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I read in the post Why worry about the Axiom of Choice ? that the existence of isomorphisms between $\overline{\mathbb{Q}_p}$, $p$ any prime, and $\mathbb{C}$, makes some worry about the Axiom of Choice. One can find another interesting discussion in Are $\mathbb{C}$ and $\overline{\mathbb{Q}_p}$ isomorphic ? (cf. also the references to Deligne's work on the Weil conjectures). My question is: are there other (highly) unexpected isomorphisms between at first sight unrelated fields ?

Thanks !

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    $\begingroup$ This is not really a question about the axiom of choice. $\endgroup$ – Asaf Karagila Jul 13 '16 at 15:58
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    $\begingroup$ This is no more unexpected than two "unrelated" sets or vector spaces being isomorphic. Algebraically closed fields are classified by their characteristic and transcendence degree over their prime subfield. $\endgroup$ – Qiaochu Yuan Jul 13 '16 at 22:05
  • $\begingroup$ It has been always amazing to me that $\mathbb C \cong \prod_{p\in \mathbb{P}}^U \overline{\mathbb{F}_p}^{{\rm alg}}$ for any non-principal ultrafilter $U$. $\endgroup$ – Tomek Kania Jul 13 '16 at 23:09
  • $\begingroup$ Any two algebraically closed $\mathrm{char}=0$ fields of same cardinality are isomorphic by abstract nonsense. I suppose you can get plenty of these "unexpected" isomorphisms this way. E.g. some of the examples are algebraic closures of pseudo-finite fields, mentioned by @TomekKania above. However for this very reason it shows that these isomorphisms don't carry any specific information and aren't very interesting. I suppose the takeaway here is "the set of algebraic automorphisms of $\Bbb C$ is too huge to study". $\endgroup$ – Anton Fetisov Jul 13 '16 at 23:53
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    $\begingroup$ @Anton your statement is incorrect for countable fields: the algebraic closures of $\mathbf Q$ and the rational function field $\mathbf Q(t)$ are both countable of char. 0 but are not isomorphic. As Qiaochu wrote, the char. and transcendence degree over the prime field determine an alg. closed field up to (abstract) isomorphism. For uncountable fields, the cardinality determines the transcendence degree over the prime field, but this is not true for countable fields. $\endgroup$ – KConrad Jul 14 '16 at 5:43

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