0
votes
0answers
38 views

Supersets of P-finite sequences and rings

P-finite sequences are a superset of C-finite sequences. While doing programming work, the question came up what generalizations or supersets of P-finite sequences have people described. In other ...
6
votes
4answers
364 views

How universal is operadic approach to studying algebras?

I have just started to read about operads, so this question might be silly. So it seems to me that any "reasonable" class of algebras can actually be defined as a class of all algebras over a certain ...
8
votes
1answer
335 views

Generalizing detropicalization

Given an identity in max,plus arithmetic, are there ways to turn it into an ordinary algebraic identity it other than by replacing addition by multiplication and replacing max by series-plus or by ...
2
votes
1answer
456 views

Commutative associative rational binary operations

What are all the nondegenerate rational binary operations that are commutative and associative? (Examples: $(x,y) \mapsto x+y$, $xy+x+y$, $xy/(x+y)$.) Feel free to re-tag if you can think of ...
3
votes
2answers
481 views

Shape of axioms in abstract algebra

When defining abstract algebraic structures (like monoids, groups, etc...), are there some constraints on the shape of the axioms, for the structure to have good properties that we implicitly use in ...
11
votes
1answer
323 views

$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?

Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by finite group. Denote the ...
0
votes
1answer
169 views

Find a special element in group algebra

Let $$G=\langle x, y, z\mid xyx^{-1}=zy, xzx^{-1}=z, yz=zy\rangle,$$ denote $l^1(G)^{\times}$ to be the set of units in $l^1(G)$, which we have considered as a ring with multiplication defined by the ...
0
votes
0answers
69 views

Example of a ring whose minimals are annihilators of idempotents?

I'm looking for examples† of rings with the property that for each $P={\rm Ann}_R(a)\in{\rm Min}(R)$ then $a\in R$ is idempotent (ie $a^2=a$) † other than domains!
1
vote
2answers
339 views

Semiring naturally associated to any monoid?

For any monoid $M$, we can naturally construct a semiring $S$ as follows: Let the additive monoid of $S$ be the free commutative monoid on $M$ Let the multiplicative monoid of $S$ be $M$ Then, if ...
12
votes
1answer
631 views

Number of idempotent $n\times n$ matrices over $\mathbb{Z}_m$ ?

Is there any known formula for the number of idempotent $n\times n$ matrices over $\mathbb{Z}_m$ ? The number of idempotent matrices over a finite field is well-known and since we can decompose $m$ ...
1
vote
1answer
294 views

Question about the shape of the dual module of a module

Hi, I am reading this paper and have the following question:. On page 5 you can see different types of $A_n$-modules ($\ A_n=k[x,y]/\langle x^2,y^{n+2},xy^{n+1}\rangle\ $)$\ $ and later in the paper ...
6
votes
3answers
1k views

Why are ring actions much harder to find than group actions?

I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia: A module is a ring action on an abelian group. ...
2
votes
1answer
161 views

Bimodule version of IBN

Hello all, Does anyone have an example in mind of a ring $R$ for which $R^n\cong R^m$ as $R,R$ bimodules for some positive integers $n\neq m$? I would be a little surprised if someone showed no such ...
3
votes
0answers
105 views

Making an algebra the uniqe maximal one-sided ideal in another unital algebra

If $R$ is an algebra without a unit, then the standard unitisation $R^\sharp$ can have maximal one-sided ideals other than $R$. Thus, it is natural to ask about the following. Let $R$ be an algebra ...
5
votes
1answer
395 views

Left ideals vs right ideals

By default, let all algebras be complex and unital. I am concerned with the non-commutative algebras. I am wondering if the following might be true (at least for some classes of algebras, like ...
1
vote
1answer
170 views

Ring of a Spectral Space

It is said, as far as I can tell that an arbitrary spectral space, i.e. a space that is $T_0$, sober and quasi-compact whose collection of quasi-compact open sets forms a basis and is closed under ...
5
votes
0answers
274 views

Can any group be realized as the multiplicative group of a ring? [duplicate]

Possible Duplicate: Ring with Z as its group of units? Given a group $G$, does there always exist a ring $R$ such that $R^\times \cong G$? I feel like this isn't true but that's just a hunch. ...
6
votes
3answers
836 views

Does “finitely presented” mean “always finitely presented”, considered in general

I'm wondering about the question, "If we have a finitely presented __, is it necessarily finitely presented with respect to any finite generating set for it?" I know this is true for groups and for ...
12
votes
4answers
3k views

Unique factorization domains

Let $c$ be an integer, not necessarily positive and $|c|$ not a square. Let $\mathbb{Z}[\sqrt{c}]$ be the set of complex numbers $$a+b\sqrt{c}, a, b\in \mathbb{Z},$$ which form a subring of the ring ...
18
votes
6answers
2k views

Consequences of not requiring ring homomorphisms to be unital?

As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1_R)=1_S$. But there has been a minority who do not require this, one prominent example ...
1
vote
2answers
425 views

Understanding the modules of semiprimitive rings

As far as I understand, a semiprimitive ring can be fully 'explored' by its simple modules, in the sense that a semiprimitive ring is the subdirect product of its simple modules (for brevity, I'll use ...
14
votes
7answers
3k views

What are the reasons for considering rings without identity?

I think a major reason is because Lie algebras don't have an identity, but I'm not really sure.
9
votes
2answers
1k views

Galois group of a product of polynomials

How can I compute the Galois group of the polynomial $fg\in K[x]$ assuming that I know the Galois groups of $f\in K[x]$ and $g\in K[x]$? Let's suppose for simplicity that the field $K$ is perfect.
39
votes
5answers
3k views

when is A isomorphic to A^3?

this is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$. is then $A$ isomorphic to $A^2$? probably no, but how construct a ...