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Results tagged with ra.rings-and-algebras
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user 22989
Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
25
votes
Accepted
Is this ring isomorphic to a quotient of a group algebra?
If $A$ is a $\mathbb{Q}$-algebra, then there is a group $G$ such that $A$ is a quotient of $\mathbb{Q}[G]$ if and only if $A$ is generated by units. For the "if" direction, take $G$ to be the group of …
- 35.2k
10
votes
Accepted
Reference request: locally erasable delta-functor is universal
This is Proposition 4.2 in Buchsbaum’s Satellites and universal functors, Annals of Mathematics 71(2), pp. 199–209 (1960).
Well, to be precise, that is the dual result (for contravariant functors). Bu …
- 12.9k
12
votes
Accepted
Infinite dimensional finitely generated algebraic division algebra
This is a fairly well known old and open (as far as I know) problem: Kurosh’s Problem for division rings. See, for example, Question 3 in Agata Smoktunowicz’s 2006 ICM talk.
- 35.2k
14
votes
Accepted
Matrix ring isomorphisms of different sizes
If $\Lambda$ is a ring, then the isomorphism classes of finitely generated
projective $\Lambda$-modules form a commutative monoid $(A,+)$, with
$[P]+[Q]=[P\oplus Q]$. This monoid contains a distinguis …
- 35.2k
6
votes
Accepted
Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?
$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need …
- 35.2k
3
votes
Minimal ideals and subalgebras of semisimple algebras
In this answer, I was assuming the naive definition of a simple module $M$ for a nonunital ring $R$ as one with no proper nonzero submodules. It seems to be common to also insist that $RM\neq0$, in wh …
- 35.2k
2
votes
Accepted
Is uniform dimension monotonic in quotients when there is a unique indecomposable injective?
Let $k$ be a field, and let $A$ be the $3$-dimensional commutative $k$-algebra $k[x,y]/(x^2,xy,y^2)$. Then in the category of $A$-modules there is a unique indecomposable injective, namely the dual $D …
- 35.2k
5
votes
Accepted
Are module finite algebras over semiperfect rings again semiperfect?
No, even if $S$ is commutative. There may be easier counterexamples, but ...
There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …
- 35.2k
5
votes
Accepted
$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
This is not a complete answer, but a construction that might give an answer.
I'll start by constructing a ring with several objects (a.k.a. preadditive
category) $\mathcal{C}$ by generators and relati …
- 35.2k
6
votes
Accepted
linear independent families in a tensor product
This answer to a related question gives a way of constructing counterexamples.
For a similar but more concrete example, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a $3$-dimensional alge …
- 35.2k
2
votes
Accepted
Condition for equality of modules generated by columns of matrices
Interpreting the various matrices as maps between free modules in the usual way, the question becomes:
If $M$ is a submodule of $R^m$, and $\alpha,\beta: R^k\to M$ are epimorphisms, then must $\alpha$ …
- 35.2k
6
votes
Accepted
Reference request for equivalent formulations of being absolutely indecomposable
This is Theorem 30.29 in
Curtis, Charles W.; Reiner, Irving, Methods of representation theory, with applications to finite groups and orders. Vol. I, Pure and Applied Mathematics. A Wiley-Interscience …
- 35.2k
6
votes
Injective modules
Yes.
Let $M$ be any $A$-module. Then its socle is a direct sum of simple modules: $\operatorname{soc}A=\bigoplus_iS_i$.
$A$ is a finite dimensional algebra, so the dual $\mathrm{Hom}_k(A,k)$ of $A$ is …
- 63.9k
11
votes
Accepted
Is any finite-dimensional algebra a sub-algebra of a finite-group algebra?
Assuming that by "sub-algebra" you mean "unital sub-algebra":
Every group algebra has a one-dimensional module (the trivial module), so any subalgebra has a one-dimensional module.
But many finite-dim …
- 35.2k
7
votes
Accepted
Dimension of division rings coming from indecomposable modules
Even if $A$ is a finite dimensional $k$-algebra, there may be no bound on the dimension of $\text{End}_A(X)/m$.
Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from …
- 35.2k