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The Surreal nummbers, $\boldsymbol{No}$, are according to Wikipidia the biggest ordered field, and the Surrcomplex numbers are the biggest field of characteristic 0. Biggest in the sense that every field (which is a set) is imbedded inside.

In light of that I was wondering whether there is a nice characterization of the biggest field of characteristic $p > 0$?

As Wojowu noted, this isn't unique to the Surreals or to $On_p$. The question then becomes is there a minimal such field?

Adi Ostrov
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    Don't those references concern the biggest ordered field, and the biggest algebraically closed field of characteristic zero? – Gerry Myerson Jan 26 '19 at 20:51
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    @GerryMyerson Well, every field embeds into an algebraically closed field. – Wojowu Jan 26 '19 at 21:14
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    @GerryMyerson Yes, but since every field has an algebraic closure, every field of char. zero can be embedded into $\mathbf{No}[i]$ as well. – Johannes Hahn Jan 26 '19 at 21:14
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    It would be a good idea to understand what is meant by "biggest", to make these quotations and this question meaningful. – YCor Jan 27 '19 at 00:53
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    The wording in the link (which maybe should be changed to match the wording? currently it goes to wikiwand, a Wikipedia viewer, rather than Wikipedia itself) seems to imply some idea of "terminal"; however, that isn't true (the field maps are nowhere near unique). – user44191 Jan 27 '19 at 04:33
  • The wikipedia link should be: https://en.wikipedia.org/wiki/Surreal_number#Arithmetic_closure – none Jan 27 '19 at 07:08

2 Answers2

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Conway's nimbers form an interesting answer for $p=2$. That every Field of characteristic $2$ embeds into it follows from the fact they form an algebraically closed Field and that they contain arbitrarily large sets of algebraically independent elements (which is immediate because the Field is proper-class-sized).

This has been generalized by DiMuro to arbitrary $p>0$ in his paper "On $\mathrm{On}_p$", see here. The resulting Field again is algebraically closed, and every characteristic $p$ Field embeds into it for the same reasons as above. The construction is more artificial than that of nimbers, but still.

By the way, let me make a remark: all the Fields mentioned in the thread, that is surreals, surcomplexes, nimbers and $\mathrm{On}_p$, are not unique maximal fields, even up to isomorphism. Indeed, for instance, taking the field of rational functions $\mathrm{On}_p(x)$, it is clearly not algebraically closed, hence not isomorphic to $\mathrm{On}_p$, yet every characteristic $p$ Field will embed into it (since they embed into $\mathrm{On}_p$). This does not change the answer, but I thought it's worth emphasizing.

To address the newly added question and the comment below, I do not know of any sense in which $\mathrm{On}_p$ is minimal - it contains proper subFields which also have the property that other Fields embed into it. However, as the other answer says, this Field is unique (up to isomorphism) characteristic $p$ Field with this property which is algebraically closed (indeed, this is a unique algebraically closed Field in characteristic $p$). This also is true of surcomplexes in characteristic $0$. I don't know of a corresponding description for surreals - there are many real closed fields of given cardinality, but see Alec Rhea's comment.

Wojowu
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    Maybe we can say that $On_p$ is minimal in the collection of (class-sized) fields $E$ such that every characteristic-$p$ field embeds into $E$? – Qfwfq Jan 27 '19 at 00:33
  • @Qfwfq In what sense would it be minimal? There are proper subFields of $On_p$ which also have this property. – Wojowu Jan 27 '19 at 07:29
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    The surreals are (up to isomorphism) the real-closed field underlying the maximal hyperreal number system in NBG set theory (see Philip Ehrlich's answer here https://mathoverflow.net/questions/91646/surreal-numbers-vs-non-standard-analysis), but this is (of course) different than being maximal in the sense that all other ordered fields embed into it. – Alec Rhea Jan 27 '19 at 07:42
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    Philip Ehrlich also gives a charaterization: No is the unique real-closed and set-saturated ordered field. But in any case, it would be better to call "universal" (or even "weakly universal" since the embeddings are not unique) such fields in which any other field in the category embeds. Or at least to say "a biggest" rather than "the biggest". – nombre Jan 27 '19 at 11:40
  • Is there an algebraically closed char-p Field $E\subset On_p$ (strict inclusion) with the property that every a.c. char-p field $k$ embeds $k\subset E$? – Qfwfq Jan 27 '19 at 12:28
  • Is it correct to say the following? $On_p$ is an a.c. char-p Field that is universal [or weakly universal, as per nombre's suggestion] for embeddings of a.c. char-p fields (i.e. every latter field embeds into On_p), but it is neither maximal nor minimal among universal a.c. char-p Fields. – Qfwfq Jan 27 '19 at 12:32
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    @Qfwfq There are such proper subFields of $On_p$, but they are necessarily isomorphic to $On_p$. To the latter question, this depends on how exactly you define maximal/minimal - the answer is "no" in the sense that $On_p$ can be embedded into as a proper subField into other such Fields, and it has proper subFields with the same property. The answer is "yes" in the sense that every such subField or superField must be isomorphic to $On_p$. – Wojowu Jan 27 '19 at 12:49
  • Very clear, thank you – Qfwfq Jan 27 '19 at 12:51
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An algebraically closed field is determined up to isomorphism by its characteristic and its transcendence degree over its prime field. So every algebraically closed field of characteristic $p$ is isomorphic to the algebraic closure of $\mathbb{F}_p(X)$, where $X$ is some set of variables.

This suggests that the "biggest field of characteristic $p$" should be constructed in the same way, but with a proper class of variables. e.g. the algebraic closure of the field of rational funtions over $\mathbb{F}_p$ in variables $(x_\alpha)_{\alpha\in \text{Ord}}$.

Under global choice, this is the unique proper class sized algebraically closed field of characteristic $p$ up to isomorphism.

It's up to you whether you view this as a "nice characterization".

As an aside: it's quite common in model theory to consider a proper class sized "monster model" for any complete first-order theory $T$. The monster model has the property that every set sized model of $T$ embeds into it elementarily. So the construction of monster models answers the analogue of your question for any complete first-order theory $T$.

Alex Kruckman
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