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This question already has an answer here:

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

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marked as duplicate by Dietrich Burde, Davide Giraudo, Namaste, apnorton, user26857 Dec 23 '14 at 14:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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?

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  • $\begingroup$ @DietrichBurde We got it man, what is it?I didn't see that answer. This is pattern. Why the downvote? $\endgroup$ – Aaron Maroja Dec 23 '14 at 14:30
  • $\begingroup$ Whatever, no reason for the downvote though. $\endgroup$ – Aaron Maroja Dec 23 '14 at 14:32
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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.

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  • $\begingroup$ Your comment here is already a good answer. $\endgroup$ – Dietrich Burde Dec 23 '14 at 14:45
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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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    $\begingroup$ 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. $\endgroup$ – hardmath Dec 24 '14 at 14:24

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