mind you, given that you came up with this way of thinking about it I'm sure you would have come up with the generalization to more primes as well :-) $\endgroup$ – Jyrki Lahtonen Jan 31 '19 at 5:01
  • $\begingroup$ And, if you study the thread I linked, do take a look at Bill Dubuque's answer as well. It gets a way with a bit more primitive tools (doesn't use Galois groups). $\endgroup$ – Jyrki Lahtonen Jan 31 '19 at 5:07
  • $\begingroup$ Thanks for your comments @JyrkiLahtonen. I felt that a generalization of this galois-type argument for more prime numbers could be possible, but I didn't write down my ideas. The answer of Bill Dubuque is an elegant and simple solution, but I like to use Galois groups since I think it is the natural place for studying fields and extensions. $\endgroup$ – DrinkingDonuts Jan 31 '19 at 16:50
  • $\begingroup$ I agree. That's why I wrote my answer in the end. Using Galois theory means that the only arithmetic fact we need to prove is that $\sqrt n\notin \Bbb{Q}(\sqrt m)$ whenever both $n$ and $m$ are square free. $\endgroup$ – Jyrki Lahtonen Jan 31 '19 at 16:56
  • $\begingroup$ @JyrkiLahtonen I've posted a new question related to field/Galois theory. I'm interested in your opinion. Here the link "math.stackexchange.com/questions/3095247/…". Thanks in advance! $\endgroup$ – DrinkingDonuts Jan 31 '19 at 18:13

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.

  • $\begingroup$ I like your solution! Clearly, is more direct than mine. Thanks for your answer. Could be extended by induction the Galois-type argument in order to solve the problem posed in the question you linked? $\endgroup$ – DrinkingDonuts Jan 30 '19 at 10:52
  • $\begingroup$ @AlgebraicallyClosed I suppose you can use such an argument to prove that the Galois group of $\mathbb{Q}(\sqrt{p_1},\cdots,\sqrt{p_n})$ is $(\mathbb Z_2)^n$. However the problem is proving that $\mathbb{Q}(\sqrt{p_1}+\cdots+\sqrt{p_n})$ is extension of order $2^n$ $\endgroup$ – Stefan4024 Jan 30 '19 at 11:13
  • $\begingroup$ Okay, now I see the problem. Thanks for the time you dedicated to my question. $\endgroup$ – DrinkingDonuts Jan 30 '19 at 11:27

I think you're overcomplicating things. The index $[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]$ is just the dimension of $\mathbb Q(\sqrt 2,\sqrt3,\sqrt5)$ as a $\mathbb Q$-vector space, which is $$\sum_{k=0}^3\binom{3}{k}=1+3+3+1=8.$$

  • $\begingroup$ How do you have directly this? Are you supposing that \sqrt{2}, \sqrt{3} and \sqrt{5} are linearly independent? I don't see trivial this part of the problem. Anyway, thanks for your answer. $\endgroup$ – DrinkingDonuts Jan 30 '19 at 10:44
  • 1
    $\begingroup$ This question has been asked (and answered!) many times : the degree over Q of Q($\sqrt p_1,..., \sqrt p_n$), where the $p_i$ are $n$ distinct primes, is equal to $2^n$. See e.g. math.stackexchange.com/a/1609061/30070S $\endgroup$ – nguyen quang do Jan 30 '19 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.