# Vakil 11.2.A: Transitivity of an equivalence relation involving composita of fields

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 $F'''$?

• If $x \in F'F'''$ and $x \in F'F''$ and $x \in F''F'''$, then clearly $x$ is algebraic over $F'$ and algebraic over $F'''$. What happens, for example, if $x \in F'$ but $x \notin F'''$? – Jeremy Gross Jun 1 '18 at 15:55
• Let $x\in F'F'''$. If $x\in F'$, $x$ is algebraic over $F'$, so it is the root of a polynomial with coefficients in $F'$, each of which is the root of a polynomial with coefficients in $F''$, each of which is the root of a polynomial with coefficients in $F'''$. So $x$ is algebraic in $F'''$. And similarly, if $x\in F'''$, then $x$ is algebraic over $F'$. What about the case where $x \notin F'$ and $x \notin F'''$? – Jeremy Gross Jun 2 '18 at 13:51
• In general, if $x\in F'F'''$, then $x$ can be expressed as a polynomial in $F'$ and $F'''$, each term of which, we have shown is algebraic in $F'$ and $F'''$. – Jeremy Gross Jun 2 '18 at 13:57
• That completes the proof. – Jeremy Gross Jun 2 '18 at 13:57

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 $$F''$$.