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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.
7
votes
Irreducible elements in endomorphism rings
Take $G = \mathbb{Q}/\mathbb{Z}$; then $\text{End}(G)$ is the profinite integers
$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$
where $\mathbb{Z}_p$ is the $p$-adic integers. The element $\pro …
6
votes
Accepted
Weyl algebra and its nontriviality
Generally the strategy for showing that some syntactic construction is nontrivial is to find a semantic model of it. E.g. the strategy for showing that some relations don't force a group to be a trivi …
17
votes
What is an example of a ring with two (or more) multiplicative right-identities?
Take the semigroup ring of a semigroup with two or more multiplicative right identities. For example, the semigroup
$$S = \langle a, b | ab = aa = a, ba = bb = b \rangle$$
works (it is the universal e …
9
votes
Ring with three binary operations
The claim that there are two binary operations on rings is misleading. Rings are actually equipped with countably many $n$-ary operations, one for each noncommutative polynomial in $n$ variables over …
11
votes
Accepted
A group algebra isomorphism problem
This is true iff $G$ is finite and abelian, the characteristic of $K$ does not divide $G$, and $K$ has all $n^{th}$ roots of unity whenever $G$ has an element of order $n$. Hopefully it is clear why $ …
19
votes
Accepted
Purely noncommutative algebra-Morita equivalence
An algebra is Morita equivalent to a commutative algebra iff it's Morita equivalent to its center, since the center is Morita invariant. So any representative of a nontrivial class in the Brauer group …
15
votes
Accepted
Left-Module Structure on the Tensor Product ofTwo Left Modules
Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product …
1
vote
Fractional powers of Dirichlet series?
Alright, let me try to justify the assertions I made in the comments yesterday. The formal argument is a bit tiresome to write out, but the basic idea of the proof goes like this: suppose we are give …
4
votes
Representation of rings
Exercise 2 in Chapter 1 of Krylov, Mikhalev, and Tuganbaev's Endomorphism Rings of Abelian Groups asks to show that $\mathbb{F}_p \times \mathbb{F}_p$ is not the endomorphism ring of any abelian group …
3
votes
Intuitive Example of a Jacobson Radical
Well, if R is a finitely generated commutative ring then J(R) is just the nilradical, so for example Z[x]/(x^2) has Jacobson radical (x).
The intuition I have about the nilradical (and by extension, …
7
votes
How to recognize a Hopf algebra?
I'm a bit late, but here's a simple observation. Consider a topological version of the question: given a topological space $X$, how can we recognize when $X$ can be given the structure of a topologica …
7
votes
2
answers
578
views
Which commutative algebras admit a nonzero Poisson bracket?
Let $A$ be a commutative algebra, not necessarily unital, over a field $k$ (of characteristic not equal to $2$, or even equal to $0$, if it helps). A second-order formal deformation of $A$ is a $k[h]/ …
12
votes
2
answers
917
views
What reasonable choices of morphisms are there for the category of Poisson algebras?
The first definition of the category of Poisson algebras that comes to mind is that a morphism between Poisson algebras is an algebra homomorphism that is also a Lie algebra homomorphism with respect …
9
votes
2
answers
617
views
Does any identity holding in all finite-dimensional Lie algebras hold in all Lie algebras?
Equivalently, is the free Lie algebra on finitely many generators over a fixed field $k$ (say of characteristic not equal to $2$) residually finite-dimensional in the sense that any nonzero element re …
13
votes
Accepted
Is it true that Nature promotes products?
You need to distinguish between "coproduct" and "comultiplication." The categorical coproduct is just a generalization of addition and is intuitive in many contexts.
Comultiplications are more inter …