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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.

6 votes
2 answers
916 views

Simple Ore extensions

Let $R[x;\sigma,\delta]$ be an Ore extension, where $R$ is an associative and unital ring and $\sigma : R\to R$ is a (not necessarily injective!) ring endomorphism. (In the literature it is often assu …
Johan Öinert's user avatar
7 votes

Definition of a Grothendieck ring

Let $\mathrm{Var}_k$ denote the category of varieties over a field $k$. Then $K_0(\mathrm{Var}_k)$ is the free abelian group generated by symbols $[X]$ for $X\in \mathrm{Var}_k$, subject to the relati …
Johan Öinert's user avatar
0 votes
1 answer
392 views

Non-simple and non-unital rings with trivial centres

Let $R$ be an associative and non-unital ring. (Suppose that $R$ is $s$-unital, i.e. for each $x\in R$ there is $u,v\in R$ such that $ux=xv=x$.) It is not difficult to show that if $R$ is a simple ri …
Johan Öinert's user avatar
1 vote

Proof a Weyl Algebra isn't isomorphic to a matrix ring over a division ring

This is not an answer to the original question. However, it is related and I think that it is worth mentioning. Assuming that ring morphisms take identity elements to identity elements, we can show t …
Johan Öinert's user avatar
2 votes
0 answers
56 views

Non-singular rings which are Rickart

A ring $R$ is said to be a right Rickart ring if the right annihilator of any element in $R$ is of the form $eR$ for some idempotent $e \in R$. It turns out that a ring $R$ is right Rickart iff every …
Johan Öinert's user avatar
64 votes
4 answers
8k views

What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group. A couple of good r …
Johan Öinert's user avatar