Search Results

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 …

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\ : …

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 …

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 …

Ok, I think I worked out the bugs in my previous answer. Let $F$ be the field with two elements (for simplicity). Let $T=F< a_{i}, x_{k,i},y_{k,i}:i\in \mathbb{N}, k\in K>$ be the free algebra over …

Let $F$ be a field, and let $f_1,f_2,\ldots, f_k\in R:=F\langle\langle x,y\rangle\rangle$ with $k\in \mathbb{N}$. Order monomials in $R$ by degree, and then lexicographically. Since the question con …

Let $Z=\mathbb{Z}$ be the ring of integers. This is a noetherian domain, and so satisfies property (a). Let $A=\mathbb{Z}\langle x,y\ :\ x^2=2x,y^2=2y,xyx=2x,yxy=2y\rangle$. This is a (non-commutat …

You cannot always find such an embedding. Consider the ring $R=\mathbb{Q}\langle x,y\rangle$ subject only to the condition that any monomial in the letters $x$ and $y$ of degree $3$ is zero. This is …

Assume (C), so that for each $a\in R$, there exists some integer $n\geq 1$ (possibly depending on $a$) such that $a^n\in Z(R)$. The equivalence of conditions (1) and (3) in Theorem 12.11 from Lam's “A …

Lam informed me that, as far as he knew, this problem was still open. However, the example below shows that the condition is not left-right symmetric. Let $$ R=\mathbb{F}_2\langle a,b,c\, :\, a^2=ab= …

Do there exist (unital, associative, noncommutative) rings $R$ and $S$, where $\mathbb{M}_2(R)\cong \mathbb{M}_3(S)$, but these matrix rings are not isomorphic to $\mathbb{M}_6(T)$ for any ring $T$?

I know this question is old, and has an accepted answer, but this excellent paper by Manny Reyes gives some further thoughts about possible Spec's for noncommutative rings.

Consider $L=\mathbb{Z}_{(2)}$ (the set of rational numbers which in reduced form have denominators coprime to $2$). This is a local ring with ${\rm rad}(L)=2L$ and $L/2L\cong \mathbb{F}_2$. However …

The answer to your first question is yes (which was very surprising to me, to be honest). I have no idea whether the second question also has a positive answer. (By the way, don't let the work below …

From the apparent lack of complete answers in the literature, I figured I'd write the whole proof of the well-definedness of multiplication since it is only boring, not hard. I will write the equival …