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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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 $ …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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, …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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]/ …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar
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 …
Qiaochu Yuan's user avatar

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