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As is well known, each natural number (except $0$) can be written uniquely as product of finitely many prime numbers (with $1$ being the empty product). My question is: Does some analogue theorem also hold for ordinal numbers?

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Well, $\omega = 2 \times \omega$, so.... Also, unique prime factorization is up to order, and it's unclear what this means for ordinal multiplication, which is noncommutative. –  Qiaochu Yuan Aug 24 '12 at 7:51
Good point. But what if we add the condition that the numbers must be multiplied in decreasing order? For natural numbers, this would not make a difference, and for infinite numbers, I believe it should resolve this specific problem. Or alternatively one could use the natural (Hessenberg) product. –  celtschk Aug 24 '12 at 8:05
Well, $(\omega + 1) \times \omega = (\omega + 2) \times \omega$, so... –  Qiaochu Yuan Aug 24 '12 at 8:17
Oh, right. So even with the ordering requirement, for the normal ordinal product, it cannot work. However, what about the natural (Hessenberg) product? –  celtschk Aug 24 '12 at 9:03
This is a very interesting question! I believe the question you are asking is whether or not the ordinals form a UFD using natural sum and product. The ordinals below $\omega^\omega$ endowed with natural sum and product are isomorphic to polynomials with natural number coefficients, so they form a UFD. I believe this remains true for higher ordinals, but I don't have a proof. –  Deedlit Sep 23 '12 at 10:18
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1 Answer 1

up vote 1 down vote accepted

Theorem 3 of Shinpei Oka's On Telgárski's formula (online summary) offers an affirmative answer.

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Sierpinski's book Cardinal and Ordinal Numbers is pretty much the bible for this kind of stuff. I don't have my copy with me where I'm at right now to give any suggestions based on it, but since no one else has yet mentioned Sierpinski's book, I thought I'd mention it in case the OP doesn't know about it. –  Dave L. Renfro Jun 20 at 14:10
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