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Right maximal ideals in skew-Laurent rings over division Rings

Let $R$ be a Noetherian domain, and let $D$ denote its division ring. Define $S = D_q[x_1^{\pm 1}, x_2^{\pm 1}]$ as an iterated skew-Laurent polynomial ring with the relation $x_1 x_2 = q x_2 x_1$. Is ...
Sky's user avatar
  • 923
0 votes
0 answers
91 views

Finitely generated module over a skew Laurent polynomial ring $\mathbb{K}[x_1^{\pm 1},x_2^{\pm1}]$

Let $A := \mathbb{K}_{\Lambda}[x_1^{\pm 1}, x_2^{\pm 1}, x_3^{\pm 1}, x_4^{\pm 1}]$, $B := \mathbb{K}_{\Lambda'}[x_1^{\pm 1}, x_2^{\pm 1}, x_3^{\pm 1}]$, and $C := \mathbb{K}_{\Lambda''}[x_1^{\pm 1}, ...
Sky's user avatar
  • 923
2 votes
0 answers
76 views

Does a matrix ring over a ring satisfy the Koethe conjecture if the coefficient ring itself satisfies the Koethe conjecture?

I just want to know whether the following statement is true or false. If $R$ is a ring satisfying the Koethe conjecture, then the matrix ring over $R$ also satisfies the Koethe conjecture. Or is it ...
Eunnaya First's user avatar
1 vote
0 answers
58 views

When is a bimodule that is projective as a right and as a left module also projective as a bimodule

Are there practical criteria for determining when a bimodule that is projective as a right and as a left module is projective as a bimodule? Some illustrative examples of what goes wrong and what goes ...
Gheorghe Bucătaru's user avatar
0 votes
0 answers
59 views

Bimodule endomorphisms of a bimodule over a noncommutative ring

Let $R$ be a noncommutative ring and $M$ an $R$-$R$-bimodule that is projective as left $R$-module. We know that the bimodule of left $R$-module endomorphisms ${}_REnd(M)$ is isomorphic to the tensor ...
Gheorghe Bucătaru's user avatar
2 votes
1 answer
210 views

Direct product of direct sum of a flat module

In the book "Rings and Categories of Modules" by Anderson & Fuller, this problem is given: If $V^A$, i.e. the direct product of the module $V$ by the index set $A$, is flat for all sets $...
karparvar's user avatar
  • 355
1 vote
1 answer
87 views

An example of a commutative ring which is not SIP

Recall that a module $M_R$ ($R$ is a unital ring) is called an SIP-module if the intersection of any two summands of $M$ is a summand. The ring $R$ is called (left) right SIP-ring if the module (${}...
Hussein Eid's user avatar
1 vote
1 answer
75 views

Lemma of Harada and Sai on sums of modules with a "chain" of monomorphisms between them

I am trying to get a contradiction from the following set of hypotheses: Let $R$ be a ring. Let $M$ be a direct sum of non-zero $R$-modules $M_1$, $M_2$, $\dotsc$. For each $i\ge1$, let $f_i:M_i\to M_{...
Tri's user avatar
  • 1,644
3 votes
2 answers
468 views

How fast does the number of "fixed" points grow compared to the size of the ball in the following group?

I have copied this question from Math.StackExchange, in the hope that some experts here can provide some relevant insight. Let $M = \oplus_{i\in \mathbb Z} V^{(i)}$ where each $ V^{(i)} \cong \mathbb ...
ghc1997's user avatar
  • 823
3 votes
2 answers
214 views

Characterize rings $R$, such that the countable product $P=R^N$ has the property that every finitely generated submodule of $P$ is free

What are the rings whose countable power has the property that every finitely generated submodule is free?
Rado's user avatar
  • 139
2 votes
2 answers
416 views

Tensor product over $\mathbb{Z}$ and p-adic integer ring $\mathbb{Z}_p$

Thanks for your reading. Suppose we have two $\mathbb{Z}_p$-modules $A,B$. Do we always have $A \otimes_{\mathbb{Z}} B \simeq A \otimes_{\mathbb{Z}_p} B$, as abelian groups or $\mathbb{Z}_p$-modules? ...
Rellw's user avatar
  • 319
1 vote
0 answers
60 views

Bounding the length in a module of evaluated skew polynomials

Let $R$ be a finite principal ideal ring, $S$ a Galois extension of $R$ of degree $m$ (so in particular $S$ is a free $R$-module of rank $m$, and we have an $R$-module isomorphism $S^n \cong \...
JBuck's user avatar
  • 223
0 votes
0 answers
71 views

When is a submodule trivial?

I am a beginner concerning module theory, but I need it for my PhD. Sorry for obvious questions therefore. Given a left $C(G)$-module $(V, \tilde{\rho})$ where $C(G)$ denotes the group algebra over a ...
relativeentropy's user avatar
0 votes
0 answers
46 views

submodules in a direct sum of semisimple modules without common simple factors

Let $A$ be an associative (unital) algebra. Let $M_1,\cdots, M_r$ be pairwise non-isomorphic simple $A$-modules and let $V=\bigoplus^r_{i=1}V_i$, where $$ V_i=M_{i,1}\oplus \cdots\oplus M_{i,n_i}\...
Yong Hu's user avatar
  • 620
6 votes
3 answers
434 views

What is known about finite dimensional modules over the nilCoxeter algebra?

Recall that the nilCoxeter algebra $\mathcal{N}_W$ for a Coxeter group $W$ is given by the $\mathbf{k}$-basis $x_w$ for each $w\in W$ and multiplication $x_ux_v=x_{uv}$ if $\ell(uv)=\ell(u)+\ell(v)$ ...
Christoph Mark's user avatar
2 votes
1 answer
88 views

Generating sets for a module and scalar extension

Let $k$ be an algebraically closed field and $K/k$ a (transcendental) field extension. Let $A$ be a finite dimensional $k$-algebra, and $M$ an $A$-module. Suppose that the $K \otimes_k A$-module $K \...
bm3253's user avatar
  • 23
4 votes
1 answer
198 views

Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?

All rings are assumed to be associative and have a 1. Let $k$ be a commutative artininan ring and $R$ a finitely generated $k$-algebra. Is it true that if $R$ is connected and hereditary, then $k$ is ...
kevkev1695's user avatar
1 vote
0 answers
242 views

Is the span of all nilpotent ideals also a nilpotent ideal?

Given a non-zero Lie algebra $\mathcal{L}$ over $\mathbb{C}$, we define $\mathcal{L}^2 = \big[\mathcal{L}, \mathcal{L} \big] = \big\{ [x, y]: x, y\in \mathcal{L} \big\}$, and for any $k\in\mathbb{N}$ ...
Sanae Kochiya's user avatar
1 vote
1 answer
84 views

Simple-direct-injective modules

A right $R$-module $M$ is called a simple-direct-injective module if it satisfies any of the following equivalent conditions: For any simple submodules $A,B$ of $M$ with $A \cong B \subseteq^{\oplus} ...
Hussein Eid's user avatar
8 votes
1 answer
191 views

Do graded-commutative rings satisfy the strong rank condition?

Let $R$ be a ring. Recall that $R$ is said to satisfy the strong rank condition if, whenever $R^m \to R^n$ is a monomorphism of right $R$-modules (with $m,n \in \mathbb N$), we have $m \leq n$. It is ...
Tim Campion's user avatar
  • 63.9k
5 votes
0 answers
302 views

Connections in non-commutative geometry

Let $K$ be a field, $A$ a unital associative $K$-algebra and $M$ a left $A$-module. A connection on $M$ is a $K$-linear map $\nabla:M\to \Omega^1A\otimes_AM$ which satisfies the Leibniz rule. ...
Qwert Otto's user avatar
1 vote
1 answer
214 views

Two exact sequences for $R$-modules: does one imply the other?

Consider the following two properties for an $R$-module $M$: For every endomorphism $f:M\rightarrow M$, there exist central endomorphisms $g, h\in \operatorname{End}_R(M)$ such that the sequence $M_{...
Najmeh Dehghani's user avatar
11 votes
1 answer
159 views

Are all indecomposable $\mathbb{Z}_+$-modules over the character ring of a group, character rings of a subgroup?

A $\mathbb Z_+$-algebra is an algebra $A$ over $\mathbb C$ with given basis $\{v_i\}$ such that $$v_iv_j=\sum_k n_{ijk}v_k,\hspace{10mm}n_{ijk}\in\mathbb Z_{\geq0}.$$ An example of such an object is ...
shin chan's user avatar
  • 301
1 vote
1 answer
130 views

A characterization of semiartinian modules

Recall that a right module $M_R$ is called semiartinian if every nonzero homomorphic image has nonzero socle. It's well known that the following two statements are equivalent: $M$ is semiartinian. ...
Hussein Eid's user avatar
2 votes
0 answers
76 views

Are the following two characterisations of symplectic modules, using the language of form rings, the same?

Page 205 of the book Classical Groups and Algebraic K-Theory defines a symplectic module to be an arbitrary quadratic module $(M,h,q)$ over a form ring $(R,\Lambda)$ with $(J,\varepsilon)$ where $J=\...
wlad's user avatar
  • 4,943
1 vote
0 answers
59 views

Is there any point in considering Form Rings when 2 admits an inverse?

In the study of quadratic spaces over general rings, there is a type of scalar which people consider called a Form ring $(R,\Lambda)$ relative to some anti-automorphism denoted $(-)^J:R\to R$ and ...
wlad's user avatar
  • 4,943
4 votes
1 answer
186 views

Exact sequences with two FL-modules

Let $R$ be a ring. An $R$-module $M$ is called FL (FP) if it has a finite resolution consisiting of finitely generated free (projective) modules. Given an exact sequence of $R$-modules, $0\to M_1\to ...
Andrei Jaikin's user avatar
3 votes
0 answers
225 views

Intersection of two modules (and sub-modules) under tensors

I have a (hopefully quick) question regarding an intersection of two tensor modules. Let $K$ be a field and $A, B$ finitely-generated modules over the Laurent series $K((X))$. Let $\tilde{A}$ be a (...
M-M's user avatar
  • 31
4 votes
1 answer
142 views

Property of simplicity and semi-simplicity under base change of base field

Suppose $K$ is a field of characteristic $0$ and $A$ is a $K$-algebra. Let $F$ be a field extension of $K$ and let $M$ be an $A$-module. What can we say about simplicity or semi-simplicity of $A_F$-...
amir hossein Ekhlasi's user avatar
5 votes
1 answer
256 views

Classifying indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$

I'm now interested in classifying the indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$ : the group ring of $\mathbb{Z}/p\mathbb{Z}$ over the ring $\mathbb{Z}/p^{2}\...
gualterio's user avatar
  • 1,013
0 votes
2 answers
387 views

Torsion of modules

Given a left module $M$ over a domain $R$, I am interested in irreducible elements $r\in R$ such that $r\cdot m=0$ for some $m\in M-\{0\}.$ I think "torsors" would be perfect name for such $...
Adam's user avatar
  • 2,390
2 votes
0 answers
102 views

When do two path algebras share an underlying graph?

Suppose $Q$ and $Q'$ are two quivers. I am curious as to what relation $\mathbb{C}Q$ bears to $\mathbb{C}Q'$ when $Q$ and $Q'$ share the same underlying graph and only differ by direction. Since ...
tox123's user avatar
  • 433
3 votes
0 answers
82 views

Example of a secondary representation of a module that is not a direct sum

Let $A$ be a commutative ring. An $A$-module $M$ is said to be secondary if $M\neq 0$ and for each $a\in A $, the endomorphism $\phi_a:M\to M$ defined by $\phi_a(m)=am$ for $m\in M$ is either ...
L. Xie's user avatar
  • 631
3 votes
0 answers
128 views

Which definitions of "local module" have gotten traction?

It seems like "local module" has been defined a lot of ways: if 𝑀 has a largest proper submodule. (This math.se post) if it is hollow and has a unique maximal submodule (Singh, Surjeet, ...
rschwieb's user avatar
  • 1,507
3 votes
0 answers
173 views

Where could a paper on a unification of matrix decompositions be published?

I've got a paper which shows that when the spectral theorem (as a statement that every self-adjoint matrix can be unitarily diagonalised) is naively generalised to $*$-algebras other than the complex ...
wlad's user avatar
  • 4,943
2 votes
1 answer
89 views

Primitive group rings and endomorphism rings

It is known that, for any group $G$, there exists a group $H$ containing $G$ such that the group ring $F[H]$ for some field $F$ is primitive, see Formanek, Edward; Snider, Robert L., Primitive group ...
user avatar
28 votes
2 answers
863 views

$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$

Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)? Edited to add: As no answers are forthcoming, does anyone ...
Pace Nielsen's user avatar
  • 18.7k
2 votes
1 answer
194 views

Structure of reflexive modules over regular local rings

Let $R$ be a regular local ring and $M$ a finitely generated reflexive $R$-module. When $R$ has dimension 2, then $M$ is a free $R$-module. This is discussed in Reflexive modules over a 2-dimensional ...
Ahmed Matar's user avatar
1 vote
1 answer
118 views

Subrings, submodules, and flatness

Let $R$ be a ring, and $S$ a subring of $R$. Let $M$ be a right $R$-module, and $N$ a right $S$-submodule of $M$. If $N$ is flat (or faithfully flat) as a right $S$-module, does it then follow that ...
Jake Wetlock's user avatar
  • 1,144
7 votes
1 answer
284 views

Does the choice of the algebraically closed field of characteristic $p$ have influence on the module category?

Let $G$ be a finite group and $p$ be a prime number dividing $|G|$. Let $k$ be the algebraic closure of $\mathbb{F}_p$. Let $K$ be another algebraically closed field of characteristic $p$ which is not ...
LSt's user avatar
  • 237
3 votes
1 answer
171 views

Modules with special properties

$\DeclareMathOperator\End{End}$Let $A$ be a finite dimensional algebra and $M$ an indecomposable (right) module with the property that every nilpotent element of $\End_A(M)$ annihilates the socle $\...
Mare's user avatar
  • 26.5k
3 votes
0 answers
104 views

Are there (co-)homological obstructions to the extendability of a homomorphism?

Let $k$ be a commutative ring and $A \subset B$ an extension of $k$-algebras. Can we associate to this extension a (co-)chain complex so that its (co-)homology $H$ allows for statement such as "...
Nicolas Cage's user avatar
4 votes
1 answer
564 views

Nondegenerate pairings versus perfect pairings for finitely generated projective modules

Let $R$ be a (not necessarily commutative) ring, $M$ a left $R$-module, and $N$ a right $R$-module. We say that a pairing $$ \langle -,-\rangle:M \otimes_R N \to R $$ is non-degenerate if, for all $n \...
Adam Bondal's user avatar
1 vote
0 answers
66 views

Prove that $f(M)=f^2(M)$ implies $f(M)$ is a direct summand of $M$ whenever $\text{End}_R(M)$ is a reduced ring

Let $M$ be a right $R$-module with the property that every homomorphism $\gamma:Sf\to M, f\in S=\text{End}_R(M)$, extends to $S\to M$. If $S$ has the property $f^2=0$ implies $f=0$ for every $f\in S$...
mariam's user avatar
  • 111
5 votes
1 answer
707 views

Absolutely irreducible representation and splitting field

Let $A$ be a finite-dimensional algebra over a field $F$. A representation $M$ of $A$ is called absolutely irreducible if $M\otimes_FE$ is irreducible as a representation of $A\otimes_FE$ for all ...
Hebe's user avatar
  • 951
1 vote
1 answer
111 views

Epimorphism going out of an inverse limit into a finite dimensional module

Let $k$ be a field and $A$ a finite dimensional $k$-algebra. Given a sequence of inclusions $M_1 \subseteq M_2 \subseteq \dots$ of $A$-modules consider the direct limit $M:= \bigcup_{i=1}^\infty M_i$. ...
kevkev1695's user avatar
3 votes
0 answers
292 views

modules over principal ideal rings

Let $R$ be a commutative principal ideal ring (not necessarily Artinian) and let $M$ be a finitely generated $R$-module. Is $M$ a direct sum of cyclic $R$-modules? (i.e. a generalization of the theory ...
Ahmed Matar's user avatar
8 votes
1 answer
520 views

For every ring R, is there a block-diagonal canonical form for a square matrix over R?

This question asks whether there exists an analogue of the Jordan decomposition for an arbitrary ring $R$. This analogue is not necessarily the Jordan-Chevalley decomposition, which is unnecessarily ...
wlad's user avatar
  • 4,943
7 votes
2 answers
166 views

Rings of finite uniserial type

If $R$ is a ring and $M$ an $R$-module, $M$ is uniserial if its lattice of submodules is a chain. Over an Artinian $R$, the chain will be finite. From what I understand, deciding when two uniserial ...
Chris Leary's user avatar
5 votes
0 answers
93 views

Structure of finitely generated $\mathbb{Z}/p^n\mathbb{Z}[[S,T]]$-modules

Let $\Omega=\mathbb{Z}/p\mathbb{Z}[[S,T]]$. $\Omega$ is a commutative, Noetherian and integrally closed domain of Krull dimension 2. According to Bourbaki's commutative algebra VII $\S 4$, if $M$ is a ...
Ahmed Matar's user avatar