extension-fieldfield-theorynonarchimedian-analysisnonstandard-analysisordered-fields
Let $F$ be a nonarchimedean ordered field. Is there always a hyperreal field $^*\mathbb{R}$ such that there is an embedding of $F$ in $^*\mathbb{R}$?
As far as I understand it, the answer here implies that the answer is positive in the special case where $F$ is the Dehn field or the Levi-Civita field.
$\DeclareMathOperator{\cof}{cof}$The Dehn field embeds naturally into the Levi-Civita field and fields $^*\mathbb{R}$ of hyperreal numbers defined by ultrafilters on $\mathbb{N}$, and the Levi-Civita field embeds into $^*\mathbb{R}$.
I'll try to explain this in some detail. I assume you are familiar with the notion of ordered Hahn series fields and I first introduce what I called Cauchy-completion. You'll find a discussion about it here.
Any linear order $X$ is equipped with a topology, called the order topology, whose open sets are the unions of open intervals. In ZFC, $X$ has a cofinality $\cof(X)$ which is the least order type of a cofinal subset of $X$. If $F$ is an ordered field, then there is also a natural uniform structure on $F$ and in particular a notion of Cauchy sequence (same as regular Cauchy sequences but indexed by $\cof(F)$). The topology is sequential, in that closed sets are sets which contains the limits of their convergent sequences (indexed by $\cof(F)$), and thus in general every topological notion can be stated in terms of sequences indexed by $\cof(F)$, with the usual properties: continuity is sequential continuity, and so on....
An ordered field $F$ is said Cauchy-complete if its Cauchy sequences are convergent in $F$, or equivalently if it has no proper dense (ordered field) extension. Every ordered field $F$ has a dense Cauchy-complete extension $(\widetilde{F},\varphi)$, with the following initial and terminal properties:
(IP:) If $(I,\mu)$ is a Cauchy-complete continuous (for the order topology, equivalently, the embedding is cofinal) ordered field extension of $F$, then there is a unique morphism $\sigma: \widetilde{F} \rightarrow I$ with $\sigma \circ \varphi=\mu$.
(TP:) If $(T,\mu)$ is a dense extension of $F$, then there is a unique morphism $\sigma: T \rightarrow \widetilde{F}$ with $\varphi=\sigma \circ \mu$.
The extension $(\widetilde{F},\varphi)$ is called the Cauchy-completion of $F$.
For instance $\mathbb{R}$ is the Cauchy completion of $\mathbb{Q}$, and the field $\mathcal{L}=\mathbb{R}((\varepsilon^{\mathbb{Z}}))$ of Laurent series is the Cauchy-completion of $\mathbb{R}(\varepsilon)$. To prove this, one needs only check that this is a dense ordered field extension of this field, and that each Cauchy sequence in $\mathbb{R}(\varepsilon)$ converges in $\mathcal{L}$.
Cauchy-completion is similar real closure in that they correspond to a specific version of reflective subcategories linked to properties of extensions: algebraic extensions and dense extensions. In this answer, I give some (quite poorly written) explanation. What matters here is that the Cauchy-completion is a functorial construction.
Using the Newton polygon method, one can prove that the real closure of $\mathcal{L}$ is the ordered field $\mathcal{P}=\bigcup \limits_{n \in \mathbb{N}^{>0}} \mathbb{R}((\varepsilon^{\frac{1}{n}.\mathbb{Z}}))$ of Puiseux series. Its Cauchy-completion, which is the Levi-Civita field $\mathcal{C}$ (sorry Levi...) can be construed as the field of Hahn series $s=\sum \limits_{n \in \mathbb{N}} s_n \varepsilon^{q_n}$ where $(s_n)_{n \in \mathbb{N}}$ is a sequence of real numbers and $(q_n)_{q \in \mathbb{N}}$ is a strictly increasing and cofinal sequence of rationnal numbers. As the Cauchy-completion of a real-closed field, $\mathcal{C}$ is automatically real-closed.
Since the Dehn field $\mathcal{D}$ is an algebraic extension of $\mathbb{R}(\varepsilon)$, by the terminal property of real closure, there is a unique embedding of $\mathcal{D}$ into the real closure of $\mathbb{R}(\varepsilon)$ extending the given real closure morphism. By the initial property of real closure, this real closure enjoys a unique embedding in $\mathcal{C}$ extending the embedding $\mathbb{R}(\varepsilon) \rightarrow \mathcal{C}$.
Now let's turn to germs of real valued functions. The ring $\mathcal{G}$ of germs of real valued functions is the quotient of the set of real valued functions defined on intervals $[a,+\infty)$ for some real number $a$ by the equivalence relation $f \sim g$ iff $f(x)=g(x)$ for sufficiently big $x$. It is a partially ordered ring under poitwise sum and product, and eventual comparison. Any (linearly) ordered subfield $F$ of $\mathcal{G}$ embeds naturally in the ultrapower $^*\mathbb{R}$ (given a free ultrafilter $U$ on $\mathbb{N}$) by sending the germ of a function $f$ defined on $[n,+\infty)$ to the class of $(0,0,...,0,f(n),f(n+1),...)$ modulo $U$ (where there are $n$ zeroes for instance). The Dehn field is such an ordered field, or more precisely it is a field or representatives of germs of real valued functions. Thus it also embeds naturally in $^*\mathbb{R}$.
The field of Laurent series, and for that matter the Levi-Civita field, are not fields of germs of real valued functions, at least not that I know of (but maybe every Laurent series is Borel summable?). Thus I don't see how to naturally embed them into $^*\mathbb{R}$. It is possible that the saying "$^*\mathbb{R}$ extends the Levi-Civita field" is better understood as a thematic remark: the Levi-Civita can be used to do analysis although it is not archimedean, and thus "not standard" analysis, with nice properties (see this here found on Wikipedia), and fields of hyperreal numbers can be seen as the completion of this goal.
In ZFC, there are embeddings of $\mathcal{C}$ into $^*\mathbb{R}$. To see this, we can use the fact that $^*\mathbb{R}$ is real-closed and countably saturated: if $L,R$ are countable sets of hyperreal numbers with $L<R$, then there is a hyperreal number $a$ with $L<a<R$. This implies that $(i)$: $^*\mathbb{R}$ contains a canonical copy of the real closure of each of its subfields, and
$(ii)$: $^*\mathbb{R}$ contains a copy of the Cauchy-completion of each of its subfields with countable cofinality.
The first statement follows from the initial property of real closure.
To prove the second one, consider a subfield $K$ of $^*\mathbb{R}$ with countable cofinality. The set of cofinal extensions of $K$ in $^*\mathbb{R}$ is inductive for the relation of inclusion, and has a maximal element $F$ by Zorn's lemma. Since any algebraic extension is cofinal and by the first statement, $F$ must be real-closed. Assume for contradiction that there is a non-convergent Cauchy sequence $u$ in $F$. We may also assume that we have $u_{2m}<u_{2n+1}$ for all $m,n \in \mathbb{N}$ (I let you figure this out). Let $L$ (resp. $R$) be the set of elements of the sequence which lie below (resp. above) infinitely many elements of the sequence. We have $L<R$ so there is a hyperreal number $a$ with $L<a<R$. Note that $a$ leis outside of $F$, otherwise $u$ would converge to it. I claim that the subfield $F(a)$ of $^*\mathbb{R}$ is a dense extension of $F$ where $u$ converges to $a$.
In fact, we only require that it is a cofinal extension of $F$, but density follows. Indeed, it will follow that $u$ converges to $a$. Then since $F$ is real closed, no polynomial in $F[X]$ annihilates $a$, and thus by continuity of fractions in $F(X)$ on $F(a)$ outside of their poles (true for any ordered field), for such fraction $f(X)$, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(a)$.
So let's prove that $F$ is cofinal in $F(a)$. It is easy to see that each number $P(a)$ for $P \in F[X]$ is bounded by elements of $F$. We must only check that for any non zero polynomial $Q \in F[X]$, the fraction $\frac{1}{Q(a)}$ is also bounded in $F$, that is, we must check that $Q(a)$ may not be infinitesimal with respect to $F$, denoted $Q(a)\prec_F 1$. We do so by valuation-theoretic arguments. Since $^*\mathbb{R}$ is countably saturated, in particular $F$ is bounded in $F$, and its convex hull in $^*\mathbb{R}$ is a proper convex valuation ring on $^*\mathbb{R}$ which contains $a$. By real closure of $^*\mathbb{R}$, the corresponding valued field is henselian (see for instance Theorem 3.5.16 in ADH). Assume towards a contradiction that there is $Q \in F[X]$ which is non zero with $Q(a) \prec_F 1$, and choose such polynomial $Q$ with minimal degree, hence $Q'(a)$ is not infinitesimal with respect to $F$. By henselianity, this means that there is $b \in ^*\mathbb{R}$ with $Q(b)=0$ and $a-b\prec_F 1$. The first relation yields $b \in F$ (by real closure of $F$), and the second implies that $u$ converges to $b$ in $F$: a contradiction.
Thus $F(a)$ is a cofinal extension of $F$, which contradicts the maximality of $F$. So $F$ must be Cauchy-complete. The initial property of Cauchy completion then implies that the Cauchy-completion of $K$ embeds in $F$ and thus in $^*\mathbb{R}$.
Applying those two results and starting with the fields-of-germs-style embedding $\mathbb{R}(\varepsilon) \rightarrow ^*\mathbb{R}$ with $f(\varepsilon)\sim (0,1,\frac{1}{2},\frac{1}{3},...)$, we get an embedding of Laurent series, then Puiseux series, then "Levi-Civita series" into $^*\mathbb{R}$.
(Using similar arguments as above, one can prove that the maximal subfields $F$ in $^*\mathbb{R}$ are in fact almost countably saturated (countably saturated but at $+\infty$ and $-\infty$), hence in particular spherically complete. This implies that $^*\mathbb{R}$ also contains a copy of the Hahn series field $\mathbb{R}((x^{\mathbb{R}}))$.)
"multiplication of numbers" is actually a relative concept, depending on the algebraic structure which you are working with. For your question, the most important example is $ \frac{\mathbb{Z}}{p\mathbb{Z}} $ which is a finite field (with mod $p$ summation and multiplication).For example when $p = 5$ we have this equalities (in $ \frac{\mathbb{Z}}{5\mathbb{Z}} $): $$ 1+1+1+1+1 = 5 = 5*1 = 0 (mod \text{ }5) $$ so you see that $5*1 = 0$ in this context.
Now, I will go through (i), (ii) and (iii):
(i): Defining axioms of an ordered field (F, +, ., <) imply that for any $0 \ne x \in F$ we have $x^2 > 0$ because if $x > 0$ we get $x*x = x^2 > 0$ and if $x < 0$ we have $x + (-x) < 0 + -x$ so $0 < -x$ and then again we have $x^2 = (-x)*(-x) > 0$ and we are done. so in a special case, we get $1 > 0$ in any ordered field (F, +, ., <). Now if we have some positive $m$ which $m*1 = 0$, we will get a contradiction, because: $$ 1>0 , 1+1>0, 1+1+1>0, ..., 0 = 1+1+...+1(m\text{ times}) > 0 $$
(ii): Assuming $F$ is a finite field with cardinality $n$, if we add up $1$: $$ 1, 1+1, ..., 1+1+...+1(n+1 \text{ times}) $$ we have to get two equal numbers in this sequence like $m*1 = p*1$ and $m \ne p$ so we get $(m-p)*1 = 0$.
(iii): This part is straight. if we look at vector space axioms, we see from that (i) and (ii) its not possible to define scalar product for a finite field $F$ over $\mathbb{R}$ and $\mathbb{C}$ as usual, like: $1_{\mathbb{R}}*1_{F} = 1_{F}$ because then we will have some $m \in \mathbb{Z}\text{: } m*1_F = 0_F$ and: $$ 0_F = \frac{1}{m} * 0 = \frac{1}{m}(m*1) = 1_{\mathbb{R}}*1_{F} = 1_{F} $$ which is absurd.
Best Answer
A definition:* A hyperreal field is an extension $^*\mathbb{R}$ of $\mathbb{R}$ equipped with an operator $*$ defined on the structure $\mathcal{S}(\mathbb{R}):=\bigcup \limits_{n \in \mathbb{N}} \mathcal{S}_n(\mathbb{R})$ where for any set $E$, we have $\mathcal{S}_0(E)=E$ and $\mathcal{S}_{n+1}(E)= \mathcal{S}_n(E)\cup\mathcal{P}(\mathcal{S}_n(E))$ for all $n \in \mathbb{N}$. This operator fixes $\mathbb{R}$ pointwise and assigns a an element $^*X \in \mathcal{S}(^*\mathbb{R})$ to each element $X$ of $\mathcal{S}(\mathbb{R})$ (with $^*\mathbb{R}$ being the value of this operator at $\mathbb{R}$), such that the following transfer principle is satisfied:
For all (set-theoretic, first order) formulas $\varphi[x_1,...,x_n]$ where each quantifier appears in a bounded form $\forall x \in y,[...]$ and $\exists x \in y,[...]$, we have that for all $X_1,...,X_n \in \mathcal{S}(\mathbb{R})$, $\varphi[X_1,...,X_n]$ holds if and only if $\varphi[^*X_1,...,^*X_n]$ holds.
(in particular, $*$ preserves the "type" of sets, e.g. sends $n$-uples of reals to $n$-uples of hyperreals and so on)
One way (in fact, the way) to obtain this type of field is through ultrapowers of $\mathbb{R}$. One can also characterize them as quotients of rings $\mathbb{R}^I$ of functions $I \rightarrow \mathbb{R}$ by maximal ideals, where $I$ is a non-empty set. By a result of Keisler-Kunen, for all cardinals $\kappa$, there exists a ultrafilter $\mathcal{U}$ on $\kappa$ such that the corresponding ultrapower $\mathbb{R}^{\mathcal{U}}$ is $\kappa^+$-saturated. This implies that any ordered field of cardinality $\leq \kappa$ embeds into $\mathbb{R}^{\mathcal{U}}$. So the answer is yes in this sense.
See A ring of homomorphisms is enough to get non-standard analysis for more details. See also A purely algebraic characterization of fields of the hyperreal numbers for a somewhat more explicit description of those fields.
*this is different from Dale and Woodin's definition of hyper-real fields in their book Super-real fields. However the result is still true with this definition, since hyperreal fields in the previous sense are hyper-real fields in their sense.