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Results tagged with ra.rings-and-algebras
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user 3199
Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
5
votes
Accepted
Finitely generated projective modules over matrix rings
The ring $\mathbb{M}_n(\mathbb{C})$ is a semisimple ring, and so every module is a sum of simple modules, and is projective. For this ring, there is only one simple module $S$, up to isomorphism. T …
- 18.7k
6
votes
Accepted
Descending chain condition in noncommutative rings
I'm somewhat surprised this question hasn't been answered previously. It turns out DCC on two-sided ideals does not imply ACC on two-sided ideals.
Let $V$ be a $k$-vector space with a basis of size …
- 18.7k
1
vote
Binary operation on subsets of rings
Any set $T$ has a binary operation on it that makes the set into a semigroup! Namely $x\ast y=x$, the "choose the left" operator. (It is easy to check that this operation is associative.)
Only slig …
- 18.7k
2
votes
Other Ring Structures on $\mathbb{Q}$
The answer is yes (unless I made a mistake somewhere).
For example, you can replace addition with the operation $a\oplus b=a+b-1$. This is a commutative, associative binary operation with identity $ …
- 18.7k
3
votes
Accepted
Ehrlich's Characterization of Unit Regular Elements
Manny's answer is absolutely correct. I wanted to add that Ehrlich's argument, which in its original formulation requires the endomorphism ring to be von Neumann regular, actually characterizes unit- …
- 18.7k
4
votes
Left quasi-inverse elements: motivation
Consider an equation like $(1-b)(1-a)=1$, which says that $1-a$ has a left inverse $1-b$. Multiplying through, and cancelling the $1$'s, we are left with $a+b=ba$. This expresses essentially the sam …
- 18.7k
3
votes
Example of noncommutative central reduced rings which is not reduced
The answer is yes, and there are many ways to do it.
Use Jeremy Rickard's direct product construction.
Let $F$ be a field, let $R=F[x\ :\ x^2=0]$, and let $S=R\langle y,z\rangle$ be the extension of …
- 18.7k
3
votes
Bimodules of fractions
Suppose $R,S,T$ are rings. Let $_RM_S$ and $_SN_T$ be bimodules. The tensor product $M\otimes_S N$ is naturally an $R$-$T$-bimodule. The "middle" $S$-structure is gone.
In your case, $E$ is an $R$ …
- 18.7k
2
votes
Proving that a semigroup is regular
If your monoid is defined in terms of generators and relations, a simple tool you can use to check whether elements have quasi-inverses of the specified form is Bergman's diamond lemma. For instance, …
- 18.7k
3
votes
Semiprime (but not prime) ring whose center is a domain
For those less familiar with Lie algebras here is a somewhat more prosaic example, with the added benefit that the ring is reduced. Let $F$ be a field, and take the free algebra $R:=F\langle a,b,c\ : …
- 18.7k
1
vote
Kaplansky's unit conjecture and unique products
In this preprint of William Carter from 2013, it appears that there are very few known classes of torsion-free groups which are not unique-product groups, and that for these groups the unit conjecture …
- 18.7k
0
votes
Artin Jacobson-semisimple rings are semisimple. Constructively, too?
I'm not as familiar with constructive proofs, so this "answer" is really just a couple questions.
Consider the ring $R=\mathbb{M}_2(\mathbb{F}_2)$, and the element $r=\begin{pmatrix} 0 & 1\\ 0 & 0\en …
- 18.7k
1
vote
When is the essential extension commutes with colimits(or push forward)
First, I want to point out that in general there is no surjection from a direct product of copies of $R$ to an arbitrary module $M$. For example, if $R=\mathbb{Z}$ and $M=\mathbb{Z}^{(\omega)}$ (a co …
- 18.7k
5
votes
Making a non-unital algebra the unique maximal one-sided ideal in a unital algebra
The answer to this question is no in general. Let $K$ be a field, and let $R=F\{s,t\ :\ st=s+t \}$, the non-unital algebra generated by the non-commuting variables $s,t$ subject to the single relatio …
- 18.7k
3
votes
How many multiplications are needed to generate a matrix algebra?
My original answer missed the word "span" everywhere, sorry!
The answer to your second question is no. Take $$A=\begin{pmatrix}0 & 0 & 0\\ 1& 0 & 1\\ 0& 0 & 0\end{pmatrix}, B=\begin{pmatrix}0 & 0 & …
- 18.7k