4
$\begingroup$

Now we fix an ultrafilter of $\mathbb{N}$ that contains the cofinite filter, consider a hyperreal field ${}^{*}\mathbb{R}$. Let $\varepsilon$ be a positive infinitesimal. We doubt that a power series field $\mathbb{R}((X))$ can be realized as a subquotient of ${}^{*}\mathbb{R}$.

In Measure, integration and elements of harmonic analysis on generalized loop spaces , Fesenko explained this in Remark 7.1. However, the meaning of some words appears unclear. For example, what does the phrase "the fraction field of approachable polynomials $\mathbb{R}[X]^{\rm ap}$'' mean? Acording to the above paper, there exists a surjective homomorphism $\mathrm{Frac}(\mathbb{R}[X]^{\rm ap})\longrightarrow \mathbb{R}((X));\varepsilon \longmapsto X$.

If you know the answer or have any idea, please tell us.

$\endgroup$
0

1 Answer 1

7
$\begingroup$

Let $A$ denote the convex subring of hyperreal numbers $y$ for which there exists an $n \in \mathbb{N}$ with $-\varepsilon^{-n}<y<\varepsilon^{-n}$. This has the set $\mathfrak{m}$ of numbers $z$ with $-\varepsilon^n<z<\varepsilon^n$ for all $n \in \mathbb{N}$ as a maximal ideal.

Since $A$ (and $\mathfrak{m}$) is convex, the quotient field $A / \mathfrak{m}$ has an induced ordering from that of ${}^{*}\mathbb{R}$, whereby $y+\mathfrak{m}$ is strictly positive if $y>\mathfrak{m}$.

Write $\overline{\varepsilon}:=\varepsilon+\mathfrak{m}$ in the quotient field. Then for every power series $f=\sum \limits_{k \in \mathbb{z}} f_k X^k \in \mathbb{R}((X))$, the sequence $(\sum \limits_{k<n} f_k \overline{\varepsilon}^k)_{n \in \mathbb{N}}$ has a unique limit $f(\overline{\varepsilon})$.

The unicity of that limit is by definition of $\mathfrak{m}$, since any two limits have their difference smaller than all $\overline{\varepsilon}^n,n \in \mathbb{N}$, and must thus be equal.

The existence follows from the fact that ${}^{*}\mathbb{R}$ is $\aleph_1$-saturated as an ordered field, so there is a $z \in {}^{*}\mathbb{R}$ satisfying the conditions expressing that $|z-\sum \limits_{k\leq n} f_k \varepsilon^k|<\varepsilon^n$ for all $n \in \mathbb{N}$, and for any such $z$, we see that $z+\mathfrak{m}$ is the corresponding limit.

You can check that $\mathbb{R}((X)) \longrightarrow A / \mathfrak{m}:f \mapsto f(\overline{\varepsilon})$ is an embedding of ordered fields.


Incidently, this subquotient can also be realized, in a non-canonical way, as a subfield of ${}^{*}\mathbb{R}$, again using $\aleph_1$-saturation and the fact that each power series is determined by a cut over a countable subset of power series.

More precisely, given an intermediate field $K$ with $\mathbb{R}(X) \subseteq K \subset \mathbb{R}((X))$, an ordered field embedding $\Psi: K \rightarrow {}^{*}\mathbb{R}$ over $\mathbb{R}$ with $\Psi(X)=\varepsilon$, and a $y \in \mathbb{R}((X)) \setminus K$, the embedding $\Psi$ can be extended to $K(y)$. There are two cases: if $y$ is algebraic over $K$, then $y$ is the $n$-th root of its minimal polynomial over $K$ in the real closure of $K$, for some $n$, and it has to be sent to the $n$-th root of that same polynomial in ${}^{*}\mathbb{R}$ (which is a real-closed field). Otherwise, the element $y$ is unique to satisfy $L<y<R$ for some countable subsets $L,R$ of $\mathbb{R}(X)$, and any $z \in {}^{*}\mathbb{R}$ with $\Psi(L)<z<\Psi(R)$ (exists by saturation) will allow one to extend $\Psi$ with $\Psi(y):=z$.

Using this and some choice (choosing each non-algebraic $z$ + a well-ordering of $\mathbb{R}((X)) \setminus \mathbb{R}(X)$), one can define an embedding $\mathbb{R}((X)) \longrightarrow {}^{*}\mathbb{R}$. But there does not seem to be a canonical way to do so because both types of choice involved require (I think) some form of axiom of choice.

$\endgroup$
2
  • $\begingroup$ Thank for your answer. I am interested in your last comment. You suggested that $\mathbb{R}((X))$ can be regarded as a subfield of hyperreal field by a non-canonical way. I would like a detailed explanation of "non-canonical way". $\endgroup$
    – M masa
    Oct 13, 2022 at 10:55
  • $\begingroup$ @Mmasa I edited my answer to address this, at least partially. $\endgroup$
    – nombre
    Oct 13, 2022 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.