0
$\begingroup$

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 !

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ This is not really a question about the axiom of choice. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 13, 2016 at 15:58
  • 1
    $\begingroup$ 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
    Commented Jul 13, 2016 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 $U$. $\endgroup$
    – Tomasz Kania
    Commented Jul 13, 2016 at 23:09
  • 1
    $\begingroup$ 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
    Commented Jul 13, 2016 at 23:53
  • 2
    $\begingroup$ @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
    Commented Jul 14, 2016 at 5:43

0

Reset to default

You must log in to answer this question.