I’d like to explain that one may find numerous nonstandard models of arithmetic as substructures of the field of complex numbers.

The issue arose yesterday at Hans Schoutens’s talk for the CUNY Logic Workshop. The main focus of the talk was the question, for a given algebraically closed field $k$ of characteristic zero and a given model of arithmetic $\Gamma\models$PA, whether $\Gamma$ and $k$ were jointly realizable as the set of powers (as he defines it) and the set of units of a model $S$ of the generalized theory of polynomial rings over fields. Very interesting stuff.

During the talk, a side question arose, concerning exactly which models of PA arise as substructures of the field of complex numbers.

**Question.** Which models of PA arise as substructures of the field of complex numbers $\langle\mathbb{C},+,\cdot\rangle$?

Of course the standard model $\mathbb{N}$ arises this way, and some people thought at first it should be difficult to realize nonstandard models of PA as substructures of $\mathbb{C}$. After some back and forth, the question was ultimately answered by Alfred Dolich in the pub after the seminar, and I’d like to give his argument here. This is a case where a problem that was initially confusing becomes completely clear!

**Theorem.** Every model of PA of size at most continuum arises as a sub-semiring of the field of complex numbers $\langle\mathbb{C},+,\cdot\rangle$.

**Proof.** Suppose that $M$ is a model of PA of size at most continuum. Inside $M$, we may form $M$’s version of the algebraic numbers $A=\bar{\mathbb{Q}}^M$, the field that $M$ thinks is the algebraic closure of its version of the rationals. So $A$ is an algebraically closed field of characteristic zero, which has an elementary extension to such a field of size continuum. Since the theory of algebraically closed fields of characteristic zero is categorical in all uncountable powers, it follows that $A$ is isomorphic to a submodel of $\mathbb{C}$. Since $M$ itself is isomorphic to a substructure of its rationals $\mathbb{Q}^M$, which sit inside $A$, it follows that $M$ is isomorphic to a substructure of $\mathbb{C}$, as claimed. **QED**

In particular, every countable model of PA can be found as a substructure of the complex numbers.

Essentially the same argument shows the following.

**Theorem.** If $k$ is an uncountable algebraically closed field of characteristic zero, then every model of arithmetic $M\models$PA of size at most the cardinality of $k$ embeds into $k$.

I’ve realized that the same collection of ideas shows the following striking way to look upon the complex numbers:

**Theorem.** The complex numbers $\mathbb{C}$ can be viewed as a nonstandard version of the algebraic numbers $\bar{\mathbb{Q}}^M$ inside a nonstandard model $M$ of PA. Indeed, for every uncountable algebraically closed field $F$ of characteristic zero and every model of arithmetic $M\models$PA of the same cardinality, the field $F$ is isomorphic to the nonstandard algebraic numbers $\bar{\mathbb{Q}}^M$ as $M$ sees them.

**Proof.** Fix any such field $F$, such as the complex numbers themselves, and consider any nonstandard model of arithmetic $M$ of the same cardinality. The field $\bar{\mathbb{Q}}^M$, which is $M$’s nonstandard version of the algebraic numbers, is an algebraically closed field of characteristic zero and same uncountable size as $F$. By categoricity, these fields are isomorphic. $\Box$

Nice! Logicians and algebraists talk a bit differently so I’d like to check some things.

1) I’m pretty sure that “a submodel of C”, in this context, is just a subfield of C.

2) I’m getting confused about whether “A is an algebraically closed field of characteristic zero, which has an elementary extension to such a field of size continuum” means the same thing as “A is an algebraically closed field of characteristic zero, which has an extension to such a field of size continuum.” The latter is true, in any case.

I think one can avoid some scary logician words like “submodel”, “elementary extension” and “categorical” by saying this: “any algebraically closed field of characteristic zero whose cardinality is less than or equal to that of C is isomorphic to a subfield of C”. I’m forgetting if the proofs of this feel more like logic or like algebra.

All your comments are fine. Since the theory of algebraically closed fields of characteristic zero admits elimination of quantifiers, every subfield is an elementary submodel, and every field extension among such fields is an elementary extension.

To my way of thinking, one shouldn’t omit the word categorical, since it is precisely this fact that unifies so much of our knowledge in this case. The theory of algebraically closed fields of characteristic zero is categorical in every uncountable cardinality, which means that there is up to isomorphism precisely one such field of each uncountable cardinality. This explains why any field of characteristic zero and size at most continuum is isomorphic to a subfield of $\mathbb{C}$.

Regarding your specific numbered comments, for (1), in general, a submodel means a substructure satisfying the same theory in question. In this case, it would mean not just a subfield, but an algebraically closed subfield. But in fact, one needs only subfield for the argument to work. For (2), those are equivalent precisely because of the quantifier elimination argument, and so amongst the algebraically closed fields, a field extension is the same as an elementary extension.

Thanks for all the clarifications! I agree that the uncountable categoricity of the theory of an algebraically closed field of characteristic zero is the engine that drives everything here. However, it’s a sad fact that lots of seemingly well-educated algebraists do not know the term “categorical” but will understand what one means by “two uncountable algebraically closed fields are isomorphic iff they have the same cardinality”. You’re running a logic blog so you shouldn’t resort to such baby-talk, but to reach a maximal audience I think one needs to… so that’s what I did on my blog. In math, alas, even popularizations require further popularization.

Thanks so much for your blog post; I appreciate it very much. I’m glad that my blog has such distinguished readers as yourself and your knowledgeable followers. It is amazing to me that someone’s observation in a pub after a seminar can subsequently reach such a huge number of interested parties.

Yes, despite the problems with quick electronic communication (which have been much in the news), it’s great how it speeds the spread of interesting mathematical and scientific knowledge. Your blog is great for feeding my hobbyist interest in mathematical logic, and when I hear something cool I want to tell everyone. And so it goes….

How much information does retain about the model ? Could it be interesting to study nonstandard models of the natural numbers using these fields?

The topic of Hans Schoutens talk was in part the idea of looking at $M$’s version of the polynomial ring $\bar{\mathbb{Q}}^M[x]$. So this field does not have only standard-finite polynomials, but nonstandard polynomials as well. And he observed that $M$ is definable inside this nonstandard polynomial ring as the semi-ring of “exponents”, corresponding to the powers $x^n$ for $n\in M$. These polynomials are definable in that ring. It is a really interesting perspective that he had about it.

Neat!

Pingback: Nonstandard Integers as Complex Numbers | Azimuth

By the way, the usual definition of ‘characteristic zero’ for fields says that

So, if we have a nonstandard model of the natural numbers, we get a nonstandard notion of ‘characteristic zero’. Does that affect anything you’re talking about? You’re using facts about algebraically closed fields.

Can there be fields that have characteristic zero if we work with the standard model of PA but finite characteristic if we use a nonstandard model? (I’m not sure what this question even means, but I can’t resist asking it.)

Regarding my last question, I guess we can try where is a nonstandard prime. This should have nonstandard characteristic.

What do people know about nonstandard primes? I guess in a nonstandard model there have to exist nonstandard primes, because none of the standard primes is ‘big enough’ to obtain a prime factorization for a nonstandard natural number.

It looks like we made the same observation at the same time! (see my next comment).

PA proves that there are infinitely many primes (since the usual proof can be undertaken in PA), or in other words, that they are unbounded, and so every nonstandard model of PA will definitely have lots of nonstandard primes. Indeed, there is always a prime between $n$ and $2n$, even for nonstandard $n$, since that fact is provable in PA.

You can consider a nonstandard version of $\mathbb{Z}/p\mathbb{Z}$, where $p$ is a nonstandard prime. For example, such a field can arise as a ultrapower of finite fields $\prod_p F_p$. This field has characteristic zero, in the usual meaning of this phrase, but it has nonstandard finite characteristic inside the nonstandard model in which it is constructed.

John and Joel,

It can also be interesting to study the Sylow p subgroups of a hyperfinite extension of a locally finite group, where p is an infinite hyperfinite prime hyperinteger. I looked at this a bit in my dissertation (1989, UH, “Nonstandard Methods and Finiteness Conditions in Algebra”). I never published the work I did on groups and rings in my dissertation. Instead, I took an interest in applying nonstandard methods to universal algebra, and lattices. Then I started my job here in Rolla Missouri, and I’ve had other collaborations since then, but I did extend some ring and field theoretic work from my M.S. thesis to topological models in a paper in Studia Logica. Back when I wrote my M.S. thesis, I had applied nonstandard methods to the Primitive Element Theorem from the theory of fields.

I’m curious what those results would inspire now in light of this result from the CUNY Logic Workshop…

Also in relation to this, if we fix a submodel A of the field of complex numbers that is a nonstandard model of PA, and let F be the subfield of C that is generated by A. It seems to me that F supports then a definable total order < that is compatible with its arithmetic (definable from the arithmetic in A?), and F is nonarchimedean relative to this order <, so that F includes infinitesimal elements. Is it clear what these infinitesimal elements look like, as elements of C?

Matt Insall

I really like your idea at the end about looking at the induced order on the field generated by the nonstandard model of PA. It seems to me that this field must be isomorphic to the rationals of the PA model, and the order will be definable from the arithmetic of the PA model.

As for what the infinitesimals “look like” as elements of $\mathbb{C}$, I believe that one can show they can be any desired element that is transcendental over $\mathbb{Q}$. The reason is that when embedding fields into $\mathbb{C}$, you map a transcendence base to a transcendence base and extend.

Commentationes Mathematicae Universitatis Carolinae

14,3 (1973)

A REPRESENTATION OP MODELS OP PEANO ARITHMETIC

J. MLCEK, Praha

Abstract: The following theorem is proved: algebraically Closed field of char. 0 is saturated if and only if every countable model of Peano arithmetic can be embedded into it.

Key words: Peano arithmetic, algebraically closed

field of char. 0 , saturated model, embedding.

AMS, Primary: 02H15 Ref. Z. 2.666

Introduction. In this paper, we shall present some results on embeddability of countable models of Peano arithmetic P into models of algebraically closed fields of characteristics 0 .

This set-theoretical result should be compared with(and was inspired by) a recent result of Vopenka saying that (under reasonable assumptions on existence of semisets) each countable model of P can be embedded into the field of real numbers by a semi-set embedding.

https://dml.cz/bitstream/handle/10338.dmlcz/105508/CommentatMathUnivCarol_014-1973-3_11.pdf

Thanks very much for this!

Harvey Friedman has pointed out his 1992 article with Robert K. Meyer, see http://www.jstor.org/tc/accept?origin=/stable/pdf/2275433.pdf?refreqid=excelsior%3A7e994c8b2e5e010c6337380cea6e2860.

Very interesting post.

Just to ensure I’m not missing anything: you wrote “we construct A, the field that M thinks is the algebraic closure of its version of the rationals.” By “M thinks” you just mean that A is the model constructed as the fraction-field of elements of M, not that A is somehow internal to M, right?

No, I had meant the internal one. For the fraction-field, this will be the same as what you think is the fraction-field, but for the “algebraic closure” part, the internal construction will be adding roots of nonstandard-size polynomials, and so this will be different from what an algebraist would do externally when taking the algebraic closure.

But actually, you can use either one and still make a successful argument, since they will all embed into the desired field, and the model of arithmetic $M$ sits inside its fraction-field.

Hi Joel, this is super interesting! I wonder, can we say anything about the (descriptive or otherwise) complexity of given model of $M \models PA$ sitting inside $\mathbb C$?

I ask because this seems relevant to the following open problem in models of PA: Recall that a Kaufmann model is an -like recursively saturated, rather classless model. Kaufmann constructed these using but then Shelah showed via a complicated absoluteness argument that such models exist in ZFC. The problem is to determine if one can actually build such a model without extra assumptions or whether the absoluteness is required in the proof. Since Kaufmann models are the right size, they embed into by this argument and I wonder if one can show that all such models, as subsets of , must outstrip any kind of natural complexity this would suggest a negative solution to the problem.

Hi Joel, the readers of your blog might be also interested in this related mathoverflow question(s) of mine, and the partial answers so far:

https://mathoverflow.net/questions/66146/nonstandard-reals-in-the-complex-plane

Thanks for the link, Ali. Very interesting questions!