Compositum of finite field extensions If $L_{1}$ and $L_{2}$ are field extensions of $F$ that are contained in a common field, show that $L_{1}L_{2}$ is a finite extension of $F$ if and only if both $L_{1}$ 
and $L_{2}$ are finite extensions of $F$. (Patrick Morandi, Field and Galois Theory, Exercise $14$ page $14$.)
Can anyone tell me a hint to prove it?
 A: For a hint, you might start out by looking at $L_1(\lambda)$ where $\lambda\in L_2$, and showing that this field is finite over $F$. From there it shouldn’t be too hard.
EDIT, in response to request for further help:
First step, $L_1(\lambda)$ is finite over $L_1$ because $\lambda$ is root of an $F$-polynomial, and thus of an $L_1$-polynomial. Second, this shows that if $\{a_a,a_2,\cdots,a_n\}$ is an $F$-basis of $L_2$, then you have the finite sequence of finite extensions
$$
L_1\subset L_1(a_1)\subset L_1(a_1,a_2)\subset\cdots\subset L_1L_2\,.
$$
A: The field $L_1L_2$ is the image of the canonical morphism of $F$-algebras $$\phi: L_1\otimes_F L_2\to L: l_1\otimes l_2\mapsto l_1\cdot l_2$$ Since the source $L_1\otimes_F L_2$ of $\phi$  is finite-dimensional so is its image $L_1L_2$.
A: The forward implication is the trivial part. For the reverse implication, the following pieces taken directly from the textbook should help complete the proof (since the question only asks for a hint I am not supplying the full solution).

Lemma 1.19. (page 9) If $K$ is a finite extension of $F$, then $K$ is algebraic and finitely generated over $F$.
Problem 13. (page 14) If $L_1 = F(a_1,\dots,a_n)$ and $L_2 = F(b_1,\dots,b_m)$, show that the composite $L_1 L_2$ is equal to $F(a_1,\dots,a_n,b_1,\dots,b_m)$.
Proposition 1.23. (page 11) Let $K$ be a field extension of $F$, and let $X$ be a subset of $K$ such that each element of $X$ is algebraic over $F$. Then $F(X)$ is algebraic over $F$. If $|X| < \infty$, then $[F(X) : F] < \infty$.

