Questions tagged [ordered-fields]

The tag has no usage guidance.

Filter by
Sorted by
Tagged with
0
votes
0answers
33 views

Tensor product of preordered rings

All rings in this post are commutative, unital, and contain $\frac{1}{2}$. To study "real" properties of a ring $R$, one is often interested in the orderings which exist on fraction fields of ...
17
votes
2answers
509 views

Are radicals dense in the real closure of an ordered field?

Let $F$ be an ordered field and let $R$ denote its real closure. It is well-known that $F$ is cofinal in $R$, but not necessarily dense. For example, consider $F=\mathbb{R}(\omega)$ with the order ...
45
votes
8answers
5k views

Cauchy reals and Dedekind reals satisfy “the same mathematical theorems”

The succinct question The conjecture of Birch and Swinnerton-Dyer (to take a random example) mentions L-functions and hence the complex numbers and hence the real numbers (because the complexes are ...
11
votes
3answers
607 views

Are archimedean subextensions of ordered fields dense?

Let $E$ be an ordered field and let $F$ be a real closed subfield. We say that $E$ is $F$-archimedean if for each $e\in E$ there is $x\in F$ such that $-x\le e\le x$. Is it true that if $E$ is $F$-...
7
votes
2answers
313 views

How do fractional tensor products work?

[I asked and bountied this question on Math SE, where it got several upvotes and a comment suggesting it was research-level, but no answers. So I'm reposting here with slight edits, but please feel ...
3
votes
1answer
117 views

'Smallest' subfield of the Surreals which is isomorphic to the Surreals as an ordered group

What is the smallest subfield $F\subset N_0$ such that $$(F,+,\times,\leq)\ncong(N_0,+,\times,\leq)$$ but $$(F,+,\leq)\cong(N_0,+,\leq)?$$ Since these are all going to be proper classes cardinality is ...
7
votes
2answers
243 views

Formally real fields with unique non-Archimedean ordering

My question is rather simple. Do there exist a formally real field that admits a unique ordering (so sums of squares are the positive elements) and such that this ordering is not archimedean? Oh, I ...
2
votes
0answers
96 views

Atomic integer parts

Let $R$ be an ordered ring (in particular, the order is linear and $R$ is a domain). Let $| \ \ |: x \mapsto \max(x,-x)$ denote its absolute value. For $(x,y) \in R \times R^{\neq 0}$ say that an ...
4
votes
0answers
235 views

Do all fields with internal absolute values arise as ordered fields or like $\mathbb{C}$ from them?

$\def\abs#1{\lvert#1\rvert} \def\Im{\operatorname{Im}} \def\Re{\operatorname{Re}}$ (Crossposted from math.stackexchange.com after 5 days with no correct answer.) Let ​ $\langle F,+,\cdot\rangle$ ​ be ...
1
vote
1answer
132 views

Archimedean completeness of some fields

I need a reference (different from Hahn's 1907 paper) for the following result. Theorem: If $G$ is a totally ordered abelian group, then the field $\mathbb{R}((G))$ is archimedean complete. $\...
1
vote
1answer
73 views

Bound for annihilating polynomials

Let $F$ be an ordered field, let $L$ be the real closure of $F$. Let $R \in L$ be strictly positive. Can one find a bound $M \geq 0$ and for each $x \in ]-R;R[_L$, an element $x' \in [x-1;x+1]_L$ ...
5
votes
1answer
144 views

Is there an exponential map on (Hahn) ordered fields?

If $F$ is an ordered field and $G$ is an ordered abelian group, one can define the Hahn product $F \boxtimes G$ to be the set of formal Laurent series with coefficients in $F$ and exponents in $G$. It ...
3
votes
1answer
201 views

Completing class-sized Fields

Let's say that an ordered Field is a class (proper or not) which satisfies the axioms of ordered fields. We work in NBG set theory with global choice. Let's say that an ordered Field is real closed ...
4
votes
2answers
331 views

Which ordinals can be embedded into an ordered field?

Let $F$ be an ordered field. What is the least ordinal $\alpha$ such that there is no order-embedding of $\alpha$ into any bounded interval of $F$?
3
votes
1answer
148 views

Least ordinal not embedded in a total order

If $(E,<)$ is a linear order, let $s(E,<)$ denote the least ordinal which doesn't embed in $(E,<)$. I am trying to prove the following: If $(M,+,.,0,1)$ is a model of open induction, (or ...
5
votes
1answer
151 views

Ordinals which embed in surreal subfields

If $k$ is an ordered field, the least ordinal $s(k)$ which doesn't embed in $(k,<)$ is regular. This is because every interval of an ordered field embeds in every infinite interval so given a ...
3
votes
1answer
95 views

Compatible total orderings of the group $\mathbb{Z}^\mathbb{N}$

Given the additive group of the module $\mathbb{Z}^\mathbb{N}$ and a total ordering of the group that is compatible with addition and where $\chi_{\{n\}} > 0$ for all $n \in \mathbb{N}$, can we say ...
2
votes
1answer
187 views

Cauchy completeness of the real closure

Let $k$ be an ordered field of cofinality $cf(k)$ whose Cauchy $cf(k)$-sequences are convergent.$^{(1)}$ Let $\mathcal{R}(k)$ be its real closure. As an algebraic extension of $k$, it has the same ...
19
votes
1answer
862 views

Differential Topology over $\mathbb{Q}$

I have a specific question in mind, but it requires some explanation and context before it can be formally stated. To summarize it in a sentence, this is it: Are every two rational manifolds of the ...
23
votes
1answer
577 views

Which ordered fields are homeomorphic to their power?

It is well known that $\mathbb{R}^2\ncong \mathbb{R}$. It is also known that $\mathbb{Q}^2\cong \mathbb{Q}$. It is a corollary to Sierpiński's theorem which states that every countable metric space ...
8
votes
2answers
478 views

Possible cardinality and weight of an ordered field

Is it true (in ZFC) that for any regular infinite cardinal $\kappa$ there exists an ordered field of weight $\kappa$ and cardinality $2^\kappa$ (or at least $>\kappa$)? The field of real numbers ...
30
votes
1answer
3k views

What did Rolle prove when he proved Rolle's theorem?

Rolle published what we today call Rolle's theorem about 150 years before the arithmetization of the reals. Unfortunately this proof seems to have been buried in a long book [Rolle 1691] that I can't ...
10
votes
3answers
1k views

Does this construction yield the surreal numbers?

There are two simple constructions for creating arbitrarily large non-Archimedean ordered field extensions of the reals. First given such a field one may consider rational functions over that field ...
0
votes
1answer
102 views

Ways to order an algebraic extension

In following post, I describe the "classical" example of $\mathbb{Q}(\sqrt{2})$ that can be ordered in two distinct ways. More generally, if $(k,P)$ is an ordered field, $R$ a real closure of $(k,P)$ ...
35
votes
6answers
2k views

On the universal property of the completion of an ordered field

I have been trying to write up some notes on completion of ordered fields, ideally in the general case (i.e., not just completing $\mathbb{Q}$ to get $\mathbb{R}$ but considering the completion via ...
5
votes
0answers
656 views

two versions of the nested interval property

There appear to be two different nested interval properties for the reals with the punchline "... then the intersection of the intervals is non-empty", and I'd like to know their respective histories (...
10
votes
2answers
901 views

Does Rolle's Theorem imply Dedekind completeness?

I think the answer to the title question is "yes", but Gerald Edgar, in his comment on Does antidifferentiability of continuous functions imply Dedekind completeness? , points out an article (actually ...
9
votes
0answers
289 views

Does antidifferentiability of continuous functions imply Dedekind completeness?

Let $R$ be an ordered field, and let $I$ be {$x \in R: a < x < b$} for some $a < b$ in $R$. Define notions of $R$-continuity and $R$-differentiability for functions $f : I \rightarrow R$ by ...
11
votes
5answers
2k views

analysis over non-Archimedean ordered fields

Can anyone suggest any good references for (or any experts on) analysis over non-Archimedean ordered fields, such as the field of rational functions in one variable (ordered at 0, or if you prefer at ...