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.
 A: 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. 
A: Here’s another elementary proof, which I used rather more advanced ideas to come up with:
Let’s look at $2\big/(\sqrt2+\sqrt3+1)=1+\frac12(\sqrt2-\sqrt6\,)$, which happens to be a root of $f(X)=X^4-4X^3+2X^2+4X-2$, Eisenstein and thus irreducible. This shows that $\sqrt2+\sqrt3$ is quartic over $\Bbb Q$, and, by degree considerations, that $\Bbb Q(\sqrt2+\sqrt3\,)=\Bbb Q(\sqrt2,\sqrt3\,)$.
So what, you say. But consider $f$ as a polynomial over $k=\Bbb Q(\sqrt5\,)$. The ring of integers of $k$ is $\Bbb Z\bigl[\frac{1+\sqrt5}2\bigr]$ , well known to be a Principal Ideal Domain, and you see easily that $2$ is an irreducible (prime) element there. Thus $f$ remains Eisenstein, and so irreducible, over $k$. As a result, the extension $\Bbb Q(\sqrt5,\sqrt2+\sqrt3\,)\supset\Bbb Q(\sqrt5\,)$ is of degree four, and the big field is of degree eight over $\Bbb Q$. Since $\sqrt2+\sqrt3$ is only quartic over $\Bbb Q$, it follows that $\sqrt5$ is quadratic over that quartic field.
