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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 $a,b \in \mathbb{Q}$

Empty
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1 Answers1

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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.

Timbuc
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  • 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? – Empty Sep 23 '14 at 21:43
  • Don't understand your question, @SayantanPanja – Timbuc Sep 23 '14 at 22:31
  • 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))? – Empty Sep 24 '14 at 03:45
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    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. – Timbuc Sep 24 '14 at 05:26
  • Sir, you have some mistake...My question is not root(7),it is cube root(7) – Empty Sep 24 '14 at 05:58
  • The principle is the same, @SayantanPanja, no matter what roots are those. – Timbuc Sep 24 '14 at 05:59
  • If it is, then how we can show that , (2^(1/2))∈Q(2^(1/2)+7^(1/3)). Further how we can show that, (7^(1/3))∈Q(2^(1/2)+7^(1/3))?..@Timbuc – Empty Sep 24 '14 at 06:55
  • That I don't know, @SayantanPanja. I suppose that you'll have to play with some nasty equations, say: $$\sqrt[3]7=a+bw+cw^2+dw^3+ew^4+fw^5;,;;w:=\sqrt2+\sqrt[3]7;,;;a,b,c,d,e,f\in\Bbb Q$$ Not very pretty... – Timbuc Sep 24 '14 at 13:31