I got an example of $\Bbb{Q}(\sqrt2,\sqrt3)$ but I don't know how to do that with 3 extensions. Any ideas? Thanks.

  • 1
    $\begingroup$ What do you mean by 2-√? Do you mean $\sqrt2$? $\endgroup$
    – Arthur
    Dec 3, 2020 at 19:07
  • $\begingroup$ Yes, sorry. It's my first question and I don't know how to use the correct characters. $\endgroup$ Dec 3, 2020 at 19:11
  • $\begingroup$ Welcome to the site, Helena. At some point (sooner rather than later) you may want to look at the guide on how to typeset math beautifully with MathJax. Possibly more urgently I encourage you to be more verbose in describing what you know about your question already. See our guide for new askers. $\endgroup$ Dec 3, 2020 at 19:15
  • $\begingroup$ The answers in this thread contain a lot of information about the field extensions of this type. It may be a bit too general to be useful to you. However, some of the answers there also gloss over the Galois group, but those are likely to be too sketchy to help you. I think we have more fitting related threads that you can use, but I'm afraid I don't have the time to look for one right now. $\endgroup$ Dec 3, 2020 at 19:17
  • $\begingroup$ Thanks!!! I'll try to find more specific information about that in this and other similiar thread. $\endgroup$ Dec 3, 2020 at 19:23

1 Answer 1


As pointed out by Jyrki Lahtonen, your question (even under a more general form) has been asked many times. For 3 radicals ($\sqrt 2, \sqrt 3, \sqrt 5$ here) it is still possible to give a direct simple solution. Consider first the extension $K=\mathbf Q(\sqrt 2, \sqrt 3)$. I don't know how you solved this case, but I'll give an approach which will be used also for $L=\mathbf Q(\sqrt 2, \sqrt 3,\sqrt 5)$. First notice that, for $a,b \in {\mathbf Q^*}$, it is straightforward that the two quadratic fields coincide iff $ab^{-1} \in {\mathbf Q^*}^2$. So here $K/\mathbf Q$ is biquadratic, with Galois group $G\cong C_2 \times C_2$. Since $G$ has exactly 3 subgroups of order, $K$ has exactly 3 quadratic subfields, which are obviously (fill in the details) $\mathbf Q(\sqrt 2, \sqrt 3,\sqrt 6)$. In particular, $L$ will be the compositum of the 2 linearly disjoint extensions $K$ and $\mathbf Q(\sqrt 5)$, hence $L/\mathbf Q$ will be Galois, with group $J \cong C_2 \times G \cong C_2 \times C_2 \times C_2$. From this you can derive easily the reticle of subgroups (resp. of subextensions). One convenient way to vizualize this is to write additively $G$ as $\mathbf F_2 \times \mathbf F_2\times \mathbf F_2$, where $\mathbf F_2$ is the field with 2 elements, so that $J$ can be viewed as an $\mathbf F_2$-vector space of dimension 3, whose strict non trivial subspaces have dimension 2 ("planes") or 1 ("lines"). The corresponding subextensions are obviously the 3 previous quadratic fields, and the 3 biquadratic fields $\mathbf Q(\sqrt 2, \sqrt 3),\mathbf Q(\sqrt 2,\sqrt 6), \mathbf Q(\sqrt 3,\sqrt 6)$.

A generalization of this approach to $\mathbf Q(\sqrt p_1,..., \sqrt p_n)$, where $p_1,..., p_n$ are $n$ distinct primes, can be adapted e.g. from https://math.stackexchange.com/a/3444776/300700


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