50
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} \...
- 3,736
39
votes
Algebraic theorems with no known algebraic proofs
Here is my favorite one (though not so elementary).
Theorem (Grothendieck). Let $X$ be a smooth projective variety over an algebraically closed field $k$. Then, the etale fundamental group $\pi^{\rm ...
32
votes
Algebraic theorems with no known algebraic proofs
An example comes from the connection of function fields and Riemann surfaces: Let $K$ be an algebraically closed field of characteristic $0$, $t$ a variable, and $L/K(t)$ be a finite Galois extension. ...
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 ...
- 35.2k
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 ...
- 18.7k
21
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 ...
- 3,928
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 ...
- 108k
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)...
- 54.6k
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} {\...
- 12.3k
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 ...
- 38.6k
19
votes
Algebraic theorems with no known algebraic proofs
Artin's theorem on positive polynomials, which solves Hilbert's 17th problem in the affirmative, apparently still has no algebraic proof.
Theorem (Artin): If $f \in \mathbb{R}[X_1,\dots,X_n]$ is ...
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$ ...
- 148k
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 \...
- 156k
16
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 ...
- 6,829
16
votes
Algebraic theorems with no known algebraic proofs
There should be many examples in algebraic number theory, as it often interacts nontrivially with analytic number theory. For example, I am pretty sure there is no algebraic proof of Bauer's theorem (...
15
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 ...
- 118k
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 ...
- 1,505
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 ...
- 35.2k
14
votes
Algebraic theorems with no known algebraic proofs
The Scott-Wiegold conjecture posits that the free product of three non-trivial cyclic groups can never be normally generated by a single element.
It was eventually proven by Jim Howie, by studying the ...
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 ...
- 50.6k
13
votes
Accepted
The state of the art on topological rings - the Jacobson topology
This is a pretty well-established and widely studied construction in the literature on topological algebra. However, in my experience it's not always easy to navigate that literature, which explains ...
- 5,407
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 ...
- 10.5k
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 ...
- 5,878
11
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 ...
- 1,690
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 ...
- 156k
11
votes
Accepted
Is being graded commutative a necessary condition on $A$ such that $H^*(A)$ is commutative?
Let $A^\bullet=C^\bullet(X;R)$, the singular cochains on a topological space $X$ with coefficients in a commutative ring $R$, endowed with the cup product of cochains. Given cochains $\varphi\in C^k(X;...
- 35.9k
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 ...
- 3,545
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 ...
- 1,244
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 \...
- 28.7k
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 ...
- 10.8k
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