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How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots \sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

Let ${p_n}$ be $n$-th prime number in $\mathbb{N}$ and ${\mathbb{Q}_n} = \mathbb{Q}\left[ {\left\{ {\sqrt {{p_i}} :i \in \left\{ {1, \ldots ,n} \right\}} \right\}} \right]$ be the least field which contains $\sqrt {{p_1}} , \ldots ,\sqrt {{p_n}} $. Is it true that ${\mathbb{Q}_n} = \mathbb{Q}\left[ {\sum\limits_{i = 1}^n {\sqrt {{p_i}} } } \right],\forall n \in \mathbb{N}$ ?

My attempt: For $n = 1$ it's obvious, for $n = 2$ obviously $\mathbb{Q}\left[ {\sqrt 2 + \sqrt 3 } \right] \subseteq \mathbb{Q}\left[ {\sqrt 2 ,\sqrt 3 } \right]$. Also, $$ - \frac{9}{2}\left( {\sqrt 2 + \sqrt 3 } \right) + \frac{1}{2}{\left( {\sqrt 2 + \sqrt 3 } \right)^3} = \sqrt 2 \in \mathbb{Q}\left[ {\sqrt 2 + \sqrt 3 } \right] \Rightarrow \left( {\sqrt 2 + \sqrt 3 } \right) - \sqrt 2 =$$

$$ \sqrt 3 \in \mathbb{Q}\left[ {\sqrt 2 + \sqrt 3 } \right] \Rightarrow \mathbb{Q}\left[ {\sqrt 2 ,\sqrt 3 } \right] \subseteq \mathbb{Q}\left[ {\sqrt 2 + \sqrt 3 } \right] \Rightarrow \mathbb{Q}\left[ {\sqrt 2 ,\sqrt 3 } \right] = \mathbb{Q}\left[ {\sqrt 2 + \sqrt 3 } \right]$$.

Now, assume that $\mathbb{Q}\left[ {\sum\limits_{i = 1}^k {\sqrt {{p_i}} } } \right] = {\mathbb{Q}_k}$ for all $1 \leqslant k \leqslant n - 1$. Obviously, $\mathbb{Q}\left[ {\sum\limits_{i = 1}^n {\sqrt {{p_i}} } } \right] \subseteq {\mathbb{Q}_n}$. We want to show that $\mathbb{Q}\left[ {\sum\limits_{i = 1}^n {\sqrt {{p_i}} } } \right] \supseteq {\mathbb{Q}_n} = {\mathbb{Q}_{n - 1}}\left[ {\sqrt {{p_n}} } \right] = \left( {\mathbb{Q}\left[ {\sum\limits_{i = 1}^{n - 1} {\sqrt {{p_i}} } } \right]} \right)\left[ {\sqrt {{p_n}} } \right]$.

And now I'm stuck. Also, I might be wrong here, they might not be equal. Any ideas?

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Alen
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    Do you want to know if they're equal or isomorphic? A subtle difference but an important one. – Mathemagician1234 Nov 21 '12 at 17:50
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    @Mathemagician1234 Since $\mathbf{Q} \left [ \sum_{i=1}^n \sqrt{p_i} \right] \subseteq \mathbf{Q}_n$, they're equal iff they're isomorphic. – bzc Nov 21 '12 at 17:52
  • If they are isomorphic, they can be embedded one into other. Maybe I'm wrong, but it seems to me that they have the same elements, so that they would be equal iff they are isomorphic – Alen Nov 21 '12 at 17:53
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    In general, isomorphism of two fields $k\subset K$ does not imply equality: with $k=\mathbb Q(x_1,x_2,\ldots)$ and $K=k(y)$ the two fields are isomorphic (a Hilbert-hotel thing), but not equal. Also, this question has been answered before: http://math.stackexchange.com/questions/93453/ – paul garrett Nov 21 '12 at 18:12
  • Thank you. Should I delete this question? – Alen Nov 21 '12 at 18:13
  • The answer below adds information to the prior discussion, so I'd think you need not delete... – paul garrett Nov 21 '12 at 20:37

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