Questions tagged [ordered-rings]

Ordered rings are (usually commutative) rings which have an additional structure, a linear order compatible with the ring structure. This tag is for questions regarding ordered rings and their properties, as well proofs related to un-orderability of certain rings.

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Visualize the Completion of (the Ordered Field) of Rational Functions

Every ordered abelian group $G$ can be completed to give a larger ordered abelian group $\bar{G}$. The original abelian group $G$ embeds into $\bar{G}$ as a dense subset, and every non-empty subset of ...
Theone's user avatar
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Dedekind completion of the integers

The Dedekind completion of any Archimedean ordered field is the real numbers. The same is true of the Dedekind completion of any Archimedean ordered integral domain whose strict order is dense. What ...
Madeleine Birchfield's user avatar
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Is the ring of integers modulo n, $\mathbb{Z_n}$, an ordered ring? [duplicate]

I'm starting to study on my own some basic group theory, maybe this is a very basic question, but I can't find any answer on the internet. I would like to know if the ring of integers modulo n, $\...
Ser Pounce of House Whiskers's user avatar
9 votes
2 answers
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Is every ordered abelian group the additive group of an ordered ring?

Let $\Lambda$ be an ordered abelian group, (there is a total order on $\Lambda$ which is compatible with addition). Is there a multiplication map on $\Lambda$ that turns it into an ordered ring? I ...
Strichcoder's user avatar
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Totally positive/negative units in preordered rings with bounded inversion

Let $A$ be a preordered ring (or $\mathbb{R}$-algebra or $\mathbb{Q}$-algebra). Say that ${a \in A}$ is totally positive if for every morphism ${f : A \rightarrow \mathbb{R}}$ of ordered algebras, ${f ...
Boogie's user avatar
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Multiplication Law for Order on Integers

I'm using the following definitiosn for addition $+$, multiplication $\cdot$, and the relation $\preceq$ on the set of integers: \begin{align*}\tag{I} [(a,b)]+[(c,d)]&:=[(a+c,b+d)] \\ \tag{II} [(a,...
PAT's user avatar
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Property of a semiring equipped with partial order relation [closed]

Let $R$ be a multiplicatively idempotent semiring with additive identity, and a partial order relation $\leq$ is defined on $R$. Then, for all $x$ in $R$, does the identity $x+2x=2x$ implies $x\leq ...
gete's user avatar
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Does the derivative of a polynomial over an ordered ring behave like a rate of change?

Suppose I have an ordered commutative ring $R$. I can define the collection $P$ of polynomial functions defined on $R$ as functions of the form $f(r) = p_0 + p_1 r + p_2 r^2 + \cdots$ where $p_0, p_1, ...
gigalord's user avatar
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Uniquely orderable subfields of $\mathbb{Q}_p$?

I have heard that, unlike $\mathbb{R}$, the field $\mathbb{Q}_p$ cannot be realised an ordered field. Is there any way to extend the natural ordering on $\mathbb{Q}$ to a larger subfield of $\mathbb{Q}...
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Are $\mathbb{R}$ and $\mathbb{Q}$ the only subfields of $\mathbb{C}$ with natural structure as ordered fields?

We know that $\mathbb{R}$ and $\mathbb{Q}$ have a unique structure as ordered fields with the usual order, and that $\mathbb{C}$ cannot be realised as an ordered field. Various non-trivial subfields ...
user829347's user avatar
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Ordering on $R[\sqrt{n}]$ for an ordered ring $R$

I'm interested in showing that if $R$ is an ordered ring (with ordering $\leq$), and $n \geq 0$, then $R[\sqrt{n}]$ is also an ordered ring. In the reals, $0 \leq a_1 + a_n\sqrt n$ iff either (1) $0 \...
mdgeorge's user avatar
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Does an infinite chain contain articulation points?

I had a question which asked whether 2-regular graphs have any articulation points. We assumed finite graphs so it's just a disjoint union of cycles. However if we allow infinite graphs how do we ...
Materia Gravis's user avatar
4 votes
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Is the number of orderings the same of the number of automorphisms in a ring?

Q: Given an ordered ring $A$ is the number of automorphisms of $A$ equal to the number of orderings in $A$? An ordering on a ring is totally defined by a subset of $A$ we call $A^+$ that satisfies ...
Tomás Pacheco 's user avatar
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Is it possible to send an element in an ordered real algebra both to a positive unit and to a negative unit?

Let ${(A, P)}$ be a preordered $\mathbb{R}$-algebra in the sense that $A$ is a $\mathbb{R}$-algebra and ${P \subseteq A}$ is a subset closed under addition, multiplication, containing the nonnegative ...
Boogie's user avatar
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An example of an ordered UFD except the ring of integers?

Are there examples of unique factorization domains which are ordered rings https://en.m.wikipedia.org/wiki/Ordered_ring except the ring of integers?
Lehs's user avatar
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If $R$ and $S$ are isomorphic rings, is $R$ an ordered ring iff $S$ is an ordered ring?

Yesterday I had my final for my introduction to abstract algebra course. One of the questions on the final asked you to prove that the field $<\mathbb{R},+, \cdot>$ was not isomorphic to $<\...
Dastur's user avatar
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1 answer
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$a^+$ Induced Order

I am studying the different orders that can be induced from the usual polynomial order in $ \mathbb R[x]$: i.e: $$ p(x) >_{+\infty} 0 \iff a_n > 0 $$ Where $a_n$ is the leading coefficient. One ...
omega-stable's user avatar
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Greateat common divisor in Z[i]

How do i use the euclidean algorithm to compute the greatest common divisor of two elements in Z[i]?
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Are there any "crazy" totally ordered rings without infinities/infinitesimals?

I was seeing this post and its nice answer. I asked myself if I could have found a more "crazy" example. I was not very succesful. So I formalized what I was trying to achieve: Question: Let $(R,+,\...
M. Winter's user avatar
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4 votes
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Metric mapping to sets other than $\Bbb{R}$ [duplicate]

A metric space is a set M together with a function $d:M \times M \rightarrow \Bbb{R} $, where $d$ satisfies: $d(x,y)\ge 0$ $d(x,y)=0 \Leftrightarrow x=y$ $d(x,y)=d(y,x)$ $d(x,z) \le d(x,y)+ d(y,z)$ ...
J J Grimes's user avatar
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Ordered ring and mutiplicative ordinal

If $(E,<)$ is a linear order, let $s(E)$ denote the supremum of the set of ordinals which (order-)embed in $(E,<)$. $s(E)$ is also the set of ordinals which embed in $E$ with a non cofinal range....
nombre's user avatar
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