[FOM] Artin-Schreier for Q, without choice?

[Colin McLarty]

  Läuchli's result is succinctly stated at


https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/lauchlis-algebraic-closure-of-q/771836F25AA1C59423DE4247AF8D2AF8

H. Läuchli (9) constructed, within a model of a weak form of set theory, an
> algebraic closure L of the field Q of rationals which had no real-closed
> subfield. Läuchli's construction is easily transferred to a model N of ZF
> (= Zermelo–Fraenkel set theory without the axiom of Choice), and it follows
> at once that neither of the two following statements is provable from ZF
> alone

:
> Every algebraic closure of Q has a real-closed subfield. (1)
>
There is, up to isomorphism, at most one algebraic closure of Q. (2)


On Sun, Jan 5, 2020 at 12:18 PM Joe Shipman <joeshipman at aol.com> wrote:

> It’s straightforward to show without choice that there is a computable
> ordered field where every odd degree polynomial has a root and nonnegative
> numbers have square roots. I think the failure of choice corresponding to
> Lauchli’s result would not be that there is no real closed field that
> satisfies all the definitions, but that a field which satisfied the same
> first order properties as the reals in the language of fields could not
> necessarily be ordered compatibly with the field operations (have to make
> infinitely many choices of which square root of each prime shall be the
> “positive” one).
>
> — JS
>
> Sent from my iPhone
>
> > On Jan 5, 2020, at 1:27 AM, Timothy Y. Chow <tchow at math.princeton.edu>
> wrote:
> >
> > In ZF, one cannot prove that the algebraic closure of Q (the rationals)
> is unique.  In fact, Lauchli has constructed an algebraic closure of Q
> which has no real-closed subfield.
> >
> > But now suppose we just want to build *an* algebraic closure L of Q with
> all the properties we know and love, and don't care that there might exist
> "exotic" algebraic closures of Q that are not isomorphic to L.  Since N
> (the natural numbers) is given to us with a well ordering, it is
> straightforward to construct L without needing to invoke AC.
> >
> > What about the Artin-Schreier theorem?  It seems straightforward to
> construct (a particular version of) the real algebraic numbers, but what
> about proving that they form a real-closed field?  Glancing at a proof of
> Artin-Schreier, I instinctively feel that AC is not needed for the specific
> example of Q, but in light of Lauchli's result, I am worried that I am
> missing some subtle point.  Also, in the absence of AC, do we need to be
> cautious about the precise statement of any of the usual properties of
> real-closed fields---such as those listed on Wikipedia?
> >
> > https://en.wikipedia.org/wiki/Real_closed_field
> >
> > Tim
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