This is a question in two parts.

Say that $\mathbf{On}$ is the proper class of all ordinal numbers in ZFC. We can define a binary operator over $\mathbf{On}$ which corresponds to the commutative version of ordinal addition; this has been called "Hessenberg addition" and "natural addition" before. It's also the operation you get by restriction of the $+$ operation from Conway's surreals to the subchain of ordinals (e.g. surreals with empty right set). I'll use the $+$ symbol for this operation over the ordinals.

$\langle\mathbf{On},+\rangle$ is a commutative monoid, which hence admits the notion of constructing a Grothendieck group $\mathrm{K}(\mathbf{On})$. The group $\langle\mathrm{K}(\mathbf{On}),+\rangle$ hence adds expressions such as $\omega$, $\omega-1$, $\omega^\omega - \omega^2 + 5$, etc. to the ordinals.

  • Question 1: is $\mathrm{K}(\mathbf{On})$ equivalent to Conway's "omnific integers" $\mathbf{Oz}$? In Conway's "On Numbers and Games," he defines an omnific integer $x$ as one which can be represented as a surreal number $\left \{ x-1 \mid x+1 \right \}$. Are these two classes isomorphic to one another?

It's also noteworthy that the field of fractions $Quot(\mathbf{Oz})$ is the full field $\mathbf{No}$ of surreal numbers. We can further turn $\mathrm{K}(\mathbf{On})$ into a ring $\langle\mathrm{R}(\mathbf{On}),+,\times\rangle$ by defining a new commutative operation called $\times$, called the "Hessenberg product", "Hausdorff product" or "natural product" of ordinals, which is commutative, associative, has an identity of 1, and distributes over the Conway normal form of the ordinal. A good definition for the Hessenberg product can be found on pages 24-25 of Ehrlich 2006.

  • Question 2: even if $\mathrm{K}(\mathbf{On})$ isn't isomorphic to $\mathbf{Oz}$, is $Quot(\mathrm{R}(\mathbf{On}))$ isomorphic to $\mathbf{No}$?

I'm tempted to answer in the negative for #1, as $\sqrt{\omega}$ is in $\mathbf{Oz}$, but is it in $\mathrm{R}(\mathbf{On})$? That is, given $\mathrm{K}(\mathbf{On})$ and ordinary commutative multiplication, is it the case that $\omega$ becomes a perfect square?

(Also, a last note - I'm aware that $\mathbf{On}$ is a proper class. I'm not sure what foundational issues arise specifically in the above question, but I don't care how you want to handle them - NBG set theory, Grothendieck universes, whatever.)


1 Answer 1


There is an obvious extension of Cantor normal form to the Grothendieck group of the ordinals. Then the standard argument that $\sqrt{x}$ does not lie in the ring $\mathbb Z[x]$ applies to $\sqrt{\omega}$. Specifically, $\sqrt{\omega}$ must have a Cantor normal form $a + b \omega + $ higher-order terms, which squares to $a^2 + 2ab \omega +$ higher-order terms. For this to equal $\omega$ we need $a^2=0$ but $2ab=1$, which is of course impossible.

So your first question has a negative answer.

For your second, the equation $a^2=b^2\omega$ is equally problematic. Again apply the standard argument:

Write $a=k \omega^x +$ higher-order terms, and $b=l \omega^y + $ higher-order terms. Then the lowest term of $a^2$ and $b^2\omega$ must be equal, so $\omega^{x+_H x}=\omega^{y+_H y+_H 1}$, so $x+_H x = y+_H y+_H 1$, which cannot be because the $1$s coefficient of the first expression must be even while the $1s$ coefficient of the second expression must be odd.

  • 3
    $\begingroup$ Hi Will, thanks for the answer. I'd considered that, but it still seemed to me that there might be some way that some extremely strange number which is some expression of ordinals might end up somehow squaring to $\omega$. Perhaps it really is as simple as you wrote, even despite the weird proper class stuff going on. As for the second part, natural multiplication is commutative, associative, has 1 as identity, and distributes over the normal form. Ehrlich has a good definition on pages 24-25 of ohio.edu/people/ehrlich/AHES.pdf, which I've added to my question for clarity's sake. $\endgroup$ Jul 25, 2012 at 4:22
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    $\begingroup$ But there isn't "some expression of ordinals". The Grothendieck group contains only differences of ordinals. Consider the ideal in the ring of ordinals generated by $\omega^2$. In this ideal, all possible Cantor normal form terms but $1$ and $\omega$ vanish. Thus, the quotient is isomorphic to $Z[x]/x^2$, which does not contain a square root of $x$. $\endgroup$
    – Will Sawin
    Jul 25, 2012 at 14:00
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    $\begingroup$ In what sense is $\omega^\omega$ a multiple of $\omega^2$? What times $\omega^2 = \omega^\omega$? $\omega^{(\omega-2)}$ isn't in the ring of ordinals, and it's also not in the Grothendieck group either. $\endgroup$ Jul 27, 2012 at 9:02
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    $\begingroup$ First, division is irrelevant. The key fact is whether the equation $a^2 = b^2 \omega$ has a solution among differences of ordinals. We can check that it does not using Cantor normal form. One is concerned merely with the lowest terms, so strange set-theoretic objects are irrelevant. $\endgroup$
    – Will Sawin
    Mar 23, 2013 at 2:27
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    $\begingroup$ The ordinals you gave me aren't in Cantor normal form. I'm not sure how to interpret that notation. Why do you expect those limits to make sense? This is just another version of the Mazur's swindle argument that infinite associativity cannot coexist with nontrivial cancellation. But my argument, does not depend on infinite associativity - only finite associativity. $\endgroup$
    – Will Sawin
    Mar 23, 2013 at 3:46

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