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I know from Wikipedia that in NBG, the surreal numbers are the largest possible ordered field (if a proper class is allowed to be a field). But then, it is written: "in theories without the axiom of global choice [...] it is not necessarily true that the surreals are the largest ordered field".
How would such a field look like? Is it even possible to define it? The surreals can be defined in various ways, e. g. axiomatically or as a Hahn series over $\mathbb R$. Is this possible without global choice?

I am not very good in logic and set theory, it is thus very probable I have made a mistake.

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    $\begingroup$ I would expect the problem not to be "the largest field is something different", but rather "there isn't a largest field", but I'm guessing a bit. $\endgroup$
    – user13113
    Nov 16, 2017 at 9:40
  • $\begingroup$ My intuition agrees with Hurkyl's; I suspect that it amounts to an inability to construct a 'largest field up to isomorphism', as the isomorphism would require a more robust notion of 'choosing a field' than we can muster without global choice. $\endgroup$
    – Alec Rhea
    Nov 16, 2017 at 9:47
  • $\begingroup$ I believe that the statement on Wikipedia is something that is expected to be true, but we don't actually yet know this. See for example my question mathoverflow.net/q/227849/1946, asking whether without global choice the surreals can fail to be a universal linear order. $\endgroup$
    – Joel David Hamkins
    Nov 16, 2017 at 13:02

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There is no problem defining the surreal field without global choice. One can define it in ZFC and considerably weaker theories, for example with the hereditary birthday construction of left-sets and right-sets, and also in other ways.

With global choice, the surreal field No is largest in the sense of model-theoretic universality: all other ordered class fields are isomorphic to a subfield of No.

One can prove this by means of the usual model-theoretic back-and-forth argument (really only just the "forth" part), combined with the fact that the surreal field No is set-saturated and homogeneous.

That is, given any class field $F$, you use global choice to well-order $F$, and then build up the embedding $j:F\to\text{No}$ in stages. At any stage, you've embedded part of $F$ into the surreals, and you consider the next point. By saturation, there is a surreal number realizing the right type, and you can extend the embedding one more step.

It is the same argument as showing that the rational line $\mathbb{Q}$ is a universal countable linear order.

Without global choice, the argument seems to break down, and one cannot seem to get started. Of course, what is left of the argument is that even without global choice, the surreal field No remains universal for all set-sized fields (can one handle well-orderable class fields?). But I don't think it is known whether there is actually or can be a proper class field that does not embed into No when global choice fails.

If universality fails, it might not mean that there is some other "larger" field. Rather, what I would expect is simply that there are some other ordered class fields that don't embed into No, perhaps none of them maximal even.

I asked an analogous question about the surreals as a universal class linear order: Is the universality of the surreal number line a weak global choice principle? Basically, it seems that most of us expect that one really needs global choice for the universality argument, but to my knowledge we don't yet have any proof of this.

I wonder how the universality of the surreals as a linear order relates to its universality as an ordered field? It has both universality properties under global choice, but can one separate these without global choice?

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  • $\begingroup$ In the "forth" argument for embedding other fields $F$ into No, you say to well-order $F$. I suppose this well-ordering should be such that each element of $F$ has only a set of predecessors. Is that part of what you mean by well-ordering, or is it an additional property that should be mentioned? (It seems that the "natural" way to get a well-ordering will also get you this property automatically, unless you take steps to void it.) $\endgroup$
    – Andreas Blass
    Nov 16, 2017 at 15:42
  • $\begingroup$ Yes, indeed, that is what I meant, since one wants to appeal to set-saturation, so it is important that at every stage, one has handled so far only set many elements. Global choice is equivalent to the assertion that the universe admits an Ord-like well order, since if there is any well-ordering of $V$, then you can order first by rank and then by that order, to get an order of type Ord. $\endgroup$
    – Joel David Hamkins
    Nov 16, 2017 at 15:44
  • $\begingroup$ Is it true (and easy to explain) that No being universal is equivalent to global choice? If so, can one construct a useful family of universality properties that are also equivalent (assuming a common base Theory) to global choice? This might lead to a better way to undergraduate teaching of choice. Gerhard "Searching For A Better Perspective" Paseman, 2019.02.28. $\endgroup$
    – Gerhard Paseman
    Feb 28, 2019 at 18:11
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    $\begingroup$ @JoelDavidHamkins I have a minor observation that is related to your last question. In ZFC, if there is a class linear order $L$ with no definable embedding into No, then there is a class field $F$ extending No with no embedding into No. Specifically, you can use $L$ to build a field $K$ of Hahn series with coefficients in No such that $L$ embeds into the value group of $K$ (and therefore into $K$ itself). I'm not sure whether this argument could be generalized to NBG. $\endgroup$
    – James Hanson
    May 14, 2023 at 22:04
  • $\begingroup$ @JamesHanson Very interesting! I don't think that is minor, but rather substantial progress. I don't see how definability would be relevant, and so I would expect this to generalize to NGB. Send me an email and let's discuss it... [email protected]. $\endgroup$
    – Joel David Hamkins
    May 15, 2023 at 6:49

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