Galois's group of $\Bbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ and its reticle of subgroups and subextensions

Question

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

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