36 votes

What do you do if you believe a problem is undecidable?

The first thing to say is that for a statement to be independent of some axioms means really two things, namely, that it is consistent with those axioms, and also that the negation of the statement is ...
user avatar
26 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} \...
user avatar
  • 1,476
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 ...
user avatar
  • 14.9k
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 ...
user avatar
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)...
user avatar
19 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} {\...
user avatar
  • 11.4k
18 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 ...
user avatar
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 \...
user avatar
16 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 ...
user avatar
15 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....
user avatar
14 votes
Accepted

Center of $k$-algebra with two generators and sole defining relation $yx - xy = 1$ when $\text{char}\,k > 0$

This is the Weyl algebra. The center is $k[x^p,y^p]$, where $p$ is the characteristic. See, for example, here for a proof.
user avatar
13 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 ...
user avatar
13 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 ...
user avatar
  • 1,446
12 votes

How to make the Capelli's identity less mysterious?

This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified ...
user avatar
12 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 ...
user avatar
12 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 $X_1,\dots, X_n$ ...
user avatar
  • 116k
11 votes
Accepted

Finitely generated projective = finitely presented flat over a noncommutative Noetherian ring

This holds over any ring, noetherian or not. See Bourbaki Algebra X, §1, no. 5.
user avatar
  • 34.5k
11 votes

non commutative polynomial which is zero for all matrix evaluation

The answer is yes. It suffices to note that for any $n$ and $d$, there are matrices $M_1,\dots,M_n$ of the same dimension $m$ such that the set of all the products $M_{i_1}\cdots M_{i_e}$ for $1\le ...
user avatar
11 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 ...
user avatar
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 ...
user avatar
10 votes

Motivation for the Preprojective Algebra

Here is one of the reasons (maybe the only reason) for the name. The preprojective algebra of the path algebra $H=K\mathcal Q$ can also be defined as follows as a graded algebra: $P(\mathcal Q)=\...
user avatar
10 votes
Accepted

nth term in the Baker-Campbell-Hausdorff formula

The Dynkin formula is somewhat cumbersome. Maybe a better choice is Goldberg's version http://projecteuclid.org/euclid.dmj/1077466673 In the commutator form Goldberg's result is reformulated in http://...
user avatar
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 ...
user avatar
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 ...
user avatar
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 ...
user avatar
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 \...
user avatar
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 ...
user avatar
  • 6,049
9 votes

What properties "should" spectrum of noncommutative ring have?

I know this question is old, and has an accepted answer, but this excellent paper by Manny Reyes gives some further thoughts about possible Spec's for noncommutative rings.
user avatar
  • 14.9k
9 votes

nth term in the Baker-Campbell-Hausdorff formula

There is a formula due to Dynkin (1947). It can for instance be found as Theorem 2.2 in these lecture notes. You may also google Baker-Campbell-Hausdorff-Dynkin formula.
user avatar
9 votes

Smooth Affine algebras are Calabi-Yau

If $X$ is smooth projective and $D = \cup D_i \subset X$ is an ample divisor so that $Y = X \setminus D$ is affine, then there is an exact sequence $$ \oplus {\mathbb Z}D_i \to Pic X \to Pic Y \to 0. $...
user avatar
  • 31.8k

Only top scored, non community-wiki answers of a minimum length are eligible