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Is it true that the degree of extension $\mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},\dotsc,\sqrt {p_n}) / \mathbb Q$ is $2^n$ where $p_n$ is the $n$th prime number. If so, how to prove this? My idea is to consider the chain of extensions $\mathbb Q\subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt{2},\sqrt{3}) \subset \dotsb \subset \mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},...,\sqrt {p_n})$ and using transitivity. I am having problems in finding degrees of intermediate extensions. Please help me.

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marked as duplicate by Ewan Delanoy, JSchlather, Henry T. Horton, Amzoti, Jyrki Lahtonen Feb 6 '13 at 17:00

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question already asked here :… –  Ewan Delanoy Feb 6 '13 at 16:34

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I guess you could argue by induction as follows. Let $$ E = \mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},\dotsc,\sqrt {p_{n-1}}) , $$ and suppose that $E/ \mathbb Q$ has degree $2^{n-1}$. Assume also by induction that the Galois group $G$ of the Galois extension $E/\mathbb Q$ is elementary abelian, sending $\sqrt{p_{i}} \mapsto \pm \sqrt{p_{i}}$. If, by way of contradiction, $\sqrt{p_{n}} \in E$, then $$ \mathbb Q(\sqrt{p_{n}}) = \mathbb Q(\sqrt{p_{i_1} \dots p_{i_k}}), $$ for some $1 \le i_1 < \dots < i_k < n$. (I am appealing to the Galois correspondence here.) So $$ a + b \sqrt{p_{n}} = \sqrt{p_{i_1} \dots p_{i_k}}, $$ for some $a, b \in \mathbb Q$. Square both sides and distinguish a couple of cases to get a contradiction.

PS Let me expand a bit on the Galois connection here. $G$ is generated by the $n-1$ elements $\sigma_i$, that fix all $\sqrt{p_{j}}$, for $j \ne i$, and send $\sqrt{p_{i}} \mapsto - \sqrt{p_{i}}$. Intermediate fields $F$ of $E/\mathbb{Q}$ such that $F/\mathbb{Q}$ has degree two correspond to maximal subgroups of $G$. Each such maximal subgroup $M$ is defined by an equation $$ a_{i_1} + \dots + a_{i_k} = 0, $$ in the exponents of an element $$ \sigma_1^{a_1} \cdot \dots \cdot \sigma_{n-1}^{a_{n-1}} \in G $$ where you may think $a_i \in \mathbb{Z}_2$. (An elementary abelian group of order a power of 2 is really a vector space over $\mathbb{Z}_2$.) Now it is easy to verify that the elements of $M$ fix $\sqrt{p_{i_1} \dots p_{i_k}}$.

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