15
votes

Accepted

### Why do we want $p$-permutation modules in splendid equivalences?

The motivation for the definition was an attempt to explain structurally the phenomenon of an "isotypy". This makes sense for arbitrary blocks, but let's stick to principal blocks for simplicity.
...

12
votes

### Moduli space of flat connections over a Riemann surface

Both of these moduli spaces are discussed in the survey "Flat connections on oriented 2-manifolds" by Lisa Jeffrey. The theme of the first section of the paper is roughly as follows. An oriented 2-...

11
votes

### Kaplansky conjecture (consequences)

Kaplansky's zero divisor conjecture: Let $\mathbb{F}$ be a field and $G$ be a torsion-free group. Then $\mathbb{F}[G]$ does not contain a zero divisor.
The existence of a nontrivial idempotent $a$ in ...

11
votes

Accepted

### Basis for free modules over an affine domain

1) In the positive direction: If any entry in $e$ is an $(n-1)!$ power, then $e$ is an element of a basis. (More generally, it's enough to have $e=(z_1^{m_1},\ldots z_n^{m_n})$ with $(n-1)!$ ...

10
votes

Accepted

### If a quotient ring is a projective module then the ideal is principal

Let $\pi:R\to R/I$ be the natural projection map. This is an $R$-module homomorphism (as well as a ring homomorphism). If $R/I$ is projective, then this map splits. Call such a splitting $\varphi:R/...

10
votes

### The projective covers of Artinian module

No: as soon as the (local noetherian) ring itself is not artinian, there exists an artinian module with no projective cover.
First recall (over any ring) that a projective cover of $M$ is $N$ ...

10
votes

Accepted

### Is every locally free module of rank $1$ over a commutative ring concretely invertible?

The answer is yes. Recall that given an invertible $A$-module $P$ and $n \in \mathbf{Z}$ there is an invertible $A$-module $P^{\otimes n}$ such that $P^{\otimes 0} = A$, $P^{\otimes 1} = P$, and $P^{\...

10
votes

Accepted

### Factoring through projective modules is an equivalence relation

This equivalence relation is just a quotient of abelian groups. It will be cleaner to show that the $\mathrm{PHom}_R(M, N)$ is a subgroup of $\mathrm{Hom}_R(M,N).$ Indeed, if $f:M\to P_1 \to N$, $g: ...

9
votes

### The projective covers of Artinian module

Take $R=\mathbb{Z}_p$, the $p$-adic integers, and $A=\mathbb{Q}_p/\mathbb{Z}_p$. Then $A$ is an Artinian $R$-module, but doesn't have a projective cover.
[Since $R$ is local, projectives are free. ...

9
votes

Accepted

### If $M \otimes -$ is continuous, why is $M$ f.g. projective? Alternative proof

I am posting the comment above as one answer. Kernels are examples of limits. So if the functor is continuous, then $M$ is flat. Next, consider the natural transformation $M\otimes_R \text{Hom}_R(-,...

Community wiki

9
votes

Accepted

### tangent bundle on noncommutative manifold

Noncommutative Riemannian (spin) geometry via spectral triples is grounded in an approach to noncommutative differential calculus that privileges the cotangent bundle over the tangent bundle: given a ...

9
votes

Accepted

### Are finite projective modules over $R[t]$ free when $R$ is DVR?

The Bass-Quillen conjecture is known to be true for principal ideal domains (that is, if $A$ is a PID, all finitely generated projective modules over $A[T_1,\dots,T_n]$ are free). This was proven in ...

8
votes

### What is the right definition of the Picard group of a commutative ring?

1) About the second definition:
$\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring ...

8
votes

Accepted

### Adjoints of scalar extension and scalar coextension

If $X$ is an $R$-module, there is a natural map $M\otimes_RX\to\text{Hom}_R\left(\text{Hom}_R(X,R),M\right)$ given by $m\otimes x\mapsto[\varphi\mapsto m\varphi(x)]$ that is easily checked to be an ...

8
votes

Accepted

### Projective modules restricted to smooth curves

Assume that $X$ is integral and smooth (as in Gaitsgory's notes). For any two points $x,y$ in $X$, there is a smooth connected curve $C$ passing through $x$ and $y$. Since $M_{|C}$ is locally free, ...

7
votes

Accepted

### When is countable direct-product of projective modules again projective ?

Under the assumption that a countable direct product of modules over $R$ means a direct product of countably many modules over $R$, I answer OP's question when $R$ is Noetherian. In addition, I ...

7
votes

### Role of stably free modules in algebraic geometry

This is just a long comment.
Stably free modules appear often in many questions regarding number of generators of modules over rings. For example, if $I\subset R$, a polynomial ring over a field in $...

7
votes

Accepted

### Are these two constructions of $K_0(A)$ isomorphic?

The two constructions give the same result, namely the universal group with a monoid homomorphism from $Proj(A)$ aka the Grothendieck group. The first definition gives this because a short exact ...

7
votes

Accepted

### Trace ideal of a projective module

The confusion is linguistic, as identified in the comments.
Lemma. Let $M$ be a projective $R$-module, and suppose $M \oplus N \cong F$ is free on a basis $\mathcal B$. For $b \in \mathcal B$, write $\...

7
votes

### Is every graded hereditary ring hereditary?

No, consider the graded ring $R_{\ast}=\mathbb{Z}[x,x^{-1}]$ with $|x|=1$. Then the functor $M\mapsto M_0$ gives an equivalence from graded $R_{\ast}$-modules to abelian groups, so $R_{\ast}$ is ...

6
votes

Accepted

### Graded and projective (but not bounded below) module that is not graded-projective?

The answer to your question is no. More generally, the properties of being graded-projective or projective are equivalent in settings far more general than yours. See C. Nastasescu, F. Van Oystaeyen, ...

6
votes

### Example of a projective module which is not a direct sum of f.g. submodules?

Pete, here is another nice example (shown to me by Bergman many years ago) answering question 1 in another way.
Let $F$ be a field, let $R=F[x]$, and let $M=F[[x]]$ (which is an $R$-module in the ...

6
votes

### Non-free projective pearls (general and Abelian)

Actuallly, I'm not sure that free pearls as defined are in fact projective; for that matter. I'm pretty sure that there are no projectives at all. I have a characterization of epimorphisms, then a ...

6
votes

Accepted

### Property of the trace on finitely generated projective modules

This is true. It's more natural to prove a generalisation:
Lemma. Let $A$ be a commutative ring, let $F_\bullet$ be a bounded complex of finite projective $A$-modules (say in degrees $[a,b]$), and ...

6
votes

### Is every locally free module of rank $1$ over a commutative ring concretely invertible?

This is a partial answer of a more general problem.
If $A$ is a commutative ring with few zerodivisors (which holds iff $\text{Quot}(A)$ is semilocal), then the answer is yes: in that case, an ...

6
votes

### Characterization of projective modules in terms of Ext groups

If $M$ is projective, then $\mathrm{Ext}^i(M,-) = 0$ for every $i>0$; this is because $\mathrm{Ext}$ can be computed by taking a projective resolution of the first argument.
For the converse, we ...

6
votes

### In a category with a projective generator, do morphisms from the generator determine the object?

It is very easy to specify an answer if $P$ is a compact projective generator (you didn't write whether it is compact). Then, we have an equivalence of categories between $\mathcal{C}$ and the ...

5
votes

Accepted

### Non-free projective pearls (general and Abelian)

Włodzimierz Holsztyński suggests, in comments to my other answer, a modification/subcategory of pearls: call a morphism of pearls $f\colon (G,S)\to (H,T)$ sharp if and only if $f(S)=T$. Now consider ...

5
votes

### Projectives in the category of modular representations of Lie algebras

It's fairly easy to answer your basic question in the second paragraph: at present there is no guaranteed method for computing these projectives, though quite a bit of work has been done in recent ...

5
votes

Accepted

### Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?

I am assuming that you meant $F$ to be a number field.
Let $A = M_n(D)$ where $D$ is a finite dimensional division algebra over $\mathbb{Q}$ and let $Z = Z(A) = Z(D)$. Then $K_1(A)\cong D^\times/[D^\...

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