40
votes

### What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $K=\mathbb{F}_2$ and virtually abelian $G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \...

22
votes

### Subtraction-free identities that hold for rings but not for semirings?

The answer to your first question is yes (which was very surprising to me, to be honest). I have no idea whether the second question also has a positive answer. (By the way, don't let the work below ...

21
votes

Accepted

### Problems concerning subspaces of $M_{n}(\mathbb{Q}) $

Let's call this maximal dimension function $\rho_{\mathbb{Q}}:\mathbb{N}\to\mathbb{N}$, i.e., $\rho_{\mathbb{Q}}(n)$ is the largest possible dimension of a subspace $N\subset M_n(\mathbb{Q})$ such ...

20
votes

Accepted

### Does Morita theory hint higher modules for noncommutative ring?

Yes. The trick is to use not just categories, but pointed categories, which are categories equipped with a choice of object (the "pointing"). Given any ring $R$, the category $\mathrm{Mod}(R)...

20
votes

### A ring for which the category of left and right modules are distinct

$\begin{pmatrix} {\mathbb Z} & {\mathbb Q}\\ 0 & {\mathbb Q} \end{pmatrix}$
is a canonical example for such things. Let us list all simple right and left modules:
$$R_p=\begin{pmatrix} {\...

19
votes

Accepted

### A ring for which the category of left and right modules are distinct

Let $Q$ be a finite acyclic quiver and $K$ a field. Let $Q^{op}$ be the opposite quiver (where all the arrows are reversed). Let $KQ$ be the path algebra. Then the category of left $KQ$-modules is ...

18
votes

Accepted

### Reason to apply the Koszul sign rule everywhere in graded contexts

A precise statement of the conventions (which Jesse is referring to) is that the authors are using the symmetric monoidal structure on graded vector spaces, where the braiding map,, $\tau: V \otimes W ...

17
votes

Accepted

### Are the trace relations among matrices generated by cyclic permutations?

The reformulation suggested by Christian Remling and Benjamin Steinberg is true (at least over a field $k$ of characteristic zero):
If $\operatorname{tr} f(X_1,\dots, X_n)=0$ for all $X_1,\dots, X_n$ ...

16
votes

Accepted

### Non-commutative Galois theory

Let $k$ be a field. Say that a $k$-algebra $A$ is separable if any of the following equivalent conditions holds (it is not obvious that they are equivalent):
$A$ is projective as an $(A, A)$-bimodule....

16
votes

### Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal{D}_X$-module?

This is a replacement for an old confused answer. There is a related context in which I know a good answer. Suppose $A$ is a ring and $S$ is a central subring. If $M$ and $N$ are $A$ modules, then $M \...

14
votes

Accepted

### Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal{D}_X$-module?

OK, I'll give it a shot. The bi-algebra structure on $D$ is something that I found very confusing too, so I will try to spell it out as best I understand. These ideas were explained to me by Pavel ...

14
votes

### Dual of a bimodule

As explained in more detail in this blog post linked by Jakob in the comments, every $(A, B)$-bimodule $M$ has two natural duals:
If $M$ is finitely generated projective as a left $A$-module, it has ...

14
votes

Accepted

### Vanishing of Hochschild homology of a category

This precise question was phrased as the vanishing conjecture in Hochschild homology and semiorthogonal decompositions. But we now know that there exist so called (quasi)phantom categories, which give ...

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 ...

13
votes

### A note on orders in quaternion algebras

Two orders need not be isomorphic.
First of all, in number fields $K$ other than $\mathbf Q$ not all orders are isomorphic rings (even if they are isomorphic abelian groups): the full ring of integers ...

12
votes

Accepted

### Is there any non-commutative ring such that every element other than the identity is a zero divisor?

[Sorry for answering my own question, and the more so because this is happening for the second time in 24 hours.]
The question might be open. In fact, a positive answer would imply an equally positive ...

12
votes

### Is there a $3$-commutative algebra?

An easy example of 3-commutative algebra is $T_3(F)$, the algebra of all strictly upper triangular 3-by-3 matrices.
Another class of examples is given by the quandle rings $F[X]$, where $F$ is a field ...

11
votes

Accepted

### Subtraction-free identities that hold for rings but not for semirings?

Tim Campion's idea works, though his example needs a little fixing. As in Tim's answer, we will find a rig with two elements $X$ and $Y$ such that $X+Y=1$ but $XY \neq YX$.
Let $(M,+,0)$ be any ...

10
votes

### Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal{D}_X$-module?

One way to think about this is that $D$ is the universal enveloping algebra $U(R,L)$ of the $(k,R)$ Lie-Rinehart algebra $\mathrm{Der}_k(R,R)$. Whenever one has such an enveloping algebra one may ...

10
votes

Accepted

### Applications of cluster algebras

One reason is that cluster algebras have motivated many recent developments in the representation theory of associative algebras. There is a lot one can say about this, so I will try to just give an ...

10
votes

Accepted

### Subalgebra of a group algebra

The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed ...

10
votes

### Brauer group of $\mathbb{Z}_{(p)}$

Lemma. Let $K$ be a number field with ring of integers $\mathcal O_K$, and let $S \subseteq \Omega_K^f$ be a set of finite places of $K$. Then there is a canonical short exact sequence
$$0 \to \...

10
votes

Accepted

### RIng that is flat over a subring as a right module but not as a left module

Take the associative algebra over a field $k$, with generators $x$ and $y$ subject to the relation $xy=0$. This admits a basis consisting of monomials of the form $y^a x^b$. It thus contains a subring ...

9
votes

### How to make the Capelli's identity less mysterious?

I am somewhat late for a year, sorry, but since the wiki-article is mainly written by me and I somehow worked on the subject, it is difficult to resist writing an answer.
Currently there is some ...

9
votes

Accepted

### Left module which cannot be made into a bimodule?

Such examples are a plenty. You are asking about non-existence of an algebra map $A\rightarrow End_AM$. Take $A$ simple, at most countably dimensional, and a simple module ${}_{A}M$. Then $End_AM={\...

9
votes

### Subtraction-free identities that hold for rings but not for semirings?

The answer to the second question is no in general.
For instance, in an associative ring, the elements $x(x+y)^{-1}$ and $y(x+y)^{-1}$ necessarily commute — in other words, if $a + b = 1$, then $a$ ...

9
votes

Accepted

### Question concerning the coefficients of block idempotents

Yes, this is true. By block orthogonality relations due to R. Brauer, it is true that $a_{g}=0$ whenever the element $g$ has order divisible by $p$. But when $g$ has order prime to $p$, it is clear ...

9
votes

### Is there a $3$-commutative algebra?

This answer may be a bit anticlimactic, but you can take free associative algebra without $1$ (same as ideal of positive degree elements in tensor algebra) and look at its quotient by all products of ...

8
votes

Accepted

### reduced norm from degree 3 division algebra

The Merkurjev-Suslin Theorem says that an element $x \in k^*$ is a norm from $D$ if and only if $[D] \cup (x)$ is zero in the Galois cohomology group $H^3(k, \mathbb{Z}/3)$. Therefore, for the ...

8
votes

### Motivation for the Preprojective Algebra

Preprojective algebras are a key ingredient in Lusztig's semicanonical basis for U(n). There is a short explanation here: http://arxiv.org/pdf/1009.4552.pdf, and references, but I am not sure what ...

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