Let $k \subset L_i \subset M$, $i \in \{1,2\}$ be fields of characteristic zero, with the three field extensions $k \subset L_i$ and $k \subset M$ transcendental of the same degree $n \geq 2$. Therefore, the two field extensions $L_i \subset M$ are algebraic.
(1) Must the (algebraic) degrees of $L_1 \subset M$ and $L_2 \subset M$ be the same (finite or infinite)? (2) If we assume that the degrees of $L_1 \subset M$ and $L_2 \subset M$ are finite, are they equal?
I really apologize if this is a trivial question; any hints are welcome.
Edit: (2) has a negative answer: $k=\mathbb{Q}$, $L_1=\mathbb{Q}(y,x)$, $L_2=\mathbb{Q}(y,x^2)$, $M=L_1$.
No, there is no relation whatsoever between the degrees of $M$ over $L_1$ and $L_2$. For instance, if $K=k(x_1,\dots,x_n)$, then $M$ can be any algebraic extension of $K$ at all, and $L_1$ and $L_2$ could be any intermediate fields between $M$ and $K$ at all.