How do I show $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\geq[\mathbb{Q}(\sqrt2):\mathbb{Q}][\mathbb{Q}(\sqrt3):\mathbb{Q}]$? I know how to show $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\leq[\mathbb{Q}(\sqrt2):\mathbb{Q}][\mathbb{Q}(\sqrt3):\mathbb{Q}]$, but don't know how to show the converse inequality. $[\mathbb{Q}(\sqrt2):\mathbb{Q}]$ and $[\mathbb{Q}(\sqrt3):\mathbb{Q}] $ both divides $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$, but they are not relatively prime, so I guess I can't the converse inequality this way. I'm considering using the fact that the intersection of $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt 3)$ is $\mathbb{Q}$, but I don't know how to proceed...
 A: The specific question in the title can be settled by elementary methods. Or as a part of more general fact about the field you get by adjoining square roots of coprime integers.
See the excellent write up by Bill Dubuque.
In a comment the question was specified to be about a general fact about the degree of the compositum of two algebraic extensions of a field $K$ that intersect trivially. It may be natural to suspect that if $K(x)\cap K(y)=K$, where $K(x), K(y)$ are both algebraic extensions of $K$, then we could have a degree formula like $[K(x,y):K]=[K(x):K][K(y):K]$. 
However, this is FALSE in general. An easy to digest counterexample is that of
$K=\Bbb{Q}$, $x=\root3\of2$, $y=\omega\root3\of2$, where $\omega=(-1+\sqrt{-3})/2$ is a primitive cubic root of unity. Both $x$ and $y$ are zeros of the irreducible polynomial $p(x)=x^3-2$, so $[K(x):K]=3=[K(y):K]$. We have $K(x)\subset\Bbb{R}$, but $K(y)\cap\Bbb{R}=\Bbb{Q}$, so $K(x)\cap K(y)=K$. Furthermore, $K(x,y)$ is the splitting field of $p(x)$ over $K$, which is known to have degree six, not nine. This follows from $\omega$ being a root of a quadratic. Anyway, the above degree formula does not hold in this case.
I gathered that you are in a context where you do need a more general result. The key concept here is that of linearly disjoint extensions. The degree formula holds, iff $K(x)$ and $K(y)$ are linearly disjoint over $K$. A not so difficult to prove general result here is that if $K(x)/K$ and $K(y)/K$ are both Galois extensions (inside a bigger field, e.g. en algebraic closure of $K$), and $K(x)\cap K(y)=K$, then they are necessarily linearly disjoint.
