Embedding of fields

Question

My question has been asked for groups I think, but there are no free fields, and fields seem more rigid than groups, so I feel like here the answer may be positive.

The question is the following : suppose we have field morphisms $L\to K\to L$ (I don't know if this is the used convention but I take my morphisms to have $f(1)=1$; and so they are embeddings), must we then have $L\simeq K$ ?

Now obviously the first thing is that, as in Cantor-Bernstein's theorem, the morphism $L\to K$ need not be itself the isomorphism (consider the field $\mathbb{Q}(X_n)_{n\in \mathbb{N}}$ which has plenty of non surjective embeddings into itself). So one would have to construct the isomorphism, not prove that the given morphisms are isomorphisms.

More generally, the question that naturally arises from this is : given a class of universal algebras (of one given type), under what conditions do we have "$A$ can be embedded in $B$ and conversely $\implies A\simeq B$" ?

The answer for the class of fields is the answer to my first question, and for the class of groups, the answer is no (see the question "When can two groups be embedded in eachother ?")

Answer 1

No. This is a famous question; and a proof may be given using the fact that algebraically closed fields with same transcendence degree over their prime field are isomorphic. Armed with this fact, let $ L = \mathbf C $ and $ K = \mathbf C(T) $. There's an obvious embedding $ L \to K $, and another obvious embedding $ K \to \bar{K} $. However, $ \bar K $ is an algebraically closed field of transcendence degree $ c = |\mathbf R| $ over $ \mathbf Q $, thus it is isomorphic to $ \mathbf C = L $. This gives us embeddings $ L \to K \to L $, and $ L $ is not isomorphic to $ K $, since one field is algebraically closed whereas the other is not.