Skip to main content

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.

Filter by
Sorted by
Tagged with
3 votes
1 answer
114 views

Is the class of linearly-orderable rings first order axiomatizable?

A linearly ordered ring is a commutative ring $R$ with unity equipped with a linear order $\leq$ that is compatible with addition, and such that the set of nonnegative elements are closed under ...
user107952's user avatar
  • 21.5k
5 votes
0 answers
69 views

Can a unique factorisation domain have a largest prime?

Suppose $R$ is a UFD and $(R,\leq)$ is an ordered ring. Is it possible that $R$ has a largest prime element? Below is my attempt so far to answer this myself, though I'm still unsure what the ...
h4tter's user avatar
  • 499
1 vote
0 answers
48 views

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
  • 333
0 votes
0 answers
21 views

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
323 views

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
  • 2,005
1 vote
0 answers
57 views

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
  • 279
2 votes
1 answer
67 views

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
  • 327
1 vote
1 answer
34 views

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
  • 1,352
3 votes
2 answers
104 views

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
  • 321
0 votes
1 answer
69 views

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}...
user829347's user avatar
  • 3,440
8 votes
2 answers
304 views

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
  • 3,440
11 votes
2 answers
192 views

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
  • 232
0 votes
1 answer
34 views

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
2 answers
342 views

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
1 vote
1 answer
55 views

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
  • 279
0 votes
2 answers
61 views

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
  • 13.9k
3 votes
2 answers
577 views

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
  • 454
3 votes
1 answer
53 views

$a^+$ 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
  • 1,235
1 vote
1 answer
2k views

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]?
user avatar
23 votes
1 answer
1k views

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
  • 30.1k
5 votes
1 answer
162 views

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
0 votes
1 answer
45 views

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
  • 5,125