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](http://en.wikipedia.org/wiki/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](http://www.ohio.edu/people/ehrlich/AHES.pdf).


* **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.)*



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


