## Nonstandard Integers as Complex Numbers

• Joel David Hamkins, Nonstandard models of arithmetic arise in the complex numbers, 3 March 2018.

Let me try to explain it in a simplified way. I think all cool math should be known more widely than it is. Getting this to happen requires a lot of explanations at different levels.

Here goes:

The Peano axioms are a nice set of axioms describing the natural numbers. But thanks to Gödel’s incompleteness theorem, these axioms can’t completely nail down the structure of the natural numbers. So, there are lots of different ‘models’ of Peano arithmetic.

These are often called ‘nonstandard’ models. If you take a model of Peano arithmetic—say, your favorite ‘standard’ model —you can get other models by throwing in extra natural numbers, larger than all the standard ones. These nonstandard models can be countable or uncountable. For more, try this:

Nonstandard models of arithmetic, Wikipedia.

Starting with any of these models you can define integers in the usual way (as differences of natural numbers), and then rational numbers (as ratios of integers). So, there are lots of nonstandard versions of the rational numbers. Any one of these will be a field: you can add, subtract, multiply and divide your nonstandard rationals, in ways that obey all the usual basic rules.

Now for the cool part: if your nonstandard model of the natural numbers is small enough, your field of nonstandard rational numbers can be found somewhere in the standard field of complex numbers!

In other words, your nonstandard rationals are a subfield of the usual complex numbers: a subset that’s closed under addition, subtraction, multiplication and division by things that aren’t zero.

This is counterintuitive at first, because we tend to think of nonstandard models of Peano arithmetic as spooky and elusive things, while we tend to think of the complex numbers as well-understood.

However, the field of complex numbers is actually very large, and it has room for many spooky and elusive things inside it. This is well-known to experts, and we’re just seeing more evidence of that.

I said that all this works if your nonstandard model of the natural numbers is small enough. But what is “small enough”? Just the obvious thing: your nonstandard model needs to have a cardinality smaller than that of the complex numbers. So if it’s countable, that’s definitely small enough.

This fact was recently noticed by Alfred Dolich at a pub after a logic seminar at the City University of New York. The proof is very easy if you know this result: any field of characteristic zero whose cardinality is less than or equal to that of the continuum is isomorphic to some subfield of the complex numbers. So, unsurprisingly, it turned out to have been repeatedly discovered before.

And the result I just mentioned follows from this: any two algebraically closed fields of characteristic zero that have the same uncountable cardinality must be isomorphic. So, say someone hands you a field F of characteristic zero whose cardinality is smaller than that of the continuum. You can take its algebraic closure by throwing in roots to all polynomials, and its cardinality won’t get bigger. Then you can throw in even more elements, if necessary, to get a field whose cardinality is that of the continuum. The resulting field must be isomorphic to the complex numbers. So, F is isomorphic to a subfield of the complex numbers.

To round this off, I should say a bit about why nonstandard models of Peano arithmetic are considered spooky and elusive. Tennenbaum’s theorem says that for any countable non-standard model of Peano arithmetic there is no way to code the elements of the model as standard natural numbers such that either the addition or multiplication operation of the model is a computable function on the codes.

We can, however, say some things about what these countable nonstandard models are like as ordered sets. They can be linearly ordered in a way compatible with addition and multiplication. And then they consist of one copy of the standard natural numbers, followed by a lot of copies of the standard integers, which are packed together in a dense way: that is, for any two distinct copies, there’s another distinct copy between them. Furthermore, for any of these copies, there’s another copy before it, and another after it.

I should also say what’s good about algebraically closed fields of characteristic zero: they are uncountably categorical. In other words, any two models of the axioms for an algebraically closed field with the same cardinality must be isomorphic. (This is not true for the countable models: it’s easy to find different countable algebraically closed fields of characteristic zero. They are not spooky and elusive.)

So, any algebraically closed field whose cardinality is that of the continuum is isomorphic to the complex numbers!

For more on the logic of complex numbers, written at about the same level as this, try this post of mine:

The logic of real and complex numbers, Azimuth 8 September 2014.

### 22 Responses to Nonstandard Integers as Complex Numbers

1. Great post, John! Thanks for spreading the word.

One minor issue: you say that the nonstandard model of the natural numbers is a subfield of the complex numbers, but that isn’t quite right, since it isn’t a field, and not even a ring, but only a semi-ring (one can’t necessarily divide or even substract natural numbers). A model of PA is a discretely ordered semiring, and so one should say that it is a subsemiring of C. Of course, the argument proceeds by observing that the model of PA constructs what it thinks is the algebraic closure of its (nonstandard) version of the rationals, and this is a field that embeds into C, and has M as a subsemiring.

• John Baez says:

Joel wrote:

One minor issue: you say that the nonstandard model of the natural numbers is a subfield of the complex numbers,

Yikes, I said that??? Where???

Here I said “it” in a potentially ambiguous way:

Now for the cool part: if your nonstandard model of the natural numbers is small enough, your field of nonstandard rational numbers can be found somewhere in the standard field of complex numbers!

In other words, it’s a subfield of the complex numbers: a subset that’s closed under addition, subtraction, multiplication and division by things that aren’t zero.

But it would be uncharitable to read “it” as referring to the nonstandard natural numbers, which aren’t a field, rather than the resulting nonstandard rationals, which are.

Is that the passage you were referring to, or did I really come out somewhere and say the nonstandard natural numbers were a subfield of something?

• I’m very sorry! I must have misread it. Of course it is fine. (And please feel free to delete my comment if you like.)

• John Baez says:

No problem: it’s best to keep your comment around: what one person misreads, others do as well.

I’m getting increasingly suspicious of pronouns as I get older. I’m not sure how much of this is a subtle sign that my memory is getting worse, and how much of it is my gradual discovery that people expect other people to read their mind when guessing the antecendent of a pronoun. My recent readings on bound and free variables in the lambda calculus make me even more doubtful that people precisely understand the scope of any pronoun they’re using.

2. zenorogue says:

In non-standard rationals there is no number $x$ such that $x^2=-1$, thus no non-standard rational must be mapped to $i$. Isn’t it the case that non-standard rationals are actually a subfield of real numbers?

• John Baez says:

Maybe. But you seem to be arguing that any subfield of the complex numbers that doesn’t contain $i$ must be a subfield of the real numbers. That’s not true: take the smallest subfield of the complex numbers containing

$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$

which is a cube root of unity. This doesn’t contain $i$, because it consists of all numbers $p + \omega q$ where $p$ and $q$ are rational.

3. Blake Stacey says:

So, if you have your favorite model of the natural numbers, and I have my favorite, and neither of us can convince the other that our model should be accepted as the “standard”, we can at least agree to meet on the playing ground of the complex numbers?

• John Baez says:

Blake wrote:

we can at least agree to meet on the playing ground of the complex numbers?

Yeah, sort of. It’s true as long as we stick to the (very limited) ‘first-order theory’ where we write down the axioms of an algebraically closed field of characteristic zero, and only let ourselves write down statements where we quantify over elements of this, not subsets. This theory is ‘complete’, so we’ll never have questions that we can’t resolve. And it’s ‘uncountably categorical’, meaning that any two models of this theory are isomorphic if they have the same cardinality. So it’s very nice. But this is mainly because this theory is very limited: we can’t even define complex conjugation, or the real numbers, in this theory!

The logic of real and complex numbers, Azimuth 8 September 2014.

But, it’s so damn cool that I’m gonna quote myself at length!

### The complex numbers

I mentioned that when we’re studying an infinite mathematical structure using first-order logic, the best we can hope for is to have one model of each size (up to isomorphism). The real numbers are far from being this nice… but the complex numbers come much closer!

More precisely, say $\kappa$ is some cardinal. A first-order theory describing structure on a single set is called κ-categorical if it has a unique model of cardinality $\kappa.$ And 1965, a logician named Michael Morley showed that if a list of axioms is $\kappa$-categorical for some uncountable $\kappa,$ it’s $\kappa$-categorical for every uncountable $\kappa.$ I haven’t worked my way through the proof, which seems to be full of interesting ideas. But such theories are called uncountably categorical.

A great example is the ‘purely algebraic’ theory of the complex numbers. By this I mean we only write down axioms involving $+, \times, 0$ and $1.$ We don’t include anything about $\le$ this time, nor anything about complex conjugation. You see, if we start talking about complex conjugation we can pick out the real numbers inside the complex numbers, and then we’re more or less back to the story we had for real numbers.

This theory is called the theory of an algebraically closed field of characteristic zero. Yet again, the axioms come in three bunches:

• the field axioms.

• the characteristic zero axioms: these are an infinite list of axioms saying that

$1 \ne 0, \quad 1+1 \ne 0, \quad 1+1+1 \ne 0, \dots$

• the algebraically closed axioms: these say that every non-constant polynomial has a root.

Pretty much any mathematician worth their salt knows that the complex numbers are a model of these axioms, whose cardinality is that of the continuum. There are lots of different countable models: the algebraic complex numbers, the computable complex numbers, and so on. But because the above theory is uncountably categorical, there is exactly one algebraically closed field of characteristic zero of each uncountable cardinality… up to isomorphism.

This implies some interesting things.

For example, we can take the complex numbers, throw in an extra element, and let it freely generate a bigger algebraically closed field. It’s ‘bigger’ in the sense that it contains the complex numbers as a proper subset, indeed a subfield. But since it has the same cardinality as the complex numbers, it’s isomorphic to the complex numbers!

And then, because this ‘bigger’ field is isomorphic to the complex numbers, we can turn this argument around. We can take the complex numbers, remove a lot of carefully chosen elements, and get a subfield that’s isomorphic to the complex numbers.

Or, if we like, we can take the complex numbers, adjoin a really huge set of extra elements, and let them freely generate an algebraically closed field of characteristic zero. The cardinality of this field can be as big as we want. It will be determined up to isomorphism by its cardinality.

One piece of good news is that thanks to a result of Tarski, the theory of an algebraically closed field of characteristic zero is complete, and thus, all its models are elementarily equivalent. In other words, all the same first-order sentences written in the language of $+, \times, 0$ and $1$ hold in every model.

But here’s a piece of strange news.

As I already mentioned, the theory of a real closed field is not uncountably categorical. This implies something really weird. Besides the ‘usual’ real numbers $\mathbb{R}$ we can choose another real closed field $\mathbb{R}',$ not isomorphic to $\mathbb{R},$ with the same cardinality. We can build the complex numbers $\mathbb{C}$ using pairs of real numbers. We can use the same trick to build a field $\mathbb{C}'$ using pairs of guys in $\mathbb{R}'.$ But it’s easy to check that this funny field $\mathbb{C}'$ is algebraically closed and of characteristic zero. Since it has the same cardinality as $\mathbb{C},$ it must be isomorphic to $\mathbb{C}.$

In short, different ‘versions’ of the real numbers can give rise to the same version of the complex numbers!

• All of this stuff is amazing! Is there any place I can find an explanation of why the theory of algebraically closed fields is uncountably categorical?

• John Baez says:

It seems the key result here is a theorem of Steinitz: if two algebraically closed fields are isomorphic if they have the same characteristic and the same transcendence degree over their prime field.

Let me explain some of the jargon here:

Every field F has a unique smallest subfield K, called its prime field; the transcendence degree of a field F over a subfield K is the least number of elements of F we need to throw in to K to generate all of F using the field operations and taking roots of polynomials.

The prime field of a field of characteristic zero is isomorphic to $\mathbb{Q}$, while the prime field of a field of characteristic $p$ is $\mathbb{F}_p.$ These are both countable.

Thus, Steinitz’s theorem implies that two uncountable algebraically closed fields with the same characteristic and the same cardinality must be isomorphic.

I don’t know where to find a proof of this theorem of Steinitz… I guess any really serious book on fields should have it. For more on mathematical logic and algebraically closed fields, this seems like a great place to start:

• Adrien Deloro, Basic model theory of algebraically closed field, 2013.

Here you’ll see a different fact: any two algebraically closed fields of the same characteristic are elementarily equivalent. This means, roughly, that the same sentences in the first-order theory of fields are true in both. One can’t capture cardinality using sentences like this, so from this point of view you can’t tell the difference between a countable algebraically closed field of characteristic zero (like the algebraic closure of $\mathbb{Q}$, or this field with finitely many transcendental elements thrown in), or an uncountable one (like $\mathbb{C}$).

• Thanks for your reply! One minor thing… when you say the transcendence degree of a field is “the least number of elements of $F$ we need to throw into $K$ to generate all of $F$ using the field operations…”, should we not also add, “and taking roots of polynomials”? For instance, I’m pretty sure $F = \overline{\mathbb{Q}[\pi]}$ has transcendence degree one, even though a finite number of elements could never suffice to generate $F$ over $\overline{\mathbb{Q}}$ using just the field operations…

• John Baez says:

silvascientist wrote:

Should we not also add, “and taking roots of polynomials”?

Yes indeed! Thanks, I’ll fix that comment of mine.

4. Bhupinder Singh Anand says:

How does the argument for non-standard models of PA (of any ilk) reconcile with Theorem 7.2—PA is categorical with respect to algorithmic computability—on p.41 of the paper that appeared in the December 2016 issue of Cognitive Systems Research 40 (2016) 35-45 (DOI: 10.1016/j.cogsys.2016.02.004):

The truth assignments that differentiate human reasoning from mechanistic reasoning: The evidence-based argument for Lucas’ Goedelian thesis

Regards,

Bhup

5. The field of complex numbers $\mathbb{C}$ itself can be viewed as a nonstandard version of the algebraic numbers $\bar{\mathbb{Q}}^M$, as computed inside a nonstandard model of arithmetic $M$ of size continuum. The reason is that $\bar{\mathbb{Q}}^M$ is an algebraically closed field of characteristic zero of the right size, and so by categoricity, it is isomorphic to $\mathbb{C}$. I added this observation and a generalized form of it to my blog post at: http://jdh.hamkins.org/nonstandard-models-of-arithmetic-arise-in-the-complex-numbers/.

• Jesús López says:

If $M$, given an ultrafilter $\mathfrak{U}$, is a countable ultrapower of the standard model of arithmetic, how would the map look like? What complex number would correspond to, say, an hyperinteger $[f(n)=n]_\mathfrak{U}$?

• John Baez says:

Given any countable model of Peano arithmetic, there’s never a specific embedding of it into the complex numbers: the theorem simply says that such an embedding exists, and the proof gives no clue about how to build it. Basically you just choose a transcendental number for each nonstandard integer, subject to the conditions that the arithmetic operations on these numbers match those of the corresponding integers. This is done with the help of the Axiom of Choice.

• Jesús López says:

Thanks for the answer and fixing the LaTeX. Wanted also to amend myself and add that I meant countable exponent ultrapower and hence countinuum power $M$, as in parent comment, though perhaps this doesn’t makes the embedding more constructive.

• John Baez says:

There’s no ‘best’ embedding of any countable model of the nonstandard integers into the complex numbers, and not even any computable such embedding. But there are lots of them. Given any embedding

$f : F \to \mathbb{C}$

of any field $F$ into the complex numbers, you can compose it with any automorphism

$g : \mathbb{C} \to \mathbb{C}$

and get a new embedding

$g \circ f : F \to \mathbb{C}$

And there are vast numbers of such automorphisms! To be precise, there are $2^c$ of them where $c = 2^{\aleph_0}$ is the cardinality of the continuum. The reason is:

1) Any transcendence basis $T$ of $\mathbb{C}$ — that is, a maximal set of numbers $T \subseteq \mathbb{C}$ that obey no polynomial equations with rational coefficients — has cardinality $c.$

2) Given any transcendence basis $T$ of $\mathbb{C}$ and any permutation $h : T \to T$ there is an automorphism $g : \mathbb{C} \to \mathbb{C}$ extending $h.$

For a bit more read Andrés Caicedo’s remark on MathOverflow. Note that in this comment I’m assuming the axiom of choice.

6. Ali Enayat says:

Hi John, the following related mathoverflow question(s) of mine, and the partial answers so far might be of interest:

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

7. Ali Enayat says:

In my previous post, I should have also brought attention to the proof (by Ax), which uses tools very similar to the those discussed in your blog, of the Ax-Grothendieck theorem, which states that every injective polynomial map from $\mathbb{C}^n$ to itself is surjective (where $n$ ranges over non-negative integers).

An exposition can be found in the Chapter 6 of the following master’s thesis by Amanda Purcell:

https://pqdtopen.proquest.com/doc/1476434703.html?FMT=AI

The model-theoretic proof should be compared to the long-winded one using standard tools of multivariable complex variables, by — guess who? — the eminent Walter Rudin in the following paper:

Rudin, W. (1995) Injective Polynomial Maps are Automorphisms, American Mathematical Monthly, 102, 6:540-543

• John Baez says:

Neat! I haven’t thought much about the Grothendieck–Ax theorem. It reminds me slightly of the apparently much harder
Jacobian conjecture.

The model-theoretic proof should be compared to the long-winded one using standard tools of multivariable complex variables, by the eminent Walter Rudin in the following paper:

• Rudin, W. (1995) Injective Polynomial Maps are Automorphisms, American Mathematical Monthly, 102, 6:540-543.

I’ll have to look at it! It’s rare to see a long-winded proof that’s only 3 pages long… though some of my students’ homeworks contain examples.

By the way, when posting a comment with LaTeX here, read the directions written in big black letters that appears directly over the comment box:

You can use Markdown or HTML in your comments. You can also use LaTeX, like this: &dollar;latex E = m c^2 &dollar;. The word ‘latex’ comes right after the first dollar sign, with a space after it.

• Ali Enayat says:

Thanks for your feedback John. I referred to Rudin’s proof as long-winded only in contrast with Ax’s terse proof, which deftly reduces the theorem to the pigeonhole principle with the help of some basic theorems of model theory and algebra.

Also sorry for not having noticed the clear Latex-related instructions in bold fonts (in these situations I am reminded of how some of us resemble some of our students).

This site uses Akismet to reduce spam. Learn how your comment data is processed.