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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.)

Todd Trimble
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Mike Battaglia
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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.

Will Sawin
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    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 http://www.ohio.edu/people/ehrlich/AHES.pdf, which I've added to my question for clarity's sake.

     – Mike Battaglia Jul 25 '12 at 04:22
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    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 $x$. – Will Sawin Jul 25 '12 at 14:00
  • Will, are you talking about the semiring of actual ordinals, or the ring constructed from the Grothendieck group I mentioned above?

    In either case, I'm not entirely sure how to interpret your reasoning here. How do I interpret the quotient of either one of these structures by $\omega^2$? It's clear that $\omega^3$ would map to $\omega^1$ under the quotient, and that $\omega^n$ for any natural number would map to n mod 2 under the quotient. What about $\omega^\omega$, what does that map to?

     – Mike Battaglia Jul 26 '12 at 00:33
  • The reason I find your analogy with $Z[x]/x^2$ to be confusing is that the exponent in $Z[x]$ must be a natural number, whereas the exponents involved with ordinal numbers are themselves ordinals. Another way to formulate my above question is to ask what $\omega$ mod 2 is. – Mike Battaglia Jul 26 '12 at 01:55
  • Actually they all map to $0$ because they are all multiples of $\omega^2$. Every ordinal is a number plus $\omega$ times a number plus an element of the ideal $(\omega^2)$. Same for the grothendieck group. – Will Sawin Jul 26 '12 at 02:16
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    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. – Mike Battaglia Jul 27 '12 at 09:02
  • Should say semiring* of ordinals above. – Mike Battaglia Jul 27 '12 at 09:02
  • Good point. Take the ideal generated by $\omega^n$ for all ordinals $n>1$. – Will Sawin Jul 27 '12 at 14:16
  • I think one immediate problem is that the rules of "natural exponentiation" are a bit hazy; in fact, I'm not even sure if Hessenberg defined such an operation. So this is a good thing to realize, I guess, because it hadn't struck me when I made this post just how tied into the nuts and bolts of that operation this question is. Things certainly get strange if you try to define it $\omega^\omega$ as "a countably infinite product of $\omega$'s", for instance. – Mike Battaglia Jul 27 '12 at 23:12
  • I've been thinking about this issue again lately, and coming back to this 7 months later, I'm still not sure if this puts the matter to rest. The reasoning laid out here initially was that if you start with $K(\mathbf{On})$, you can sort of treat it as $\mathbb{Z}[x]$ and mod by $(x^2)$, showing that no $\sqrt(x)$ remains. As above, that won't work, because $\omega^\omega$ isn't in the ideal generated by $\omega^2$. – Mike Battaglia Mar 23 '13 at 02:10
  • The suggestion was the instead to mod by the ideal generated by $\omega^n$ for all ordinals $n > 1$. But that's definitely not a trivial statement - this has to be one of the strangest ideals in all of ring theory. There are things like ordinal-linear combinations of measurable cardinals, minus a few Mahlo cardinals each to the power of aleph-aleph-aleph-three, and if you're lucky you can then perhaps divide by the Church-Kleene ordinal and subtract the first uncountable ordinal, squared. (cont'd) – Mike Battaglia Mar 23 '13 at 02:18
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    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. – Will Sawin Mar 23 '13 at 02:27
  • (contd) Or maybe not, since ZFC isn't strong enough to say if large cardinals exist, and hence if these elements are in the ring to begin with (and if they are, whether they'd be in the ideal you mention). So while it's true that $\sqrt{\omega}$ isn't in $K(\mathbf{On})$ iff the quotient by this enormous ideal doesn't contain $\sqrt{\omega}$, the foundational issues here leave a lot of lee-way for what exactly is in that ideal. That being said, I wonder if the existence of $\sqrt{\omega}$ could be undecidable. – Mike Battaglia Mar 23 '13 at 02:30
  • Will: I see I missed a post of yours mid-type. I'll deal with the divisibility thing in a second, but first, I don't see how you're expecting Cantor normal form to work here, either with the Grothendieck group or the commutative operations. Perhaps the easiest way to demonstrate this is to give you the example of the ordinal $… + \omega^2 + \omega + 1$, where the exponents in the sum go through all natural numbers. Now say that we subtract the ordinal $… + \omega^52 + \omega^32 + \omega*2 + …$ from it. We're left with the ordinal $… + \omega^4 - \omega^3 + \omega^2 - \omega + 1$. – Mike Battaglia Mar 23 '13 at 03:22
  • What element in the ring is this - is it positive or negative? And also, if you expect that multiplication distributes over addition, then multiplication by $\omega + 1$ gives you $1$. I didn't realize this when I made my initial post, but I'm not sure if this means that somehow, despite $\mathbf{On}$ being well-ordered, that $K(\mathbf{On})$ isn't - or if Cantor normal form doesn't extend as well to the Grothendieck group as you might expect. – Mike Battaglia Mar 23 '13 at 03:29
  • Typo - that's $… + \omega^52 + \omega^32 + \omega*2$ we're subtracting; the ellipsis on the right hand side should have been gone. But you get the point; we end up with $\sum_{n\in\mathbb{N}} \omega^n(-1)^n$. So something's gotta give: either the ring isn't an ordered ring (which I think it should be) or the Grothendieck group + commutative operations don't play nice with the Cantor normal form. – Mike Battaglia Mar 23 '13 at 03:34
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    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. – Will Sawin Mar 23 '13 at 03:46
  • Sorry, I'm getting all screwed up here. You're talking about Cantor normal form, and my example uses Conway normal form. OK, it all makes sense now. – Mike Battaglia Mar 23 '13 at 04:46
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    I posted a follow-up question to this at http://mathoverflow.net/questions/188430/transcendence-degree-of-the-surreals-over-the-subfield-generated-by-the-ordinals – Jesse Elliott Nov 30 '14 at 09:16
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    @MikeBattaglia "I think one immediate problem is that the rules of "natural exponentiation" are a bit hazy; in fact, I'm not even sure if Hessenberg defined such an operation." Related: https://math.stackexchange.com/questions/3447931/hessenberg-power-of-ordinals (see comments under the question). – user76284 May 16 '20 at 20:02