What is the smallest subfield $F\subset N_0$ such that $$(F,+,\times,\leq)\ncong(N_0,+,\times,\leq)$$ but $$(F,+,\leq)\cong(N_0,+,\leq)?$$ Since these are all going to be proper classes cardinality is not sufficient to delineate what we mean by 'smallest', so we use the following notion. For two proper class sized ordered fields $F,F'$ we will say that $F$ is smaller than $F'$ iff there exists an injective order preserving homomorphism of the natural value group of $F$ into the natural value group of $F'$, but the reverse does not hold.

I asked this question to a friend recently and he pointed out that the class of all Conway normal forms with rational coefficients satisfies the above isomorphism identities, and since the natural value group of an ordered field is determined by the multiplicative structure of the field I believe this means that we will have such a one sided injection between their value classes. He also pointed out that the group $\mathbb{J}$ of all purely infinite surreal numbers is the classical example of a subgroup of $N_0$ which is isomorphic to it as an ordered group but not as a field, but this structure does not form a field as it does not contain $1$.

Are there any subfields 'smaller' than this with the desired property? Can we determine what the 'smallest' one is up to an appropriate type of morphism?

  • I am not sure that this is the best possible notion of 'smaller' for this question -- suggestions are appreciated. – Alec Rhea Jun 2 at 8:31
  • I wrote JJ because on my text editor it looked most like $\mathbb{J}$, which is the real usual notation. The group $\mathbb{J}$ is stable under product but not does not contain $1$. – nombre Jun 3 at 10:10
  • @nombre I had assumed that, for example, $\omega$ and $\frac{1}{\omega}$ were both in $\mathbb{J}$, no? – Alec Rhea Jun 3 at 10:21
  • @nombre Ah, nevermind, I completely misread the definition of $\mathbb{J}$ in your email. Thank you again haha. – Alec Rhea Jun 3 at 10:25
up vote 3 down vote accepted

$\DeclareMathOperator{\Noo}{\mathbf{No}}$This might actually be a dead end.

This is because if $F$ is isomorphic as an ordered group to $\Noo$, then their value classes under natural ordered group valuation, that is, the underlying orders of their value groups under natural ordered field valuation, must be isomorphic. So the value group $(v F,+,\leq)$ of $F$ is isomorphic as a linear order to $\Noo$. If it is additionnaly divisible, then it is saturated as an ordered group, hence isomorphic to $(\Noo,+,\leq)$ - possibly by a different isomorphism. Thus in general $(\mathbb{Q}.vF,+,\leq)$ and $(\Noo,+,\leq)$ are ismorphic. Of course $F$ could still be "much smaller" than $\Noo$, for instance it might be the field generated by all monomials. I don't know if this is the type of field you are looking for.

  • I think you’re correct once again. Thank you for this — if I can think of no workaround to edit the question, I’ll accept this before I pass out tonight. – Alec Rhea Jun 3 at 10:19
  • 2
    I'd like to point out that arguments for universality via saturation with proper-class-sized objects generally make use of the global axiom of choice (to run the back-and-forth construction). This axiom is not provable in ZFC, although it is a standard axiom of GBC and KM, and GBC with global choice is conservative over ZFC for first-order assertions about sets. For example, we don't seem to know in ZFC even that No is universal as a linear order; see my question mathoverflow.net/q/227849/1946. – Joel David Hamkins Jun 3 at 15:48
  • @JoelDavidHamkins: Yes, maybe "hence isomorphic to" is a little quick. Although here it must not be too difficult to recover a bijection $\mathbb{Q}. v F \rightarrow \mathbf{Ord}$ from the given group isomorphism. – nombre Jun 5 at 5:56

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