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I have the following exercise :

Let $p_1,...,p_n$ be distinct primes.Consider $L= \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$.

Find all the intermediate subfields $K$ such that $[K:\mathbb{Q}]=2$

I have already proved that $[L:\mathbb{Q}]= 2^n$

I know that if $K$ is an intermediate subfield such that $[K:\mathbb{Q}] =2$ then $[L:K] = 2^{n-1}$.

I also know that $\sqrt{p_i}\notin \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{i-1}})$ so :

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

But I am not seeing how to find the ones that have degree 2 over $\mathbb{Q}$

Physmath
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    Each subgroup $H \subseteq G$ corresponds to an intermediate field $F$, and if $\vert H \vert=2$, then $[F: \Bbb Q]=2$. But for any element $1 \neq g \in G$, the set ${ 1, g } = H$ is a subgroup of $G$. That tells you how many intermediate subfields you need to find. – Robert Shore Nov 19 '19 at 16:24
  • I prove this result by induction on $n$ here. I'm afraid you may find it a bit sketchy. Do observe that I needed the induction hypotheses of part c) to prove the induction step of part b). In that same thread Bill Dubuque gives a proof of part b) that does not need Galois theory for the induction step. – Jyrki Lahtonen Nov 19 '19 at 17:55

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