All Questions
1,123 questions
4
votes
0
answers
172
views
When a ring is a polynomial ring?
In the paper (2.11) the authors show that if $k^*$ is a separable algebraic extension of $k$ and $x_1,x_2, \ldots, x_n$ are indeterminates over $k^*$ and a normal one dimensional ring $A$ with $k \...
- 271
4
votes
0
answers
85
views
An order in $\mathbb Q[G]$ which is a maximal $\mathbb Z_p$-order in $\mathbb Q_p[G]$ for finitely many primes $p$
Let $G$ be a finite group and $S$ a finite set of prime numbers. I know that every separable $\mathbb Q$-algebra $A$ contains a maximal $\mathbb Z$-order but I wonder if the following is true.
Is ...
- 255
0
votes
1
answer
198
views
Example of noncommutative central reduced rings which is not reduced
A ring $R$ is called central reduced if every nilpotent element is central. Ungor et al. math.RA 14 Dec 2013 has given an example of a commutative ring which is central reduced but not reduced. Can we ...
- 585
3
votes
1
answer
190
views
What is known about semigroups that are generated by (cyclic) subgroups?
A semigroup $(A,\cdot)$ that is idempotent (i.e. $a^2=a$ for every element $a\in A$) is naturally generated by its subgroups (every element on itself constitutes a trivial group). I would like to know ...
6
votes
0
answers
129
views
Center Picard group non-commutative algebra
I am wondering if there is a way to describe the center of the Picard group of a non-commutative algebra.
Namely, let $A$ be a finitely generated algebra over a field $k$. Denote by $\mathrm{Pic}(A)$...
- 7,300
2
votes
0
answers
448
views
Completion of an algebra
Based on arXiv:math/9802041v1, there is a definition for $NC$-filtration and $NC$-completion of an associated algebra over the complex numbers:
Let $R$ be an associative algebra and $R^{\rm Lie} = (R,...
- 51
3
votes
1
answer
404
views
How many monoids with $n$ arrows exist?
How many monoids with strictly $n$ arrows exist? Is this known? I ask this only out of curiosity. Looking at $n=1,2,3,4$, this number doesn't appear to be very large relative to $n$.
- 41
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 ...
- 18.7k
2
votes
0
answers
178
views
Monoid prime ideals and prime congruences
I was wondering what the connection is between the notion of "prime congruence" on a monoid, and the notion of "prime ideal" in a monoid. Starting from a prime ideal $P$ in a monoid $M$, one can ...
- 4,547
6
votes
1
answer
220
views
Trivial algebras given by generators and relations
Let $V$ be a finite dimensional vector space over a field $K$ of characteristic zero. Assume that we are given a set of (not necessarily homogeneous) elements $f_1,\ldots f_n$ in the tensor algebra $T(...
- 5,039
17
votes
1
answer
1k
views
Non-commutative Galois theory
Recall that an finite-dimensional algebra $A$ over a field $k$ is central simple iff there is an iso
$A \otimes_k A^{op} \cong M_n(k)$
where $A^{op}$ is the opposite ring and $M_n(k)$ is the matrix ...
- 2,040
8
votes
2
answers
708
views
Quantum Grassmannians?
In noncommutative algebraic geometry a commonly studied family of objects are quantum projective spaces. Theses are certain deformations of the homogeneous coordinate ring of $\mathbb{CP}^n$. For ...
- 449
0
votes
0
answers
125
views
Embedding a cancellative monoid into another in such a way that $|X-x|=|X|$, where $X$ is a fixed finite set and $x\in X$
Preliminaries.
Let $\mathbb A = (A, +)$ be a possibly non-commutative semigroup. For $X, Y \subseteq A$ we set
$$
X - Y := \{a \in A: a + y \in X\text{ for some }y \in Y\},
$$
which is just the usual ...
- 10.5k
6
votes
1
answer
371
views
Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible
Let $\mathbb A = (A, +_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ ...
- 10.5k
13
votes
0
answers
251
views
Is every simply connected finite complex the classifying space of a finite monoid
On page 323 of Fiedorowicz, "Classifying Spaces of Topological Monoids and Categories" it was stated that "it seems likely that any finite simply connected complex should [have the same weak homotopy ...
- 665
0
votes
1
answer
83
views
Congruences of abelian monoids which can be extended to (ideal) congruences of polynomials
Some weeks ago I asked the same question at [math.stackexchange][1] but I have not gotten any feedback. The flavour of the question (but see the details later) is about whether to understand ...
- 173
1
vote
1
answer
68
views
Under what assumptions can endomorphisms of $M/IM$ be realized as a subquotient of endomorphisms of $M$?
Suppose we have an algebra $A$ (unital, associative), with an ideal $I \leq A$ and a finitely generated module $M$ over $A$.
It is possible to obtain both $\mathrm{End}_A(M)$ and $\mathrm{End}_A(M/IM)...
- 33
5
votes
1
answer
383
views
Projective resolutions for commutative monoids
What is the right notion of a projective resolution of a commutative monoid?
The category Mon of commutative monoids has plenty of projective (and even free) objects. Indeed, for every set $X$ one ...
- 3,497
1
vote
1
answer
203
views
Filtered Colimit of associative $k$-algebras that are domains
Let $C$ be a filtered subcategory of the category of commutative algebras over a fixed field $k$ whose objects are all integral domains.
Then the colimit of the obvious diagram is an integral domain.
...
user78817
5
votes
5
answers
2k
views
Elementary linear algebra over a (possibly skew) field $K$
I have a number of questions which seem linked to me, about basic (?) linear algebra:
Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $...
- 1,555
2
votes
0
answers
193
views
Factorisation of twisted polynomials
Let $K=\mathbb{C}((t))$ and let $K_m=\mathbb{C}((t^{1/m}))$. let $K\{x\}$ denote the ring of twisted polynomials. The addition in this ring is defined as usual, but the multiplication is adjusted by ...
- 2,751
8
votes
2
answers
1k
views
Sets of matrices which are irreducible but not strongly irreducible
A set of $d \times d$ real or complex matrices is commonly called irreducible if those matrices do not jointly preserve a linear subspace with dimension strictly between zero and $d$. A stronger ...
- 6,206
2
votes
1
answer
199
views
Quaternion algebra in characteristic $p$
Given a prime number $p$, can you give me concrete examples of fields $\mathbf F$ of characteristic $p$ and quaternion algebras $\mathbb H(\mathbf F)$ over $\mathbf F$ such that $\mathbb H(\mathbf F)$ ...
- 1,555
7
votes
0
answers
291
views
Differentially closed fields
Let F be a field. Recall that an additive map $d: F\rightarrow F$ is said to be a derivation if $d(ab)=ad(b)+d(a)b$.
Now let $F$ be a ring and let $d$ be a derivation of $F$. Examples I have in mind ...
- 2,751
7
votes
1
answer
361
views
For a new operation on a finite group of odd order giving a loop structure, when does this also gives a group
For finite groups $G$ of odd order, as $x \mapsto x^2$ is bijection (but no automorphism in general) then, we can define for each $g \in G$ the element $x^{1/2}$ by requiring $(x^{1/2})^2 = x$. Then ...
- 798
5
votes
1
answer
284
views
Directed homotopy in the Cayley graph of a monoid
There is a the notion of the Cayley graph $C(G)$ of a group $G$ (which depends on a given presentation $G \cong \mathcal F(S) / \sigma$ where $\mathcal F$ is the free group functor and $\sigma$ some ...
- 190
4
votes
1
answer
171
views
Quaternion orders such that every proper ideal is invertible
Let $B$ be a quaternion algebra over $\mathbb{Q}$ and let $\mathcal{O} \subset B$ be an order.
A lattice in $B$ is (left) proper over $\mathcal{O}$ if its left order is equal to $\mathcal{O}$. We ...
- 3,009
8
votes
1
answer
261
views
A generalisation of $C_0$-semigroups
A $C_0$-semigroup is a strongly continuous family $\{T(t)\}$ of bounded linear operators on a normed space $X$, indexed in $\mathbb R_+$ and with two additional properties that make it look like an ...
- 3,388
12
votes
2
answers
914
views
non commutative polynomial which is zero for all matrix evaluation
Let $K$ be a (commutative) field.
We can define the free $K$-algebra of polynomials in non commutative variables $x_1, \cdots, x_n$. It is usually denoted by $K\langle x_1, \cdots, x_n \rangle$.
Fix a ...
- 135
18
votes
1
answer
771
views
Koszul complex for non-Koszul algebras
Let $A$ be a graded, connected, locally finite, quadratic algebra over a field $k$; that is, $A$ may be presented as $T(V)/I$, where $V = A_1$ is a finite dimensional $k$ vector space, and the ideal $...
- 4,133
0
votes
0
answers
308
views
Basic question about power series and complete group algebras
This is a pretty basic question, but I suspect it might be too exotic for math.stackexchange.
Let $\mathbb{Z}_p$ be the $p$-adic integers. For free pro-$p$ group $F_r$ of rank $r$, we can consider ...
- 10.7k
5
votes
1
answer
204
views
Center of $k$-algebra with two generators and sole defining relation $yx - xy = 1$ when $\text{char}\,k > 0$
Let $A(k)$ be a $k$-algebra with two generators, $x$, $y$, and one defining relation: $yx - xy = 1$. What is the center of the algebra $A(k)$ in the case $\text{char}\,k > 0$?
user78704
2
votes
1
answer
515
views
dg-resolution of the polynomial algebra
I am intersted in constructing a cofibrant resolution of the commutative polynomial algebra in some number of variables in the category of dg-algebras(not necceserily commutative).
The resolutions ...
- 165
8
votes
1
answer
229
views
Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topological groups
An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a ...
- 10.5k
3
votes
2
answers
476
views
Rings all of whose torsion modules are cyclic
Let us call a (possibly non-commutative) ring $R$ "very good" if every finitely generated torsion left $R$-module is cyclic. Here is an example of such a ring:
Let $k=\mathbb{C}((t))$ and let $R=k[\...
- 2,751
6
votes
1
answer
661
views
Graded Hopf algebras and H-spaces
Let $k$ denote an algebraically closed field of characteristic $0$. Suppose $K=\bigoplus_{i\geq 0}K(i)$ is a Hopf $k$-algebra which admits a connected Hopf-grading (that is, a grading which is both ...
- 426
4
votes
1
answer
169
views
Is this notion of 'closed subset' of a semigroup action something people have thought of?
Suppose $S$ is a semigroup (or a monoid, or a category), and $X$ is an $S$-set -- that is, a set with an action of $S$. Say that a sub-$S$-set $Y$ is "downward closed" (or maybe "well-generated") if ...
- 3,149
1
vote
1
answer
158
views
Graded category O for for rational Cherednik algebras, but at t=0
The paper [1] introduced the category $\mathcal{O}$ for rational Cherednik algebras $H_{t,c}(W)$. This construction is tailored for the $t=1$ case (equivalently, the $t\neq 0$ case). The general setup ...
- 1,066
3
votes
1
answer
191
views
Is every $n$-ary semigroup a subalgebra of an algebra derived from a binary semigroup?
Let $(A,f)$ be an $n$-ary semigroup ($n \ge 2$). Then there exists a ($2$-ary) semigroup $(\overline A,*)$ with an inclusion homomorphism $A \hookrightarrow \overline A$ such that that the restriction ...
- 190
9
votes
0
answers
373
views
Embedding $\beta\mathbb{N}$ into a product of Cantor sets
Let us consider $\beta\mathbb{N}$, the Stone-Čech compactification of the natural numbers (where we do not take $0$ to be a natural number, so the only idempotent elements are nonprincipal ...
- 297
6
votes
0
answers
294
views
Laurent and power series over the field with one element?
Question. Is there a suitable notion of the Laurent series ring $\mathbb{F}_1((t))$ and power series ring $\mathbb{F}_1[[t]]$ in some framework for the field with one element $\mathbb{F}_1$?
For ...
user1437
11
votes
1
answer
630
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
4
votes
1
answer
246
views
group completion theorem of homology as Hopf algebras
Let $M$ be a topological monoid with product $\mu$. Then $H_*(M)$ is a Hopf algebra with product $\mu_*$ and coproduct $\Delta_*$. The group-completion theorem by McDuff-Segal, 1976 gives that as a ...
- 2,223
4
votes
0
answers
112
views
Centers of Noetherian Algebras and K-theory
I'll start off a little vauge: Let $E$ be a noncommutative ring which is finitely generated over its noetherian center $Z$. Denote by $\textbf{mod}\hspace{.1 cm} E$ the category of finitely ...
- 161
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 ...
- 201
4
votes
0
answers
77
views
Catenarity of monoid algebras
Let $R$ be a commutative ring, let $M$ be a commutative monoid, and let $R[M]$ denote the corresponding monoid algebra. Suppose further that $R$ is universally catenary. One may ask for conditions on $...
- 6,700
9
votes
2
answers
945
views
Must a finitely generated projective module over a group ring with vanishing coinvariants be trivial?
Let $G$ be a (possibly infinite) group. Let $\mathbb{Z}[G]$ be its integral group ring and let $P$ be a finitely generated projective module over $\mathbb{Z}[G]$. Suppose that the coinvariants of $P$ ...
- 2,310
2
votes
0
answers
87
views
Terminology for torsion semigroups where the order of elements is uniformly finite
A (multiplicatively written) semigroup $\mathbb A = (A, \cdot)$ with the property that ${\rm ord}_\mathbb{A}(a) := |\{a^n: n \in \mathbf N^+\}| < \infty$ for every $a \in A$ is called a periodic (...
- 10.5k
0
votes
1
answer
381
views
Maximal group image
How does one prove: if $S$ is a finitely generated Clifford semigroup its maximal group image is actually $S_{e_{n}}$?
- 1
3
votes
1
answer
199
views
group completion theorem by using homology fibrations
In the paper Homology fibrations and group completion theorem, McDuff-Segal (www.maths.ed.ac.uk/~aar/papers/mcdsegal.pdf), page 281:
Let $M$ be a topological monoid such that $\pi_0M$ is generated by ...
- 2,223