All Questions
1,123 questions
12
votes
3
answers
1k
views
construct scheme from quivers?
I heard from some guys working in noncommutative geometry talking about the idea that one can construct the noncommutative space from quivers. I feel it is rather interesting. However, I can not image ...
- 1,305
12
votes
1
answer
624
views
Stone–Čech compactification as a semigroup
Let $G$ be a topological group (we can assume that $G$ is countable and discrete) and let $\beta(G)$ be the Stone–Čech compactification of $G$. It is known that $\beta(G)$ can be turned into a left ...
12
votes
2
answers
785
views
Is the Petersen graph a "Cayley graph" of some more general group-like structure?
The Petersen graph is the smallest vertex-transitive graph which is not a Cayley graph. Is it the "Cayley graph" of some slightly more general group-like structure?
- 1,947
12
votes
1
answer
494
views
Tensor products and two-sided faithful flatness
Let $f: R \to S$ be a morphism of Noetherian rings (or more generally $S$ can just be an $R-R$ bimodule with a bimodule morphism $R \to S$). Suppose $f$ is faithfully flat on both sides, so $M \to M \...
- 37k
12
votes
1
answer
744
views
Is the following construction of the 0-Hecke monoid (well) known?
Let W be a Coxeter group with Coxeter generators S. The corresponding 0-Hecke monoid H(W) has generating set S, the braid relations of W and the relations that each element of S is an idempotent. If ...
- 38.6k
12
votes
1
answer
1k
views
A problem in commutative algebra whose solution requires algebraic geometry (resp., noncommutative algebra)?
One can argue that commutative algebra is affine algebraic geometry. However, a great deal of commutative algebra generalizes to non-commutative algebra, and in that setting there is little geometry, ...
- 4,985
12
votes
0
answers
542
views
Does Wedderburn's Little Theorem hold constructively?
Wedderburn's Little Theorem states that every finite division ring is commutative. Perhaps even more surprising, this implies that every finite reduced ring is commutative.
The proofs that I am aware ...
- 63.1k
12
votes
0
answers
321
views
Combinatorial proof of invertibility of a symmetric matrix associated to the ring of matrices over a finite field
Let $F$ be a finite field of $q$ elements with characteristic $p$. Let $M_n(F)$ be the ring of $n\times n$ matrices over $F$. We define a $q^{n^2}\times q^{n^2}$ symmetric matrix $L$ over the ...
- 38.6k
12
votes
0
answers
185
views
Hopf-Galois extensions where the "extension" is a module?
For $H$ a Hopf-algebra, an $H$-Hopf-Galois extension is a map of rings $\phi\colon\thinspace A\to B$ such that $H$ coacts on $B$ over $A$, $B\otimes_AB\cong B\otimes H$, and the cofixed points, or the ...
- 10.4k
12
votes
0
answers
267
views
Finitely generated skew-fields
There is a well known theorem saying that a commutative field that is finitely generated as a ring has to be finite (Kaplansky).
Is the same true for non-commutative "fields" (usually called ...
12
votes
0
answers
533
views
Does there exist a Noetherian ring of finite injective dimension but higher Krull dimension?
Definition: a (not necessarily commutative) left and right Noetherian ring $R$ is said to be Auslander-Gorenstein if
(i) $R$ has finite left and right injective dimension (in which case it turns out ...
- 221
12
votes
0
answers
443
views
Nullstellensatz for quaternionic plane curves?
By a quaternionic plane curve I mean the zero locus of a noncommutative polynomial in two variables, $x$ and $y$ say, over ${\Bbb H}$, Hamilton's quaternions. It is evidently well-known that, after ...
- 17.6k
11
votes
6
answers
1k
views
Computing the structure of the group completion of an abelian monoid, how hard can it be?
Cherry Kearton, Bayer-Fluckiger and others have results that say the monoid of isotopy classes of smooth oriented embeddings of $S^n$ in $S^{n+2}$ is not a free commutative monoid provided $n \geq 3$. ...
- 44.4k
11
votes
3
answers
1k
views
The concept "conjugate class" in monoids.
Is there any concept in monoids that is similar to the concept "conjugate class" in groups? For example, are there any such similar concept in symmetric inverse monoids? Thank you very much.
- 6,201
11
votes
3
answers
908
views
Does Morita theory hint higher modules for noncommutative ring?
Two possibly noncommutative rings are called Morita equivalent if their left-module categories are equivalent. In the commutative case, Morita equivalence is nothing more than ring isomorphism. ...
- 5,230
11
votes
3
answers
942
views
What is the smallest variety of algebras containing all fields?
A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative ...
- 2,464
11
votes
1
answer
4k
views
Are there any finitely generated artinian modules that are not Noetherian?
It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example ...
- 2,027
11
votes
1
answer
520
views
Problems concerning subspaces of $M_{n}(\mathbb{Q}) $
Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.Then if all the non-zero matrices in $N$ are invertible, what is ...
- 923
11
votes
2
answers
574
views
Identifying a group without 2-torsion
Suppose we have a finitely presented group $G$ with solvable word problem. (For instance, the command RWSGroup in Magma terminates giving us a finite [but possibly gigantic] rewrite system.) Is there ...
- 18.7k
11
votes
2
answers
950
views
Define Turing machine with algebraic concepts/structures
Usually, during lectures Turing Machines are firstly introduced from an informal point of view (for example, in this way) and then their definition is formalized (for example, in this way).
Is it ...
user60665
11
votes
3
answers
1k
views
Existence of non-commutative desingularizations
Let $R$ be normal, local ring of dimension at least $2$. Let $M$ be a reflexive $R$-module and let $A=Hom_R(M,M)$. Suppose $A$ has finite global dimension. Then one can view $A$ as a weak non-...
- 30.5k
11
votes
1
answer
629
views
Are all separable algebras Frobenius algebras?
Let $\mathcal C$ be a [added later: semi-simple] tensor category, and let $A=(A,m:A\otimes A\to A,i:1\to A)$ be an algebra object in $\mathcal C$.
The algebra is...
Separable if there is an $A$-$A$-...
- 43.2k
11
votes
1
answer
949
views
Magma "actions" (or alternatively, "What is the Yoneda lemma for magmas?")
Arguably the most import thing about groups, semigroups and more generally categories, is that they can act on sets (or even collections of sets in the case of a category). This is the basis for all ...
- 2,392
11
votes
1
answer
740
views
Determinants of octonionic hermitian matrices
For quaternionic hermitian matrices (i.e. quaternionic square matrices $(a_{ij})$ satisfying
$a_{ji}=\bar a_{ij}$) there is a nice notion of (Moore) determinant which can be defined as follows.
...
- 21.8k
11
votes
0
answers
427
views
Is there a theory of completions of semirings similar to $I$-adic completions of rings?
Let $L = \text{Con } (\mathbb{N}, 0, +) \setminus \Delta$ be the lattice of monoid congruences on the naturals, excluding the trivial congruence. As it happens, every $\theta \in L$ is the meet of ...
- 621
11
votes
0
answers
286
views
Does every finite poset have a rigid endomorphism?
Crossposted on Mathematics.
In this post, an order-preserving self-map of a poset $X$ will be called an endomorphism of $X$, and such an endomorphism $f$ will be called rigid if the only automorphism ...
- 4,416
11
votes
0
answers
265
views
Criteria for a map of rings to induce an equivalence on K-theory?
Algebraic $K$-theory is Morita invariant, but surely it does not detect Morita equivalence. What are some examples of rings (or ring spectra) $R$ and $S$ that are not Morita equivalent, but ...
- 331
11
votes
0
answers
214
views
Is it decidable if a tree-presented semigroup contains an idempotent?
A semigroup presentation $\langle A | R\rangle$ is called tree-like if every relation has the form $ab=c$, $a,b,c$ are in $A$ and if two relations $ab=c, a'b'=c'$ belong to $R$, then $c=c'$ if and ...
user6976
10
votes
5
answers
661
views
Is there a $3$-commutative algebra?
Let $k$ be a field of characteristic $0$. If $m\ge2$, I denote $P_m$ the standard polynomial in $m$ non-commutating indeterminates:
$$P_m(X_1,\dotsc,X_m)=\sum_{\sigma\in\mathfrak S_m}\epsilon(\sigma)...
- 52.3k
10
votes
2
answers
1k
views
Connective spectra versus simplicial abelian groups - very basic question
Hello,
I have very general , "introductory" questions (It is quite hard for me to seek for specific things in the algebraic topology literature).
I guess that connective spectra have a model ...
- 5,562
10
votes
3
answers
2k
views
nth term in the Baker-Campbell-Hausdorff formula
I am trying to prove a result for which I need the nth term of the Baker-Campbell-Hausdorff formula. I came at this particular result (which is not of significance for the question, but mentioning for ...
- 245
10
votes
4
answers
2k
views
Strongly Noetherian property. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian rings $A$ and $B$?
Let $k$ be a field. It is well-known that $A\otimes_{k}B$ is not necessarily Noetherian even if $k$-algebras $A$ and $B$ are Noetherian. For example $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$.
When ...
- 1,663
10
votes
3
answers
1k
views
Dual of a bimodule
For a noncommutative ring $R$, and an $R$-$R$-bimodule $B$, is there a "correct/natural" notion of a dual bimodule? I am interested, really, when $B$ is projective as a left $R$-module.
Note: ...
10
votes
1
answer
422
views
Generalized cancelation properties ensuring a monoid embeds into a group
Context: an obvious necessary condition for a monoid to embed into a group (as submonoid) is to satisfy the left and right cancelation rules:
$$xy=xz \quad\Longrightarrow y=z;$$
$$yx=zx \quad\...
- 63.9k
10
votes
1
answer
221
views
Matrix ring isomorphisms of different sizes
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 $T$?
- 18.7k
10
votes
1
answer
2k
views
Who invented Monoid?
I was trying to find (and failed) the original author of either
the concept of Monoid (set with binary associative operation and identity)
the name (which sounds french ? and also Dioid (for what ...
- 203
10
votes
1
answer
409
views
Does every set have a rigid self-map?
The question was asked on Mathematics Stackexchange
but has remained unanswered so far.
A self-map is a map $f:X\to X$ from a set $X$ to itself. There is an obvious notion of morphism, and thus of ...
- 4,416
10
votes
1
answer
746
views
Division algebras in which every proper subfield is maximal
I have a (noncommutative) division algebra D which is finite dimensional over its center F. I know that every subfield of D which contains F properly is a maximal subfield of D. What can we say about ...
- 279
10
votes
2
answers
716
views
On functors preserving monoid objects
If $C$ is a monoidal category, we can define the category $Mon(C)$ of monoids in $C$; call $U_C : Mon(C) \to C$ the forgetful functor. I'm interested in functors between categories of monoids:
...
- 613
10
votes
1
answer
579
views
Group completion of topological monoids
Let $M$ be an abelian monoid. For sake of simplicity we shall assume that in $M$ the cancellation law holds true. With this last assumption we define the group completion $G$ of $M$ as $$G:=M\times M/\...
- 1,252
10
votes
1
answer
1k
views
Explicit isomorphism for quaternion algebras over $\mathbb{Q}$?
It is known that the isomorphism class of a quaternion algebra $A=\binom{a,b}{K}$ over a number field $K$ is determined by the finite set of places $v$ of $K$ where $A\otimes_K K_v$ is a division ...
- 2,851
10
votes
5
answers
1k
views
On the notion of partial semigroup
A partial binary operation on a set $X$ is just a (partial) function $\varphi: X \times X \rightharpoonup X$ (I'm using \rightharpoonup for partial maps), and a partial magma is a pair $\mathbb M = (M,...
- 10.5k
10
votes
1
answer
673
views
Given any finite relation $R$ what is the cardinality of $\langle R\rangle=\{\underbrace{R\circ R\cdots \circ R}_{n\text{ times}}:n\in\mathbb{N}\}$?
Given any finite relation $R$ if we let $\circ$ denote relation composition and define $R^n=\underbrace{R\circ R\cdots \circ R}_{n\text{ times}}$ then does there exist an explicit formula for the ...
- 2,506
10
votes
1
answer
274
views
A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids
In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following.
...
- 38.6k
10
votes
1
answer
300
views
NCG with all noncommutativity in a nilpotent ideal
While in general non-commutative geometry behaves rather differently from commutative geometry when it comes to local-to-global properties (descent), there are versions of "mild" noncommutative ...
- 19.8k
10
votes
1
answer
440
views
Reference for a generalization of Γ-spaces to monoidal model categories
Γ-spaces were introduced by Segal in 1969 as models for what can be now described
as commutative ∞-monoids and ∞-groups in cartesian symmetric monoidal ∞-categories, e.g., E_∞-spaces and connective ...
- 37.8k
10
votes
2
answers
752
views
Adding a formal inverse of an element to a free monoid
Let $FM_2=\langle a,b\rangle$ be the free monoid of rank 2. If we add a formal inverse to the word $aba$, we get the free group $F_2$ (because both $a$ and $b$ will have inverses).
Question: For ...
user6976
10
votes
1
answer
355
views
Is Zariski closure of finitely generated matrix semigroup computable?
In general, can the Zariski closure of the semigroup of matrices $\langle M_1, \ldots, M_k \rangle$ be algorithmically computed (at least in theory)?
For this purpose I'm happy to assume the ...
- 163
10
votes
1
answer
243
views
Can a semigroup with zero be globally isomorphic to a semigroup without zero?
This is not a great question for sure and it may even be trivial for all I know, but a couple of years ago, when I still thought I'd be a mathematician, I spent quite a lot of time thinking about it ...
- 1,445
10
votes
2
answers
444
views
Iterated sumset inequalities in cancellative semigroups
This question is motivated by the following well-known theorems:
Thm (Plünnecke): If $A$ is a finite nonempty subset of an abelian group, then for every $n$ we have $|A^n| \le \frac{|AA|^n}{|A|^n}|A|$...
- 8,688