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Tagged with localization noncommutative-rings
9 questions
3
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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 (...
- 2,452
13
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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 ...
- 63.9k
6
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1
answer
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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 ...
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4
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1
answer
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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. ...
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1
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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 $R$ ...
- 595
4
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1
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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, ...
- 149k
2
votes
1
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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 ...
- 3,545
0
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1
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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\...
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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 $X$. ...
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