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Is $\mathbb{Q}(\sqrt{2}, \sqrt{5})=\mathbb{Q}(\sqrt{10})$ in field extension?

We know that, $\mathbb{Q}(\sqrt{2}, \sqrt{5})=\mathbb{Q}(\sqrt{2}+ \sqrt{5})$. Also, $\mathbb{Q}(\sqrt{2}, \sqrt{5})=\mathbb{Q}(a\sqrt{2}+ b\sqrt{5})$ where $a,b \in \mathbb{Q}$.

If the question is true, then for which values of $a$ & $b$ can $\sqrt{10}$ be expressed in the form $a\sqrt{2}+ b\sqrt{5}$ where $a,b \in \mathbb{Q}$

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    $\begingroup$ Look at degrees. $\endgroup$ – user14972 Sep 23 '14 at 18:06
  • $\begingroup$ No. $\mathbb{Q}[\sqrt{2}, \sqrt{5}] = \{ a+b\sqrt{2}+c\sqrt{5}+ d\sqrt{10} : a,b,c,d, \in \mathbb{Q}\}$. $\endgroup$ – Crostul Sep 23 '14 at 18:10
  • $\begingroup$ For no values $(a,b)$. And the question has answer No. $\endgroup$ – orangeskid Sep 23 '14 at 18:16
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    $\begingroup$ You can show $\sqrt{10} \in \mathbb{Q}(\sqrt{2}, \sqrt{5})$ but that $\sqrt{2} \notin \mathbb{Q}(\sqrt{10})$ and $\sqrt{5} \notin \mathbb{Q}(\sqrt{10})$. Can you see why? Also, as someone else noted, look at the degrees of the field extension. $\endgroup$ – Islands Sep 23 '14 at 18:18
  • $\begingroup$ Possible duplicate of Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? $\endgroup$ – Watson Nov 26 '18 at 12:44
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As comment suggested, look at degrees over the rationals:

$$[\Bbb Q(\sqrt2,\sqrt5):\Bbb Q]=[\Bbb Q(\sqrt2)(\sqrt5):\Bbb Q(\sqrt2)]\cdot [\Bbb Q(\sqrt2):\Bbb Q]=2\cdot2=4$$

since $\;x^2-5\in\Bbb Q(\sqrt2)[x]\;$ is irreducible here (proof?)

On the other side

$$[\Bbb Q\sqrt{10}:\Bbb Q]=2$$

since x$\;x^2-10\in\Bbb Q[x]\;$ is irreducible here.

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  • $\begingroup$ OK. I understand it. Due to the degree, Q(square root(2),cube root(7))=Q(square root(2)+cube root(7)) as well as =Q(square root(2).cube root(7)).As the degrees are equal. Am I right? $\endgroup$ – Empty Sep 23 '14 at 21:43
  • $\begingroup$ Don't understand your question, @SayantanPanja $\endgroup$ – Timbuc Sep 23 '14 at 22:31
  • $\begingroup$ I want to say that, [(Q(square root(2),cube root(7)):Q]=6=[(Q(square root(2).cube root(7)):Q]=[(Q(square root(2)+cube root(7)):Q]. So can we write that, Q(square root(2),cube root(7))=Q(square root(2).cube root(7))=Q(square root(2)+cube root(7))? $\endgroup$ – Empty Sep 24 '14 at 3:45
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    $\begingroup$ Well, yes we can...but only if you prove something else. For example, $\;[\Bbb Q(\sqrt2):\Bbb Q]=[\Bbb Q(\sqrt5):\Bbb Q]=2\;$ , yet $\;\Bbb Q(\sqrt2)\neq\Bbb Q (\sqrt5)\;$ . Equality follows if for example one of the fields contains the other one. In your example, since clearly $\;\Bbb Q(\sqrt2+\sqrt7)\subset\Bbb Q(\sqrt2,\sqrt7)\;$ , if the degrees over $\;\Bbb Q\;$ are equal then the fields are the same. $\endgroup$ – Timbuc Sep 24 '14 at 5:26
  • $\begingroup$ Sir, you have some mistake...My question is not root(7),it is cube root(7) $\endgroup$ – Empty Sep 24 '14 at 5:58

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