6

John Conway proved in his book, On Numbers and Games (ch6, theorem 49) that the set of all ordinals smaller than $\omega^{\omega^\omega}$ form a field of characteristic 2 that is isomorphic to $\bar{\mathbb{F}_2}$, the algebraic closure of $\mathbb{F}_2$. The arithmetic of this field is described in more detail in Lenstra's note, On the Algebraic Closure of Two Additionally, there's another good note on this field at lieven le bruyn's blog, neverendingbooks

In doing so, Conway also proved that the finite numbers that form fields in the same way are of the form $2^{2^k}$. Let $[n]$ denote one of such fields (exempli gratia, $[4]=[2^2] ={0,1,2,3}$, the set of all smaller ordinals. Hence we have all extensions $[2^{2^n}]$ contained in the algebraic closure.

Is there a choice (and explicit construction) of monic irreducible polynomials $m_n(x)$ of degree $n$ so that the field isomorphism $$ \mathbb{F}_2[x]/(m_n(x)) \cong [2^n] $$ is natural for all finite $n=2^k$?

I suppose I need to say what I mean by "natural". Let me try to explain, and please let me know if this makes no sense: We can always construct such an isomorphim $\phi$, if we choose a root $\mu$ of $m_n(x)$ and a corresponding $\phi(\mu) \in \{3, 4, \ldots, n\}$, then the axioms of fields and isomorphisms will force the other choices (I think). I want a set of $m_n(x)$'s so that the isomorphism doesn't depend on any arbitrary choices I make. (I really don't know how to make precise what I mean by natural. Suffice to say that I'm asking if there is a preferred set of polynomials that make the isomorphism obvious!)

Update* Of course, other fields, $\mathbb{F}_{2^k}$ for k not a power of two, must also live inside $\omega^{\omega^\omega}$. As already noted, $[\omega]$ is a quadratic closure. Conway's method is then to take a cubic closure, by noting that $[\omega^3], [\omega^9], \ldots$ are also fields; and then taking a quintic closure through the fields $[\omega^\omega], [\omega^{5 \omega}], \ldots$, et cetera. Hence a field like $\mathbb{F}_8$ can be embedded into one of the cubic fields, exempli gratia $[\omega^3]$. Can we describe how $\mathbb{F}_{2^k}$ fits into these fields, and can we also find an irreducible polynomial that gives us a natural embedding?

*Perhaps this should be a separate question? Let me know...

Vincent
  • 10,614
mebassett
  • 1,290
  • 1
    Isomorphic how? One set is naturally well-ordered and the other cannot be linearly ordered in coherence with its field structure. – Asaf Karagila Jul 04 '12 at 23:17
  • 4
    The field with 8 elements is not an extension field of the field with 4 elements. –  Jul 04 '12 at 23:19
  • @AsafKaragila field isomorphisms here. My suspicion from the literature, actually, is that there isn't any preferred field extension, but I wanted to ask the clever folks on here before I run off with that conclusion. – mebassett Jul 04 '12 at 23:20
  • @Hurkyl a very good point! I'll correct that. – mebassett Jul 04 '12 at 23:21
  • But the ordinals do not make a good field. Their addition and multiplication is not commutative is hardly invertible. How do you define the addition on the ordinals? – Asaf Karagila Jul 04 '12 at 23:23
  • @AsafKaragila See Conway's book or Lenska's note for that, or even this link: http://www.neverendingbooks.org/index.php/on2-conways-nim-arithmetics.html

    the addition is binary arithmetic (for finite numbers) without carrying the extra digit. multiplication is a lot more complicated.

     – mebassett Jul 04 '12 at 23:29
  • You may want to add that link to the post. – Asaf Karagila Jul 04 '12 at 23:32
  • 2
    +1: Thanks for telling me about this! I didn't know about this construction by Conway (even though I'm relatively familiar with finite NIM-addition). Playing with the multiplication rule a bit quickly leads to the observation that with finite ordinals you only get the fields $F_k=GF(2^{2^k})$ as the intervals $F_k={0,1,\ldots,2^{2^k}-1}$. This family of finite fields has other recursive constructions (trying to reformulate one in a way that helps here). So for example the unique field of 8 elements has the elements 0,1 and then 6 infinite ordinals (that I dare not try and guess yet:-) – Jyrki Lahtonen Jul 05 '12 at 07:22

0 Answers0