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Is it obvious that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)$ ? If not how can I show it ?

idm
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    Thisis a well-known question on this site - see here. Replace $2$ and $3$ by $3$ and $5$. See also http://math.stackexchange.com/questions/598262/bbb-q-sqrt2-sqrt3-bbb-q-sqrt2-sqrt3?lq=1. – Dietrich Burde Dec 23 '14 at 14:20

3 Answers3

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The first inclusion is obvious because $\mathbb{Q}(\sqrt{3},\sqrt{5})$ is closed under addition. So $\mathbb{Q}(\sqrt{3} +\sqrt{5}) \subseteq \mathbb{Q}(\sqrt{3} ,\sqrt{5})$.

To second inclusion observe that $$(\sqrt{3} +\sqrt{5})(\sqrt{3} -\sqrt{5}) = 3 - 5 = -2 \Rightarrow \sqrt{5} -\sqrt{3} = \frac{-2}{\sqrt{5} +\sqrt{3}}$$

then $\sqrt{5} -\sqrt{3} \in \mathbb{Q}(\sqrt{3} +\sqrt{5})$. Now use sum and subtraction to find that $\sqrt{3}, \sqrt{5} \in \mathbb{Q}(\sqrt{3} +\sqrt{5})$.

Can you take it from here?

Aaron Maroja
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It is obvious if the facts below are obvious:

  • $\mathbb Q(\sqrt 3,\sqrt 5)$ has degree 4

  • $\mathbb Q(\sqrt 3+\sqrt 5)$ has degree 4

  • $\mathbb Q(\sqrt 3+\sqrt 5) \subseteq \mathbb Q(\sqrt 3,\sqrt 5)$

Of these, only the last one is really obvious.

lhf
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$(\sqrt3+\sqrt 5)^2\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies 8+2\sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies \sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies (\sqrt3+\sqrt 5).\sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies 3\sqrt 5+5\sqrt3\in \mathbb Q(\sqrt 3+\sqrt 5)$

$3(\sqrt3+\sqrt 5)\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies 2\sqrt 3\in \mathbb Q(\sqrt 3+\sqrt 5)$

$\implies \sqrt 3\in \mathbb Q(\sqrt 3+\sqrt 5)$

similarly $\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt 5)$

Thus $\mathbb Q(\sqrt 3+\sqrt 5)\subseteq \mathbb Q(\sqrt3,\sqrt5)$

The other part is obvious

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    I'm sure you meant to have the inclusion in the next to last line going in the other direction. You have indeed tackled the nonobvious half of the problem. – hardmath Dec 24 '14 at 14:24