<s>[INST] Find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$ We want to find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$.
My first thought is to find the minimal polynomial of $\sqrt5$ over $\Bbb{Q}(\sqrt2+\sqrt3)$. And from this, to say that $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]=\deg m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x).$
We take the polynomial $m(x)=x^2-5\in\Bbb{Q}(\sqrt2+\sqrt3)[x].$ This is a monic polynomial which has $\sqrt5\in \Bbb{R}$ as a root. We have to show that this is irreducible over $\Bbb{Q}(\sqrt2+\sqrt3)$, in order to say that $m(x)=m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x)$. The roots of $m(x)$ are $\pm \sqrt5\in \Bbb{R}.$ So,
$$m(x) \text{ is irreducible over } \Bbb{Q}(\sqrt2+\sqrt3) \iff \pm \sqrt5 \notin \Bbb{Q}(\sqrt2+\sqrt3)=\Bbb{Q}(\sqrt2,\sqrt3)$$
because $\deg m(x) =2.$ 
And this is the point I stack. I tried with the use of the basis of the $\Bbb{Q}$-vector space $\Bbb{Q}(\sqrt2+\sqrt3)$:
$$A:=  \{1,\sqrt2,\sqrt3,\sqrt6 \}$$ 
in order to claim that $\nexists a,b,c,d\in \Bbb{Q}:\sqrt5=a+b\sqrt2+c\sqrt3+d\sqrt6$ but this doesn't help.
Any ideas please? 
Thank you in advance.
 [/INST] Here is an idea that avoids most of the theory. 
The correspondence:
$$a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6} \mapsto a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6}$$ preserves sums and products. So if 
$$\sqrt{5} = a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6}$$ then
$$(a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})^2= 5$$ so
$$(a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6})^2 = 5$$ so
$$a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6} = \pm (a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})$$ so either $a = b = 0$ or $c = d = 0$.  Now it should be easy, but really, the same trick applies: If $\sqrt{5} = c\sqrt{3} + d\sqrt{6}$ then $c\sqrt{3} - d\sqrt{6}= \pm( c\sqrt{3} + d \sqrt{6})$, so again, $c= 0$ or $d=0$.  In the end we get a contradiction. 
Note that this procedure can be generalized, for more radicals, provided we have the uniqueness of the writing. We haven't really used $5$, it really shows which square roots are in such a field. 
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