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von Neumann regular ring homomorphisms

Let us call a ring homomorphism $f\colon R\rightarrow S$ von Neuman regular if it has the property that for every left $S$-module $M$, the left $R$-module $f^*M$ is flat. In particular, $\mathrm{id}...
nikola karabatic's user avatar
5 votes
0 answers
219 views

Constructing a noncommutative algebra from a commutative algebra

I was told at a conference that one way to construct a noncommutative algebra from a commutative one is to "replace the product of finite spaces (which on the level of continuous functions corresponds ...
TerryL's user avatar
  • 111
5 votes
0 answers
104 views

Finitely generated submodules of projectives lie inside f. g. projectives?

Let $R$ be a (not necessarily commutative) ring. If $M$ is a finitely generated submodule of a projective module $P$, is there a finitely generated projective submodule $P'$ such that $M \subseteq P'...
user124388's user avatar
5 votes
0 answers
208 views

Which groups can occur as the group of units of finite-dimensional noncommutative algebras?

This is a continuation of a previous question: Connectedness of groups of units in finite-dimensional commutative algebras. Let $k$ be an algebraically closed field of characteristic $0$. Which ...
M.G.'s user avatar
  • 7,127
5 votes
0 answers
64 views

Characters on monotone functions

Characters on the semigroup $(C_{+}^{b}(\mathbb{R}^{d}),+)$, i.e. on bounded positive continuous functions with the ususal pointwise addition, are known to be of the form $C_{+}^{b}(\mathbb{R}^{d})\ni ...
Tobsn's user avatar
  • 289
5 votes
0 answers
395 views

Derived tensor products and Tor of commutative monoids

Two commutative monoids $M,N$ have a tensor product $M\otimes N$ satisfying the universal property that there is a tensor-Hom adjunction for any other commutative monoid $L$: $$\text{Hom}(M\otimes N,L)...
John Berman's user avatar
5 votes
0 answers
264 views

Do almost commutative flat degenerations induce equality in K-theory? (Or: Is the characteristic variety actually a support of a class in $K$-theory?)

I intentionally phrased the title to match a different question which is almost identical to the one i'm asking. However similar, the answer there, which uses commutative algebraic geometry, is not ...
Saal Hardali's user avatar
  • 7,789
5 votes
0 answers
99 views

Zappa-Szép products of the monoid of integers with itself

Question What are all the functions $\alpha , \beta : \mathbb{N} \to \mathbb{N}$ satisfying the following functional equations? $\bullet ~~~~ \alpha(0)=0, \quad \beta(0)=0\\ \bullet ~~~ \...
HeinrichD's user avatar
  • 5,482
5 votes
0 answers
637 views

Unique product groups (and semigroups)

A group $G$ is called a u.p.-group (short for unique product group) if for all nonempty finite subsets $A,B\subseteq G$, there exists an element $g\in A \cdot B$ which can be uniquely written as a ...
Pace Nielsen's user avatar
  • 18.7k
5 votes
0 answers
317 views

Hochschild Cohomology of the Quantum Torus

I would like some advice on how to compute directly, or by a higher powered method the Hochschild Cohomology groups of the quantum torus using the stated complex I have found. I think there are ...
No1729's user avatar
  • 201
5 votes
0 answers
134 views

Projective dimension of ring over its center

If $A$ is a ring and $Z(A)$ is its center then what is a sufficient condition for the projective dimension of $A$ over $Z(A)$ (ie: $pd_{Z(A)}(A)$) to be finite? (Assuming that $A\neq Z(A)$).
ABIM's user avatar
  • 5,405
5 votes
0 answers
245 views

Orders of Clifford algebra

Let $C_n$ be the Clifford algebra over $\mathbb{Q}$ associated to negative definite quadratic form $-I_n$ (i.e. $-x_1^2-\dots-x_n^2$). Let $\mathcal{O}$ be a $\mathbb{Z}$-order of $C_n$. Q1) Is it ...
Subhajit Jana's user avatar
5 votes
0 answers
137 views

Pseudovarieties of monoids

All (pseudo)varieties considered here are (pseudo)varieties of monoids. It is known that any (finite or infinite) monoid that satisfies the identities \begin{equation} xhxyty = xhyxty, \quad xhytxy=...
E W H Lee's user avatar
  • 563
5 votes
0 answers
295 views

Orbit-Stabilizer theorem for continuous groups

The orbit-stabilizer relationship (also known as the orbit-stabilizer theorem) is very clear for finite groups. Is there an equivalent relation for continuous groups? Also, is there a similar notion ...
Arnab's user avatar
  • 615
5 votes
0 answers
843 views

Chinese remainder theorem

For non-commutative rings, we have this generalization of the Chinese remainder theorem (CRT). I wonder if there is another statement involving only left or right ideals; do you know any?
Exodd's user avatar
  • 59
5 votes
0 answers
246 views

Injectivity criterion for surjective coalgebra maps: does it hold in full generality?

Let $\mathbf{k}$ be a commutative ring. Let $C$ be a filtered $\mathbf{k}$-coalgebra. This means a $\mathbf{k}$-coalgebra equipped with an increasing $\mathbf{k}$-module filtration $C^0 \subseteq C^1 \...
darij grinberg's user avatar
5 votes
0 answers
578 views

Central Element in Sklyanin Algebras?

I'm interested in Sklyanin Algebras or Artin-Shelter regular algebras of type A. These are generated in degree 1 by three variables x,y,z, and have three defining relations in degree 2, which you can ...
euklid345's user avatar
  • 807
5 votes
0 answers
442 views

A reference on semisimple linear algebra

Is there any literature where the tools familiar from (multi)linear algebra are systematically transferred to the setting of semisimple modules over noncommutative rings? In fact this question is a ...
Alexander Shamov's user avatar
5 votes
0 answers
918 views

Commutator Baker-Campbell-Hausdorff formula

Consider the Baker-Campbell-Hausdorff formula $\Phi(X,Y)\in\mathbb{Q}\langle\!\langle X,Y\rangle\!\rangle$ in non-commutative variables. Define $X*Y:=\Phi(X,Y)$ and $[X,Y]=(-X)*(-Y)*X*Y$, and then (as ...
Diego Sulca's user avatar
5 votes
0 answers
2k views

Is the radical of a homogeneous ideal homogeneous?

Let $S$ be an $M$-graded $R$-algebra, where $M$ is some monoid, and $I\subset S$ an homogeneous ideal. The original, naïve, question, was: is it true that $\sqrt{I}$ is homogeneous? In this generality,...
quim's user avatar
  • 1,811
5 votes
0 answers
350 views

Chain/Hierarchy of Monoids

Let's assume that we have the following collection of structures: Some space $P$. Monoids $(M_{i+1},\circ_{i+1})$, and Actions $\bullet_{i+1}:M_{i+1}\times M_i\to M_i$, for $i\ge 0$ And $\bullet_{0}:...
supercooldave's user avatar
5 votes
1 answer
226 views

Examples of noncommutative Bezout domains

I would like to see some (or many!) examples of noncommutative Bezout domains (one-sided principal ideals sum to one-sided principal ideals). I've read somewhere that it's not easy to find an example ...
4 votes
4 answers
1k views

Why do we choose the standard total order on the integers?

I understand why the set of natural numbers $\mathbb N = \{ 0, 1, 2, \cdots \}$ is equipped with a total order. Indeed, every monoid has a pre-order, where $$n' \succeq n \quad \mathrm{if~and~only~if} ...
Tom LaGatta's user avatar
  • 8,512
4 votes
4 answers
961 views

Homological dimension of a graded ring which is like polynomial ring

Let $k$ be a field of characteristic $0$. Consider the following $k$-algebra $R$, which is the quotient of a tensor algebra generated by elements $x_i$ in degree $1$ with the relation $x_ix_j=-x_jx_i$...
Daniel Pomerleano's user avatar
4 votes
3 answers
3k views

Finitely generated projective = finitely presented flat over a noncommutative Noetherian ring

Let $R$ be a possibly noncommutative left Noetherian ring and $M$ an $R$-module. I am looking for a reference or a proof for the following fact: $M$ is finitely generated and projective if and only if ...
Question Mark's user avatar
4 votes
2 answers
1k 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 ...
Mike Battaglia's user avatar
4 votes
3 answers
567 views

Gröbner/SAGBI bases for non-commutative setting

It is well known that SAGBI/Gröbner bases are important for commutative and non-commutative algebra. The references for commutative scenery is ample and vast, but I am in trouble to find a good ...
Binai's user avatar
  • 829
4 votes
1 answer
417 views

Is a solvable group satisfying a semigroup law?

Let $S$ be the free semigroup on the set $\{x_1,\ldots ,x_n\}$, where $n$ is a positive integer. Suppose that $\mu=\mu (x_1,\ldots ,x_n)$ and $\nu = \nu (x_1,\ldots ,x_n)$ are two elements in $S$. We ...
mahdi meisami's user avatar
4 votes
2 answers
586 views

Brauer group of $\mathbb{Z}_{(p)}$

This may be a well known result but I could not find it in the standard references. What is the Brauer group of the local ring $\mathbb{Z}_{(p)}$ (the ring of integers localized at $p$)?
user123's user avatar
  • 81
4 votes
3 answers
1k views

Set of invertible operators in B(H) is connected. Is it true? Is there a reference?

Suppose $H$ is a Hilbert space, $B(H)$ is the algebra of bounded linear operators on it, $K(H)$ is ideal of compact operators in $B(H)$, $Inv(B(H)/K(H))$ is the topological group of invertible ...
Fiktor's user avatar
  • 1,284
4 votes
2 answers
1k views

strong nilpotent elements

An element x in a noncommutative ring R is strongly nilpotent if any chain $x_1=x, x_2, ... $, with $x_{n+1}\in x_n R x_n$ terminates at zero. It becomes clear why this is a good definition once one ...
Roman 's user avatar
  • 43
4 votes
2 answers
544 views

Membership problem in monoids

What is the simplest example of a monoid with undecidable membership problem? In other words, I'm looking for a concrete monoid $S$ such that there is no algorithm which takes elements $s_1,...,s_n$ ...
dan's user avatar
  • 41
4 votes
1 answer
222 views

Addition and Rudin-Keisler ordering in $\beta \omega$

$\DeclareMathOperator{\RK}{\mathrm{RK}}$Let $\beta\omega$ be the Stone-Cech compactification of $\omega$ with the discrete topology. We can endow $\beta\omega$ with an addition operation that extends ...
Dominic van der Zypen's user avatar
4 votes
2 answers
1k views

Semiprime (but not prime) ring whose center is a domain

The center of a prime ring is a domain and the center of a semiprime ring is reduced. Now I have no evidence to believe that if the center of a semiprime ring R is a domain, then R has to be a ...
carlos's user avatar
  • 279
4 votes
1 answer
270 views

Kaplansky inverse element theorem on group C-star algebra

In a class talking about $C^*$ algebra and (higher) index theory, I heard a theorem (related to Kaplansky, proved?), that is Suppose $\Gamma$ is a group (admitting Haar measure if necessary) while $\...
YOTAL's user avatar
  • 193
4 votes
3 answers
1k views

Is there a general notion of semigroup action?

The analogous result to Cayley’s theorem or Yoneda lemma in semigroup theory represents semigroups as semigroups of functions from a set to itself. This suggests that a semigroup action consists of a ...
Thomas Klimpel's user avatar
4 votes
1 answer
423 views

What is the formula for the commutative multiplication on CP(infinity)?

There is a classic formula for maps $\mathrm{CP}(r) \times \mathrm{CP}(s) \to \mathrm{CP}(r+s)$ or maybe $r+s+1$ using Plücker coordinates - IF memory serves. In the limit we get the abelian ...
Jim Stasheff's user avatar
  • 3,880
4 votes
1 answer
239 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, ...
Salvo Tringali's user avatar
4 votes
1 answer
196 views

Is the monoid of taking iterated images and inverse images freely generated by the image and inverse image operation?

Let $\mathcal{F}$ denote the class of all functions. Let $U,L:\mathcal{F}\rightarrow\mathcal{F}$ denote the mappings where if $f:X\rightarrow Y$, then $U(f):P(X)\rightarrow P(Y),L(f):P(Y)\rightarrow P(...
Joseph Van Name's user avatar
4 votes
1 answer
314 views

A semigroup with the property that $x^n = a$ has at least one solution

Is there a standard name for a (multiplicatively-written) semigroup $(A, \cdot)$ such that, given an arbitrary $a \in A$, the equation $x^n = a$ has at least one solution $x \in A$ for each $n \in \...
Salvo Tringali's user avatar
4 votes
1 answer
151 views

Three preprints and one manuscript of Tamura on power semigroups

I'm reading Takayuki Tamura's article "On the recent results in the study of power semigroups", pp. 191-200 in Goberstein & Higgins' Semigroups and Their Applications, Kluwer, 1987 (the ...
Salvo Tringali's user avatar
4 votes
1 answer
146 views

When is semigroup algebra local?

Let $G$ be a finite semigroup (or monoid if that helps) and $K$ a field. Question: When is the semigroup algebra $KG$ local? Here local means that there is a unique maximal right (or left) ideal. ...
Mare's user avatar
  • 26.5k
4 votes
1 answer
178 views

Can all finite-dimensional noncommutative algebras with trace be trace-preserving embedded into matrix rings?

Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I always find an embedding $f: R \to M_r(k)$ compatible ...
Asvin's user avatar
  • 7,746
4 votes
1 answer
307 views

Characterization of Archimedean linearly ordered monoids

In this question, it is shown that all Archimedean ordered groups are isomorphic to an ordered subgroup of $\mathbb R$. Additionally, it is shown that if such a group is complete, then it is ...
eyeballfrog's user avatar
4 votes
1 answer
243 views

number of indecomposable summands of an extension of two modules

I have the following question : in a Krull-Schmidt category (say the category of finite length left modules over a ring, this is the case which interests me), is it possible to relate the number of ...
hennlu's user avatar
  • 333
4 votes
1 answer
189 views

Some questions about homogroups

Every semigroup containing an ideal subgroup is called a homogroup. Let $(S,\cdot)$ be homomgroup, hence it contains an ideal $I$ that is also a subgroup. It is easy to see that $I$ is the least ideal,...
M.H.Hooshmand's user avatar
4 votes
1 answer
297 views

Reference for subsemigroups of $\mathbb{N}^n$

A well known result about the natural numbers $\mathbb{N}$ says that for any finite subset $A \subset \mathbb{N}$ there exists $R \ge 0$ such that if $n$ is in the subgroup of $\mathbb{Z}$ generated ...
Lee Mosher's user avatar
  • 15.4k
4 votes
1 answer
434 views

Regarding a new algebraic structure

By "left semigroup-joined-semigroup" I mean an algebraic structures $(S,\cdot,*)$ such that both $\cdot,*$ are associative, and the following property holds (see this ) $$ x*(y\cdot z)=x*y*z\;\; ; \;...
M.H.Hooshmand's user avatar
4 votes
1 answer
855 views

Left- and right-sided principal ideals of quaternions have same index?

One fact about the Lipschitz integers (quaternions of the form $a + bi + cj + dk$ where $a, b, c, d$ are integers) is that the left-sided ideal generated by any element $Q$ has the same index in the ...
Sridhar Ramesh's user avatar
4 votes
1 answer
454 views

The Jordan Plane and Enveloping Algebras

Let $k$ denote a field of characteristic $0$ (assume algebraically closed for convenience). Define $J=k\langle x,y|[x,y]=y^{2}\rangle$. This noncommutative algebra (which can be viewed as a derivation ...
Jordanblock's user avatar

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