I am trying to prove the following result from an elementary field extensions tutorial:

Let $F$ be a field, $K$ a field extension of $F$ and $L$ a field extension of $K$.

Assume that $K$ is algebraic over $F$, and that element $a \in L$ is algebraic over $K$. Deduce that $a$ is algebraic over F.

So far I am not even convinced that it is true. Confirmation that it is true would be a great help, then any hints would be welcome. I have tried two approaches:

(1) We know that as $a$ is algebraic over $K$, so $[K[a] : K]$ is finite (and $K[a]$ is a subfield of $L$). If we knew that $[K:F]$ was finite, then we might argue that $[K[a] : F]$ is finite (since $[K[a] : F] = [K[a] : K][K : F]$), and so somehow combine that with claim: $F[a] \subset K[a]$ implies $[F[a] : F]$ is finite, which by a theorem implies $a$ is algebraic over $F$. However we are not given that $[K:F]$ is finite, and from my reading, this is not implied by $K$ algebraic over $F$, so I don't know how to proceed in this direction.

(2) Given that $a$ is algebraic over $K$, there exists an irreducible polynomial of $a$ in $K[X]$ (call it $p$). From here we need to construct the irreducible polynomial of $a$ in $F[X]$ (call it $q$) (This will show that $a$ is algebraic in $F$). As I understand it, I need to show the existence of polynomial $g$ in $K[X]$ such that $gp = q$, meaning that both $g$ and $p$ have coefficients in $K$, but $q$ has only coefficients in $F$. Right now I have no idea how to show that such a $g$ exists -- presumably I would need to somehow use the fact that K is algebraic over F.

Are either of these approaches worth pursuing further or is there a more illuminating way to think about it?

  • 2
    $\begingroup$ It's definitely true. $\endgroup$ – D_S Jun 15 '17 at 3:49
  • 3
    $\begingroup$ Hint: $a$ is the root of a polynomial $f(X) = c_0 + c_1X + \cdots + c_nX^n$ with coefficients in $K$. So $a$ is algebraic over $F[c_0, ... , c_n]$. Instead of using $K$, use the field $F[c_0, ... , c_n]$, and then you can invoke finiteness. $\endgroup$ – D_S Jun 15 '17 at 3:55

Your Answer

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

Browse other questions tagged or ask your own question.