Isomorphic fields of finite degree have same dimension over base field

Question

Let $K/F$ be a field extension and $L_1,L_2$ subfields of $K$ such that $L_1$ and $L_2$ have finite degree over $F$.

Does $L_1 \cong L_2$ imply $[L_1 : F ]=[L_2 : F]$? Obviously, if the isomorphism fixes $K$ (which isn't always necessarily true) the result holds. The result even holds if $F$ is of finite degree over its prime field.

When trying the usual proof, we get that $[L_1 : F] = [L_1^{\theta} : F^{\theta} ]=[L_2 : F^{\theta}]$ when $\theta \,\colon L_1 \rightarrow L_2$ is the given isomorphism. But I don't see how to relate $F^{\theta}$ to $F$ in an easy way. Any help or counterexample?

Many thanks in advance.

Answer 1

This is indeed a very subtle question. If you assume the standard $F$-vector space structures on both $L_1$ and $L_2$, this need not be true in general. For example, let us take $L_1=\mathbb{C}(X^2)$, $L_2=\mathbb{C}(X)$, and take $F=\mathbb{C}(X^2)$. Clearly the map $X^2 \to X$ gives you an isomorphism from $L_1$ to $L_2$. Note that under the usual $F$-vector space structures, we have, $$[L_1:F]=[\mathbb{C}(X^2):\mathbb{C}(X^2)]=1$$

On the other hand, $$[L_2:F]=[\mathbb{C}(X):\mathbb{C}(X^2)]=2$$

The idea here, is that under this isomorphism, $\mathbb{C}(X)$ gets endowed with a different $\mathbb{C}(X^2)$-vector space structure, one in which you do have $[\mathbb{C}(X):\mathbb{C}(X^2)]=1$. The structure, as I'm sure you'll notice instantly, is the following: $$\forall h \in \mathbb{C}(X^2),g\in \mathbb{C}(X),~~h(X^2).g(X)=h(X)g(X) $$

So, you can possibly conclude that $[L_1:F]=[L_2:F]$, provided you are considering a non-canonical $F$-vector space structure.

Answer 2

No.

A simple counterexample is $K=\Bbb{Q}(\pi)$, $F=\Bbb{Q}(\pi^6)$, $L_1=\Bbb{Q}(\pi^2)$ and $L_2=\Bbb{Q}(\pi^3)$. Then

1. All the fields $K,F,L_1,L_2$ are simple transcendental extensions of $\Bbb{Q}$ and thus they are all isomorphic to each other.
2. $[K:F]=6$, $[L_1:F]=3$, $[L_2:F]=2$.