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Can you find an example of non-isomorphic fields which embed inside each other?

Most probably we can't but I am looking for extraordinary answer...

Watson
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user175632
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    Answer: Yes, you can. Hint: Try something with infinite transcendence degree over its prime field... – Zhen Lin Sep 12 '14 at 21:04
  • Yes, $\overline{\Bbb C(t)}\cong\Bbb C$. So $\Bbb C(t)$ embeds into $\Bbb C$ and vice versa. – PVAL-inactive Sep 12 '14 at 21:06
  • See: http://math.stackexchange.com/questions/465631, http://math.stackexchange.com/questions/803332, http://math.stackexchange.com/questions/257650 – Watson Aug 17 '16 at 18:41

2 Answers2

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A start: Let the two fields be the complex numbers $\mathbb{C}$ and the field $\mathbb{C}(X)$ of rational functions with complex coefficients.

These are non-isomorphic, since one is algebraically closed and the other isn't.

Embedding in one direction is trivial. In the other direction, take a transcendence base for $\mathbb{C}$ and map $X$ to an element of the base, and map the remaining elements of the transcendence base appropriately.

Remark: The "construction" of the embedding uses the Axiom of Choice. My shortcut approach would be to use the fact that the theory of algebraically closed fields of characteristic $0$ is $\kappa$-categorical for every uncountable $\kappa$.

It would be interesting to know whether one can produce an example without using AC.

André Nicolas
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  • An alternative in the same spirit is to replace $\mathbb{C}$ an algebraic closure of $F(X_1, X_2, \ldots)$ for some field $F$. If $F$ is nice, I think you don't need AC to construct a closure. –  Aug 12 '16 at 10:34
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This is not something I understand well, but I believe you can construct examples as follows.

Let $E$ be an elliptic curve, and let $E'$ be another elliptic curve that is isogenous to $E$, but not isomorphic to $E$.

Then the isogeny should provide an embedding $k(E') \to k(E)$, and its dual should provide an embedding $k(E) \to k(E')$.

  • A related example is the one from David Speyer: "Let E be an elliptic curve over C, and let E –> E’ be a degree two isogeny. (In other words, E=E’/(G) where G is an order two subgroup of E.) If we choose E generically, E and E’ are not isomorphic. There is also a degree two isogeny from E’ to E: quotient E’ by the image in E’ of the 2-torsion of E. Now use the equivalence of categories between smooth algebraic curves over C and finitely generated fields of trans. degree 1 over C." (taken from here). – Watson Aug 17 '16 at 18:48
  • Another comment, from Scott Carnahan, suggests another example using "compact manifolds without boundary, or compact manifolds rel some fixed boundary manifold." – Watson Aug 17 '16 at 18:49