Let $F\_0 \subset F\_1 \subset F\_2 \subset \cdots$ and $K\_0 \subset K\_1 \subset K\_2 \subset \cdots$ be two towers of fields. Also, let $F = \cup\_{i=0}^\infty F\_i$ and $K = \cup\_{i=0}^\infty K\_i$.


Now suppose for each $i$ we have injective homomorphisms from $F\_i$ to $K\_{\sigma(i)}$ and from $K\_i$ to $F\_{\mu(i)}$ where $i \leq \sigma(i)$ and $i \leq \mu(i)$. In other words, each field $F\_i$ is isomorphic to a subfield of some $K\_j$ where $j \geq i$ and each field $K\_i$ is isomorphic to a subfield of some $F\_j$ where $j \geq i$. [Think of the two towers sitting next to each other with arrows pointing diagonally upward.]


My question, can we conclude that $F \cong K$?


A colleague asked me this question some time ago. I came up with a sketch of a proof for the case when $F\_{i+1}$ is an algebraic extension of $F\_i$ and $K\_{i+1}$ is an algebraic extension of $K\_i$ for each $i$. I suspect it's false in general [something to do with the fact that injective and surjective aren't equivalent for maps between infinite dimensional spaces.]


Does anybody know a counterexample for the general case? 


I would also appreciate a reference for the algebraic case (where I'm 99% sure it's true).


Thanks!



Given 2 towers of fields, when are these fields isomorphic?
This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. That said, we can take the standard counter example for the claim of CSB for fields as a counter example for your claim. Let $F\_i = \mathbb{C}(X)$ and $K\_i = \mathbb{C}$ for all $i.$ Then $K\_i$ injects into $F\_i$ and $F\_i$ injects into $K\_i$ for all $i,$ but the unions $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed. 


