Skip to main content

All Questions

Filter by
Sorted by
Tagged with
3 votes
0 answers
96 views

Cohn's localization for rings with enough idempotents

I am in the following situation: I have a non-unitary (associative) ring $R$ with enough idempotents or, if you prefer, a small pre-additive category. Actually, I even know that $R$ is right coherent (...
Simone Virili's user avatar
4 votes
1 answer
298 views

What is the extended centroid of a free algebra?

For a prime ring $R$, you can define its "Martindale ring of quotients" $Q(R)$. See for example: Martindale, Wallace S. III, Prime rings satisfying a generalized polynomial identity, J. ...
Nick's user avatar
  • 213
6 votes
1 answer
372 views

Cohn localization examples

I'm working on my master's thesis, part of which involves an exposition on Cohn localization. (nlab discussion) In Free ideal rings and localization in general rings, Sec 7.4, Cohn gives a ...
Tyler's user avatar
  • 161
1 vote
0 answers
82 views

Localizing prime ideals over Noetherian rings

Let $R$ be a prime Noetherian ring which is not necessarily commutative. Consider the two natural ways to "extend" $R$: $R[x]$ and $M_n(R)$, polynomials over $R$ and $n$ by $n$ matrices over $R$ ...
user304582's user avatar
4 votes
1 answer
113 views

If $\{f\in R[x]\:|\:f\text{ monic}\}$ is a right denominator set, is $\{f^i\:|\:i\geq 0\}$ a right denominator set also?

Let $R$ be a right (and left) Noetherian ring and $T=R[x]$ its polynomial ring. It was shown by Stafford that the set $S=\{f\in T\:|\:f\text{ monic}\}$ is a right denominator set. So my question is, ...
Sam Williams's user avatar
0 votes
1 answer
183 views

Right localization of $R[x,x^{-1}]$ at monic $f\in R[x]$

Let $R$ be a right Noetherian ring and $S=\{f\in R[x]\;|\;f\text{ monic}\}$. It is a result of Stafford that $S$ is a right denominator set in $R[x]$, so in particular we can localize $R[x]$ at any $f\...
Sam Williams's user avatar
2 votes
1 answer
105 views

Localising a right Noetherian ring at a set of regular elements

Let $R$ be a right Noetherian ring, and $S$ a multiplicative set consisting of regular elements where $1\in S$ and $0\not\in S$. Does the right ring of fractions $RS^{-1}$ exist? This is what I know ...
Sam Williams's user avatar
0 votes
0 answers
87 views

When does an automorphism extend to a localisation (noncommutative rings)

Let $R$ be a (not necessarily commutative) ring. Let $\tau$ be an automorphism of $R$. Consider the localisation of $R$ at a set of multiplicative elements which satisfy the ore condition, say $X$. ...
No1729's user avatar
  • 201
13 votes
5 answers
3k views

Noncommutative localization of a ring: complete construction

I've been looking for the following construction in the literature, but I've only been able to find (very) partial proofs or proofs of special cases. Let $R$ be a non-commutative ring and $S$ a ...
Steve's user avatar
  • 465