Pretty simple question. Does there exist a ordered field smaller than (i.e. is a strict subset of) $\mathbb{Q}$? It seems like we can't go any smaller than $\mathbb{Q}$. Is this true? Why?

  • 4
    $\begingroup$ Well, you have to argue that no finite field can be ordered. More broadly, try to argue that no field of finite characteristic $p$ can be ordered. Beyond that, if $\mathbb F$ is a subfield of $\mathbb Q$ it must contain $1$ hence it contains $\mathbb Z$, hence it contains $\mathbb Q$. $\endgroup$ – lulu Nov 9 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.