# Hyperreal field extension

In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension?

Is it algebraic? Probably not, right? Transcendental? Normal? Finitely generated? Separable?

For instance, I was thinking: Would it be enough to adjoin an infinitesimal to $\mathbb{R}$ and get $\mathbb{R}^*$? The axioms of hyperreals in Keisler's Foundations of Infinitesimal Calculus seem to suggest so, but I'm not sure.

I don't know how to approach this question, since infinitesimals and such things don't "result" from polynomials (like, say, complex numbers do).

Is $\mathbb{R}^*$ maybe isomorphic to the field of fractions of some polynomial ring? (This occured to me after thinking about $\mathbb{R}(\epsilon)$ and such things, where $\epsilon$ is an infinitesimal).

• I just realised it's quite easy to show it isn't algebraic. But what about the rest?
– FPP
Nov 10, 2012 at 22:24
• There actually exists no single field of hyperreals. Different fields of hyperreals may fail to be isomorphic. Nov 10, 2012 at 22:24
• Let's assume the continuum hypothesis, then. If I remember correctly, that implies that all are isomorphic, right?
– FPP
Nov 10, 2012 at 22:25
• @FFP Yes.${}{}$ Nov 10, 2012 at 22:26
• Well, exactly what should $\epsilon$ satisfy? How is (an) $\Bbb R^*$ constructed? Nov 10, 2012 at 23:18

There isn't any trouble constructing $\mathbb{R}(\epsilon)$ as a formally real field that is isomorphic (as a field) to $\mathbb{R}(x)$, and with $\epsilon$ a positive infinitesimal. (it may be easier to see the ordering by writing a rational function as a formal Laurent series in $\epsilon$ about 0)

It's easy to see that this is not a hyperreal field: it only has finite powers of $\epsilon$. In particular, if there were a positive transfinite positive integer $H$, then there would be an element $\epsilon^H$ that is smaller than every power of $\epsilon$ appearing in $\mathbb{R}(\epsilon)$.

($\mathbb{R}(\epsilon)$ also fails to have a square root of $\epsilon$. This can be fixed by passing to its real closure, but that would still fail the above property)

If you tried adjoining a second infinitesimal, a similar argument proves it can't be hyperreal.

I don't think it would be enough to adjoin one infinitestimal. To construct the hyperreals via an ultrapower requires the Axiom of Choice, but you could certainly adjoin one infinitesimal without Choice.

• I would be happy to make changes/elaborate if the downvoter would like to explain what they didn't like. If you're not comfortable commenting publicly, feel free to email me (see my profile). May 5, 2017 at 12:25

Briefly, one way to approach this question would be to try to construct alternative models of the hyperreals via the Compactness Theorem that either satisfy or fail to satisfy the properties you want. The Enderton text, A Mathematical Introduction to Logic, uses this kind of construction and has a very rigorous treatment of nonstandard analysis that you can use as a guide.

Since I only have 44 minutes left on this question, it could easily contain errors so I'm retagging your question to include logic and set theory so it can be properly checked.

A small correction to the formulation of the original question: the continuum hypothesis is not needed to construct a hyperreal field extension of the reals. The alternative simpler field extension you are proposing is related to the Levi-Civita field, which was of intense interest to Abraham Robinson. However, it does not have the crucial properties that make the hyperreals useful in analysis.