I'm working on Vakil's "Rising Sea" notes for an independent study. In Exercise 11.2.A, he gives a relation between intermediate field extensions. If $E/F$ is a field extension, and $F'$ and $F''$ are intermediate field extensions, then $F'\sim F''$ if $F'F''$ is algebraic over both $F'$ and $F''$. Here $F'F''$ is the compositum of $F'$ and $F''$, the smallest field extension in $E$ containing $F'$ and $F''$.
He asks to show that $\sim$ is an equivalence relation. Reflexivity and symmetry are obvious, but I'm having trouble showing transitivity. I'm not sure about how knowing that $F'F''$ is algebraic over both $F'$ and $F''$, and knowing that $F''F'''$ is algebraic over both $F''$ and $F'''$ gives us that $F'F'''$ is algebraic over both $F'$ and $F'''$.
A nudge in the right direction would be appreciated. If $x\in F'F'''$, how can we show that $x$ is algebraic over $F'$ and algebraic over $F'''$?
One way to interpret this is by noting that $F' \sim F''$ is equivalent to the statement "every element of $F'$ is algebraic over $F''$, and vice versa."
To see this, suppose first that $F' \sim F''$. Then if $a \in F'$, we know $a \in F'F''$, which we are assuming is algebraic over $F''$, and so $a$ is algebraic over $F''$; similarly if $b \in F''$ then $b$ is algebraic over $F'$. Conversely, suppose every element of $F'$ is algebraic over $F''$ and vice versa. We know any element $c \in F'F''$ can be written as a rational function of elements of $F'$ and elements of $F''$, and that a rational function of elements algebraic over a field is again algebraic (since the set of algebraic elements over a given base field is a field), and since all elements of $F', F''$ are algebraic over both $F', F''$, we have that $c$ is algebraic over $F'$ and $F''$.
Once we've interpreted the equivalence relation like this, transitivity becomes easy.