All Questions
23 questions
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 ...
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 $f\in S$...
- 111
1
vote
0
answers
134
views
Composition of faithfully flat ring extensions
Let $R$ be a not necessarily commutative, unital, ring, and for simplicity let module always mean right module. We say that a unital ring extension $R \hookrightarrow S$ is flat, or faithfully flat, ...
- 435
3
votes
1
answer
277
views
Faithful flatness for rings
Let $R$ be a ring and let $M$ be a right module over $R$. We say that $M$ is faithfully flat as a right module if the functor $M \otimes_R -$ from left $R$-modules to abelian groups that preserves ...
- 435
2
votes
2
answers
256
views
An algebra which is a direct sum of simple sub-bimodules over a subalgebra
Let $A$ be an infinite-dimensional noncommutative algebra over a field, let $B$ be an infinite-dimensional subalgebra of $A$, and let $A$ be a direct sum of projective simple $B$-sub-bimodules. Then ...
- 393
6
votes
1
answer
190
views
Finitely presented modules admitting projective covers
A ring $R$ is called semi-perfect if every finitely generated $R$-module has a projective cover, and it can be proved that this is equivalent to say that the category consisting of the finitely ...
user144185
5
votes
1
answer
173
views
Projective module which splits off sequence of submodules, but not the sum
Does there exist an example of a module $X$ over some ring $R$ together with submodules $T_i$ such that:
$X$ is projective,
$X$ splits as an internal direct sum $X\cong T_1\oplus T_2\oplus \ldots \...
- 858
2
votes
0
answers
56
views
Non-singular rings which are Rickart
A ring $R$ is said to be a right Rickart ring if the right annihilator of any element in $R$ is of the form $eR$ for some idempotent $e \in R$.
It turns out that a ring $R$ is right Rickart iff every ...
- 1,065
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 ...
1
vote
0
answers
151
views
Lifting idempotents and projective coverings --- reference request
Let $R$ be a (non-commutative, unital) ring with Jacobson radical $J$, write $\overline{R}=R/J$ and denote the quotient map $R\to \overline{R}$ by $a\mapsto \overline{a}$.
It is easy to see that if ...
- 2,928
8
votes
1
answer
2k
views
Adjoints of scalar extension and scalar coextension
Let $h\colon R\rightarrow S$ be a morphism of commutative rings. We consider the following functors (I am aware that the notations may be different in other contexts):
$h^*$: Scalar extension by ...
- 6,700
8
votes
1
answer
335
views
Graded and projective (but not bounded below) module that is not graded-projective?
Let $A = \bigoplus_{n = 0}^\infty A_n$ be a graded algebra over a field $k$ that is locally finite: each $A_n$ is a finite-dimensional $k$-vector space. We say that a graded left $A$-module $P = \...
- 5,407
3
votes
2
answers
227
views
Operations on semi-hereditary rings
I am learning about left (right) semi-hereditary rings, and got stuck with the following questions. Just to recall that being left semi-hereditary means that every finitely generated submodule of a ...
- 31
2
votes
1
answer
172
views
Morphisms in K-theory: comparison of two pictures
Algebraic $K_0$ group for an algebra $A$ may be defined in terms of stable isomorphism classes of idempotents in $M_n(A)$ or equivalently in terms of isomorphism clasess of finitely generated ...
- 9,330
6
votes
1
answer
302
views
Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?
Let $A$ be a finite dimensional semisimple $F$-algebra and $K_1(A)$ the Whitehead group of $A$.
By splitting $A$ into its Wedderburn components, the reduced norm map $\operatorname{nr}_A:K_1(A)\to Z(...
- 255
3
votes
1
answer
303
views
An invariant submodule of a projective module
This is a basic question (not research level) which has already been asked on SE by someone else but doesn't yet have an answer so I'd like to repost it on MO.
Let $R$ be a commutative ring with ...
- 255
9
votes
1
answer
472
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 ...
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
6
votes
1
answer
491
views
Homomorphisms from projective modules
Let $B$ be a $A$-algebra which is free of finite rank as $A$-module. Let $X$ be a finitely generated projective left $B$ module. (So $X$ is also a f.g. projective $A$ module.) Are these homomorphism ...
- 4,484
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 ...
- 618
5
votes
0
answers
195
views
Sum of projective submodules of a projective over a semihereditary ring
Sorry in advance if this is too silly. Let $R$ be a right semihereditary ring and $P$ a projective right $R$-module. It is well-known that finitely generated (thus projective) submodules of $P$ form a ...
- 409
14
votes
0
answers
1k
views
Kaplansky's theorem and Axiom of choice
Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow:
Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...
- 1,788
1
vote
1
answer
600
views
Unimodular column property
Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property.
I would like to know if there is a ring $R$ that doesn't ...
- 113