Can two fields which are extensions of one another be non isomorphic

Question

Let $\mathbb L$ and $\mathbb K$ be two fields. We say that $\mathbb L$ is an extension of $\mathbb K$ if there exists a ring homomorphism $\varphi$ from $\mathbb K$ to $\mathbb L$. In that case, $\varphi$ is necessarily one-to-one and it is customary to identify $x\in\mathbb K$ with its image $\varphi(x)\in\mathbb L$, so that we can think of it as an inclusion $\mathbb K\subset\mathbb L$.

I am not really clear with the caveats of such an identification. In particular there is a question I could not find an answer to: if $\mathbb K\subset\mathbb L$ and $\mathbb L\subset\mathbb K$ for the real inclusion, we obviously have $\mathbb K=\mathbb L$. But what about with the aforementioned identifications? That is, if $\mathbb L$ is an extension of $\mathbb K$, and $\mathbb K$ is an extension of $\mathbb L$, do we necessarily have that $\mathbb K$ and $\mathbb L$ are isomorphic fields?

What I have tried: Suppose that $\mathbb L$ and $\mathbb K$ are extensions of one another. Then there exist ring homomorphisms $\varphi:\mathbb K\to\mathbb L$ and $\psi:\mathbb L\to\mathbb K$. So $\varphi\circ\psi$ is a ring homomorphism from $\mathbb L$ to itself. To conclude it would suffice to prove that its image is $\mathbb L$, or at least isomorphic to $\mathbb L$. Why would that be true? I do not know. Is there a counter-example? I do not know.

Answer 1

That is, if $\mathbb L$ is an extension of $\mathbb K$, and $\mathbb K$ is an extension of $\mathbb L$, do we necessarily have that $\mathbb K$ and $\mathbb L$ are isomorphic fields?

No. For example, $\mathbb{C}(x)$ is clearly an extension of $\mathbb{C}$. Much less obviously, $\mathbb{C}$ is also an extension of $\mathbb{C}(x)$, because for general field-theoretic reasons the algebraic closure $\overline{\mathbb{C}(x)}$ must be abstractly isomorphic to $\mathbb{C}$ (assuming the axiom of choice). And $\mathbb{C}$ and $\mathbb{C}(x)$ are not isomorphic because the latter is not algebraically closed, e.g. the polynomial $t^2 - x$ doesn't have a root.

This implies that $\mathbb{C}$ embeds nontrivially into itself, although this embedding can't really be written down explicitly and is very poorly behaved, e.g. it is nowhere measurable.

In the positive direction a field which is finite-dimensional over its prime subfield clearly can't embed nontrivially into itself, so you are safe if $L$ and $K$ are either finite fields or number fields.

Answer 2

Let me copy my answer from MO:

Here is another counterexample for fields: If $K=\overline{\mathbb{Q}(x_1,x_2,...)}$, then there are monomorphisms $K(x_0) \to K \to K(x_0)$, but no isomorphism since $K(x_0)$ is not algebraically closed.