$[\mathbb{Q}(A,B): \mathbb{Q}(A)]=[\mathbb{Q}(B):\mathbb{Q}]$? Suppose that I have $A,B \not \in \mathbb{Q}$ and $A \neq B$. Is it always true that
$[\mathbb{Q}(A,B): \mathbb{Q}(A)]=[\mathbb{Q}(B):\mathbb{Q}]$?
Example:
$[\mathbb{Q}(5^{1/6},\zeta_6): \mathbb{Q}(\zeta_6)]=[\mathbb{Q}(5^{1/6}):\mathbb{Q}]=6$
My intuition is that it is the case, but Im not sure why.
 A: I'm going to talk about the question I think you hoped to ask. Your interested in an equation that follows from another interesting one:$$[\mathbb{Q}(A,B):\mathbb{Q}]=[\mathbb{Q}(A,B):\mathbb{Q}(B)]\cdot[\mathbb{Q}[B]:\mathbb{Q}]=[\mathbb{Q}(A,B):\mathbb{Q}(A)]\cdot[\mathbb{Q}(A):\mathbb{Q}]$$  For example, if $A=\sqrt{2}$ and $B=\sqrt{5}$ or if $A=\sqrt{71}$ and $B=\zeta_2$ this second equation holds, and so the one in the OP does as well. Since $[K:F]=[K:L][L:F]$ always holds, it is sufficient for the two field extensions to "not interact" in a way that causes reduction of degree when looking at the extension $\mathbb{Q}(A,B)/\mathbb{Q}(A)$ and $\mathbb{Q}(A,B)/\mathbb{Q}(B)$. One notion of "not interacting" that is sufficient to prevent degeneracy and therefore imply all of the equations here is that the minimum polynomials $p_A(x)$ and $p_B(x)$ are irreducible over $\mathbb{Q}(B)$ and $\mathbb{Q}(A)$ respectively.
This condition turns out to be both necessary and sufficient, which can be proven by playing around with the equation $[K:F]=[K:L][L:F]$ in various combinations.
A: No, say $B\not\in\Bbb Q$ and $A=1+B$ then

$$\Bbb Q(A,B)=\Bbb Q (A)$$
  $$\implies[\Bbb Q(A,B):\Bbb Q(A)]=1< [\Bbb Q(B):\Bbb Q)].$$

And this is not even unique to $\Bbb Q$, this works for any field, $F$ since all of them have $1$.
