All Questions
27 questions
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 ...
CommunityBot
- 1
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 ...
- 44.4k
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 ...
- 38.6k
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 $\...
- 1,244
5
votes
1
answer
713
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 ...
- 12.9k
3
votes
1
answer
272
views
Thin representations for quiver algebras
A representation $M$ of a quiver is called thin when $M$ has a dimension vector consisting only of 0 or 1 entries.
When $A=kQ$ is a path algebra for a tree $Q$, then there is the nice result that ...
- 356
2
votes
1
answer
197
views
Top and bottom composition factors of $M$ are isomorphic
Let $k$ be a field and $N$ a finite group. Let $M$ be a projective indecomposable $kN$-module. Since the algebra $kN$ is symmetric, it follows that the top and bottom composition factors of $M$ are ...
CommunityBot
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2
votes
1
answer
160
views
MAGMA-question concerning the transformation of a $kG$ -module $M$ into a right ideal of the group algebra
Let $G$ be a finite group and $k$ be a finite field of characteristic $p>0$ such that $p\mid |G|$.
Let $M$ be a $kG$-module which has an embedding $M\hookrightarrow kG^{reg}$ into the regular $kG$-...
- 37.4k
4
votes
1
answer
135
views
Global splitting field for algebras
Let $A$ be a finite dimensional algebra.
A field $K$ is a splitting field for an indecomposable $A$-module $M$ in case the local algebra $End_A(M)/(rad(End_A(M))$ is 1-dimensional.
$K$ is called a ...
- 35.2k
4
votes
1
answer
159
views
Effect of extending scalars on maps of modules
Let $k$ be a field and $R$ be a $k$-algebra. Let $M$ and $N$ be left $R$-modules. Finally, let $\ell$ be a field extension of $k$. We thus have an $\ell$-algebra $\ell \otimes R$, and both $\ell \...
- 26.5k
4
votes
1
answer
382
views
Non-existence of projective covers
I didn't get the argument of Example 7.3.11, page 123, from the representation theory book of Peter Webb, available also online at:
http://www-users.math.umn.edu/~webb/RepBook/RepBookLatex.pdf
In ...
- 1,726
3
votes
1
answer
244
views
Left module which cannot be made into a bimodule?
Let $A$ be a noncommutative unital algebra, defined over $\mathbb{C}$ say. What is an example of a left $A$-module $M$ that does not admit a right $A$-module structure giving $M$ the structure of a ...
- 12.3k
2
votes
1
answer
143
views
A weak Schur's lemma for non-semisimple finite dimensional algebras
Let $B \subseteq C$ be an inclusion of finite dimensional (associative) algebras over a field $k$. Assume that $C$ is a free $B$-module. Let $\bigoplus_i U_i$ be
a decomposition of $B$ into ...
- 35.2k
7
votes
1
answer
262
views
Description of modules over self-injective algebras of finite representation type
Is there any description of indecomposable modules and irreducible morphisms over self-injective algebras of finite representation type? I am interested mainly in such a description for nonstandard ...
- 21
5
votes
1
answer
192
views
Given a representation-infinite algebra, when is every AR component infinite?
Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. The Auslander-Reiten quiver $\Gamma_A$ of $A$ is a means of presenting the category of finitely generated right $A$-...
- 482
9
votes
1
answer
473
views
Why do we want $p$-permutation modules in splendid equivalences?
First Rickard (in Splendid Equivalences: Derived Categories and Permutation Modules ) and then Rouquier (Block theory via stable and Rickard equivalences, Appendix A.1) define splendid equivalences ...
CommunityBot
- 1
6
votes
3
answers
446
views
Is the category of symmetric bimodules over a commutative ring closed under extensions?
Let $A$ be a commutative ring, and consider the category of bimodules over $A$.
An $A$-bimodule $M$ is called symmetric if $a\cdot m = m \cdot a$ for all $a \in A$, $m \in M$.
Is the category of ...
- 26.5k
1
vote
1
answer
149
views
Finding modules to check for finite global dimension
Given a finite dimensional algebra $A$ and a generator-cogenerator $M$ and let $B:=End_A(M)$. $B$ has finite global dimension iff every $A$-module has finite $add(M)$-resolution dimension, which is ...
- 356
6
votes
1
answer
302
views
Endomorphism ring of bimodules
Given an algebra $A$ with a right $A$-module $M$ with $End_A(M) \cong A$.
Then we can view $M$ as a natural $A$-bimodule. When is $M$ as a bimodule indecomposable and what is its endomorphism ring as ...
- 9,650
2
votes
1
answer
263
views
Nonnegatively graded algebra $A$ finitely generated as $k$-algebra iff $A_0$ finitely generated, $A_{>0}$ finitely generated as $A$-module?
This is related to my question here. My question is as follows. How do I see that a nonnegatively graded algebra $A$ is finitely generated as a $k$-algebra if and only if $A_0$ is finitely generated ...
CommunityBot
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5
votes
1
answer
394
views
Classification of indecomposable modules in tame hereditary algebras
An algebra $A$ is said to be tame if the isomorphism classes of indecomposable $A$-modules in each dimension occur in a finite number of 1-parameter families. $A$ is said to be of finite ...
- 3,686
1
vote
1
answer
271
views
Rank of a locally free $\mathbb Z[G]$-module
This is a basic question. Let $G$ be a finite group, $M$ a finitely generated $\mathbb Z[G]$-module so that the $\mathbb Z_p[G]$-module $M_p$ is free for all prime numbers $p$, i.e. is locally free.
...
- 5,039
7
votes
1
answer
456
views
Hopfian modules
My question is slightly motivated by basic results in linear algebra, for example, that if $F$ is a field then a surjective linear map $F^n \rightarrow F^n$ is injective. More generally, any ...
- 35.2k
8
votes
1
answer
387
views
Jordan-Hölder-like statements for modules with $\Delta$-filtrations over a quasihereditary algebra
Definitions
Let $A$ be an Artin algebra (for instance, take $A$ to be a finite dimensional algebra over some field) and label the isomorphism classes of simple $A$-modules by the elements of a ...
- 425
2
votes
1
answer
404
views
Example of a Frobenius algebra that is not projective over a Frobenius subalgebra
I'd like to know an example of a Frobenius algebra $A$, with a subalgebra $B$ that is itself a Frobenius algebra, such that $A$ is not projective as a left $B$-module. I don't require any ...
- 3,176
18
votes
2
answers
1k
views
Does base extension reflect the property of being isomorphic?
Let L/K be a (separable?) field extension, let A be a finite dimensional algebra over K, and let M and N be two A-modules. Let $A' = L \otimes_K A$ be the algebra given by extension of scalars, and ...
CommunityBot
- 1
28
votes
4
answers
5k
views
When are modules and representations not the same thing?
I've been trying for a while to get a real concrete handle on the relationship between representations and modules. To frame the question, I'll put here the standard situation I have in mind:
A ring $...
- 14.5k