I am struggling with the following:
Let's have the field extension $R/Q$. How can I find $[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q]$? (*)
So far all I have is that:
$[Q(\sqrt 6):Q] = 2$, because $minpoly_{\sqrt6}(X) = X^2 - 6$
Samely $[Q(\sqrt 10):Q] = [Q(\sqrt 15):Q] = 2$.
Another thing I read somewhere (couldn't find it anywhere else), is that:
$[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q] = [Q(\sqrt 2 \cdot \sqrt 3,\sqrt 2 \cdot \sqrt 5,\sqrt 3 \cdot \sqrt 5):Q] = [Q(\sqrt 2,\sqrt 3,\sqrt 5):Q]$
Is that true? Can I conclude that way? And if so, what can I do afterwards to find (*)?
My first idea was to calculate it this way:
$[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q] = [Q(\sqrt 6,\sqrt 10,\sqrt 15):Q(\sqrt 6,\sqrt 10)] \cdot [Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)] \cdot [Q(\sqrt 6):Q]$
but I don't know how to compute the terms $[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q(\sqrt 6,\sqrt 10)]$ and $[Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)]$ either.
What I also found is that $\sqrt 6, \sqrt 10$ and $\sqrt 15$ are linearly dependent over $Q$, since $\sqrt 6 \sqrt 10 - 2 \cdot \sqrt 15 = 0$.
Is there some general way to find $[Q(\alpha, \beta):Q]$ and $[Q(\alpha, \beta, \gamma):Q]$?
Thank you in advance.
$\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15)=\mathbf Q(\sqrt 6,\sqrt 10)$, since $\sqrt 15=\frac12\sqrt 6 \sqrt 10$, whence $$[\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15):\mathbf Q(\sqrt 6,\sqrt 10)]=1.$$ $x^2-10$ is irreducible over $\mathbf Q(\sqrt 6)$, sin you can easily check no $\alpha=a+b\sqrt 6$ is a root of $x^2-10$ just writing down the equations for $a$ and $b$.. Thus $$[\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15):\mathbf Q]=[\mathbf Q(\sqrt 6,\sqrt 10:\mathbf Q]=[\mathbf Q(\sqrt 6,\sqrt 10):\mathbf Q(\sqrt 6)]\cdot[\mathbf Q(\sqrt 6):\mathbf Q]=4.$$