Consider the Grothendieck ring $K[\Omega]$ of the semiring $\Omega$ of all ordinals under the operations of natural sum and product. Its quotient field $K(\Omega)$ is naturally a subfield of the ordered field No of surreal numbers. Is the field No algebraic over the field $K(\Omega)$ (and therefore equal to the real closure of $K(\Omega)$)? If not, what is the transcendence degree of No over $K(\Omega)$?

EDIT: I think on second thought that the real closure of $K(\Omega)$ intersected with $\mathbb{R}$ in No is possibly just the real closure of $\mathbb{Q}$, rather than all of $\mathbb{R}$. Perhaps then I need to consider the real closure of the compositum in No of $K(\Omega)$ and $\mathbb{R}$ and ask if No is algebraic over that. If that's not true, then is there some way of bootstrapping such a construction to some sort of algebraic construction of No from $K(\Omega)$?

Related mathoverflow post:

Will Sawin's answer to another mathoverflow question shows that No is a proper extension of $K(\Omega)$. See Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addition?

It is perhaps relevant to note that $K[\Omega]$ can be identified with the polynomial ring $\mathbb{Z}[\omega^{\omega^\alpha}: \alpha \in \Omega]$ generated by the (algebraically independent) delta numbers $\omega^{\omega^\alpha}$ for $\alpha \in \Omega$, and the field $K(\Omega)$ is then identified with the field $\mathbb{Q}(\omega^{\omega^\alpha}: \alpha \in \Omega)$ of rational functions in the delta numbers.


Since in NBG the class $\Omega$, and therefore also the class $\{\omega^{\omega^\alpha}: \alpha \in \Omega\}$, maps onto every class, it follows that the ring $K[\Omega]$ maps homomorphically onto every commutative ring, even those with an underlying proper class. In particular, there is a surjective ring homomorphism $K[\Omega] \longrightarrow \operatorname{No}$, or equivalently a maximal ideal $M$ in $K[\Omega]$ such that $K[\Omega]/M$ is isomorphic to $\operatorname{No}$. Is there a nice way to define such a homomorphism and/or maximal ideal?

YET ANOTHER ADDITION: Is $K[\Omega]$ isomorphic to a quotient ring of the ring Oz of omnific integers?

  • $\begingroup$ I believe that the answer to the additional question is that whether or not the surjective ring homomorphism $K[\Omega] \longrightarrow \operatorname{No}$ of NBG is first-order definable in ZFC is independent of ZFC and in fact is equivalent to $V$ = HOD, because of Hamkins' answer to a previous question here: mathoverflow.net/questions/93468/… $\endgroup$ Dec 1, 2014 at 0:35

1 Answer 1


The transcendence degree of the surreals over the ordinals is proper class sized. There are many ways of seeing this; here's one. Recall that there is an order-preserving map $x\mapsto \omega^x$ from the surreals to the positive surreals which extends the usual ordinal exponentiation and satisfies $\omega^{x+y}=\omega^x \omega^y$, and the numbers of the form $\omega^x$ are linearly independent over $\mathbb{Q}$ (or $\mathbb{R}$). Now consider the elements $\omega^{\omega^{-x}}$ for all positive surreals $x$. It is easy to see that these are algebraically independent over the ordinals.

  • $\begingroup$ How does the proof of independence go? Are they independent over the composite of $\mathbb{R}$ and $K(\Omega)$? $\endgroup$ Dec 1, 2014 at 0:09
  • $\begingroup$ @JesseElliott Assuming by 'the composite' of $\mathbb{R}$ and $K(\Omega)$ you mean the subfield of the surreals whose Conway normal form has exponents in $K(\Omega)$, the transcendence degree is still proper class sized. I'm not sure the example Eric gives above is sufficient off the top of my head (it probably is but I'd have to think about it), consider the elements $\omega^{\omega^{-\sqrt{x}}}$ for all positive surreals $x$ -- there are still a proper class, and they're algebraically independent over the above defined 'composite'. I am currently looking into how to use the Galois (cont.) $\endgroup$
    – Alec Rhea
    Feb 6, 2019 at 7:05
  • $\begingroup$ @JesseElliott theory of Borceux and Janelidze to categorically construct the surreals from $K(\Omega)$, but it is more involved than standard Galois theory and requires dropping all standard Galois assumptions on the extension. $\endgroup$
    – Alec Rhea
    Feb 6, 2019 at 7:06

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