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I'm wondering if there's been any work done on prime factorizations of Omnific integers as products of prime Omnific integers.

I suspect that each Omnific integer has a unique prime factorization, and that this can be lifted from the fact that every ordinal has a unique decomposition into other prime ordinals with respect to Hessenberg multiplication (not recursive multiplication) -- I was just wondering if any work had been done on this subject that perhaps delved a bit deeper than this surface level observation.

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  • $\begingroup$ Isn't $\omega$, $\omega^{1/2}$, $\omega^{1/4}$, $\omega^{1/8}$, ... an infinite descending chain of divisors in omnific integers? Are there any prime omnific integers (other than usual integers) in the first place? $\endgroup$ Commented Jun 19, 2017 at 9:07
  • $\begingroup$ In An introduction to the theory of surreal numbers, Harry Gonshor does a quick study of arithmetic properties of $Oz$. He does not find any non trivial (i.e. non integer) prime omnific integer, and proves some results giving little hope that they exist. $\endgroup$
    – nombre
    Commented Jun 19, 2017 at 9:22
  • $\begingroup$ I believe that there is a correspondence between Omnific Integers and cantor normal forms with possibly negative integer coefficients (think Grothendieck group over $O_n$ with Hessenberg operations) -- I'll be publishing a paper in the next couple of weeks that clearly elucidates this relationship, however these normal forms factor much the same as polynomials in base $\omega$. If $\omega^{\frac{1}{2}}$ is an Omnific integer however, I have no idea what normal form this would correspond to -- perhaps I am mistaken that every Omnific integer corresponds to such a normal form. $\endgroup$
    – Alec Rhea
    Commented Jun 19, 2017 at 13:33
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    $\begingroup$ But the Grothendick group of the ordinals under natural arithmetic is not isomorphic to the underlying group of $Oz$ which for instance is "almost divisible" in the sense that for $z \in Oz$ and $n \in \mathbb{Z} - \{0\}$, there is $z' \in Oz$ and $m \in \{0;...n-1\}$ such that $n.z' = z+m$. The other group is not. So you can't expect the rings to be isomorphic. $\endgroup$
    – nombre
    Commented Jun 19, 2017 at 21:47
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    $\begingroup$ Indeed, Oz is an integer part of No: every surreal number is within distance 1 of an omnific integer. And No is the fraction field of Oz. $\endgroup$ Commented Jun 20, 2017 at 12:03

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