Consider the Galois extension $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})\vert\mathbb{Q}$ where $p_1,...,p_n$ are distinct prime numbers. Find all the intermediate subfields $K$ such that $[K:\mathbb{Q}]=2$. I know that:

1) $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ is the splitting field of $f(x)= (x^2-p_1)...(x^2-p_n)$

2) $[\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}]= 2^n $

3) Since $\sqrt {p_i}\notin\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{i-1}})$ we have that


4) By Galois Correspondence the subfields with degree 2 over $\mathbb{Q}$ corresponds to subgroups of index 2 of the Galois group(that has order $2^n$),that are subgroups of order $2^{n-1}$.

I am not seeing how can I find and write these subgroups.

PS : I did a numerical example with $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ in this case I found that the intermediate subfields of degree 2 are of the form $\mathbb{Q}(\sqrt{q})$ where $q$ is a element (not 1) from the basis of $ \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ over $\mathbb{Q}$

Thanks in advance

  • $\begingroup$ Welcome to mathematics SE. People here like to see what you have tried. $\endgroup$ Nov 20, 2019 at 17:25
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    $\begingroup$ Think of the Galois group $G$ as a vector space of dimension $n$ over $\Bbb F_2.$ Each non-zero element of $G$ is perpendicular to a subspace of dimension $n-1$, which is also a subgroup of index $2$. $\endgroup$ Nov 20, 2019 at 17:44
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    $\begingroup$ math.stackexchange.com/questions/3442307/… $\endgroup$ Nov 20, 2019 at 18:21
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    $\begingroup$ Do you know what the Galois group of, for example, $\Bbb{Q}(\sqrt2,\sqrt3,\sqrt5)/\Bbb{Q}$ looks like? Can you see it as a 3-dimensional vector space over $\Bbb{F}_2$, when the maximal subgroups would automatically be the distinct 2-dimensional subspaces that, in turn, can be listed as the "orthogonal complements" ot the 1-dimensional subspaces? I know I'm asking quite a bit, if you are relatively new to groups. That's why concentrating on a specific case like $n=3, p_1=2,p_2=3.p_3=5$ may help at first. $\endgroup$ Nov 21, 2019 at 5:04
  • $\begingroup$ Anyway, we have a lot of material here. $\endgroup$ Nov 21, 2019 at 5:07

1 Answer 1


I think that the quickest proof comes from linear algebra. I give all the details. You know that your Galois group $G=Gal(K/\mathbf Q)$ has order $2^n$. Any $s\in G$ is determined by its action on the roots $\sqrt p_i$, and since $s(p_i)=p_i$, necessarily $s(\sqrt p_i)=\pm \sqrt p_i$, which means that any $s$ has order $2$, and so $G$ is abelian, isomorphic (in additive notation) to $(\mathbf Z/2\mathbf Z)^n$. In other words, $G$ may be viewed as a vector space of dimension $n$ over the field $\mathbf F_2$ with $2$ elements. A basis of $G$ consists of the $s_j$ defined by $s_j(\sqrt p_i)/ \sqrt p_i = \delta_{ij}$ (Kronecker's symbol). By the Galois correspondance, you are looking for all the subgroups $H$ of $G$ of index $2$. In terms of linear algebra, $H$ is a hyperplane of $G$, or equivalently, $H$ is the kernel of a linear form $f:G\to \mathbf F_2$. In general, two linear forms with the same kernel are proportional, but here, because the base field is $\mathbf F_2$, they must coincide. In other words you are simply looking for the dual $\hat G$ of the vector space $G$, which has also dimension $n$. Actually, a dual basis consists of the linear forms $f_i$ determined by $f_i(s_j)=\delta_{ij}$. Since the linear forms $\pi_i$ defined by $\pi_i(s_j)=s_j(\sqrt p_i)/ \sqrt p_i$ share the same property, $\hat G$ can be identified with the subspace $R$ of $\mathbf Q^*/{\mathbf Q^*}^2$ generated by the classes $[p_i]$ of $p_i$ mod ${\mathbf Q^*}^2$, usually called the "Kummer radical" of $K$. The duality above is then presented as a non degenerate pairing $G\times R\to\mathbf F_2, (s,[a])\to s(\sqrt a)/\sqrt a$, and the fixator of $\mathbf Q(\sqrt a)$ is the hyperplane orthogonal to $[a]$, as pointed out by @Robert Shore.

NB: If you know about Kummer theory, see https://math.stackexchange.com/a/1609061/300700, where everything (including your beginning) can be given a unified proof.


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