1
$\begingroup$

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 $F'''$?

$\endgroup$
  • $\begingroup$ 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 $x \notin F'''$? $\endgroup$ – Jeremy Gross Jun 1 '18 at 15:55
  • $\begingroup$ 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 $x \notin F'''$? $\endgroup$ – Jeremy Gross Jun 2 '18 at 13:51
  • $\begingroup$ 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 $F'''$. $\endgroup$ – Jeremy Gross Jun 2 '18 at 13:57
  • $\begingroup$ That completes the proof. $\endgroup$ – Jeremy Gross Jun 2 '18 at 13:57
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.