Given 2 towers of fields, when are these fields isomorphic? - MathOverflow

Given 2 towers of fields, when are these fields isomorphic?

2011-09-06T00:31:45Z

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!

Answer by JSpecter for Given 2 towers of fields, when are these fields isomorphic?

2011-09-06T01:06:33Z

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.

Answer by KConrad for Given 2 towers of fields, when are these fields isomorphic?

2011-09-06T10:10:38Z

Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.

By the way, this question is similar to http://mathoverflow.net/questions/39460/schroeder-bernstein-for-rings and my answer is similar to one given there.