# Finding intermediate subfields of an extension

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

$$[(\mathbb{Q}(\sqrt{p_1},..,\sqrt{p_{i}}):\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{i-1}})]=2$$

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 $$\mathbb{Q}$$

• 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 $2$. Nov 20, 2019 at 17:44
• 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. Nov 21, 2019 at 5:04
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.