Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addition?

Question

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.)}

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.