28 votes

Why is the bicategory viewpoint useful?

You believe the 1-category is interesting but somehow not the bicategory, yet one could say the exact same thing one dimension below: Why is this category even interesing while you could just consider ...
Simon Henry's user avatar
  • 39.9k
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 ...
Pace Nielsen's 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)...
Theo Johnson-Freyd's user avatar
14 votes

Algebra with a certain abelian group as the multiplicative group

I am going to assume that by "algebra" you simply mean a ring. The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose ...
Alex B.'s user avatar
  • 12.8k
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 ...
Salvo Tringali's user avatar
12 votes
Accepted

Simple component that is not a two-sided ideal

I'm not sure why you believe it depends on $R$ being semisimple. The usual argument goes like this and makes no reference to semisimplicity: If $r\in R$ and $L'\cong L$ where $L'$ is a left ideal of $...
rschwieb's user avatar
  • 1,593
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 ...
David E Speyer's user avatar
10 votes

Center of a monoid ring

This was originally two questions, one asking about the center of a group ring and one asking about the center of a monoid ring. The answer for groups is quite simple but the answer for monoids is ...
Benjamin Steinberg's user avatar
10 votes
Accepted

Ring in which $x^n-x$ is central for every $x$

Herstein, "A generalization of a theorem of Jacobson" Amer. J. Math. 73 (1951), 756–762 proves this. Also part III, Amer. J. Math. 75 (1953), 105–111 proves something a bit more general. ...
Dave Benson's user avatar
  • 11.2k
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 $a$ ...
Tim Campion's user avatar
  • 60.2k
8 votes
Accepted

Categories of modules generated under coproducts by a small set?

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book Prest, ...
Jeremy Rickard's user avatar
8 votes
Accepted

If the Grothendieck ring of a semiring on a free commutative monoid is unital, is the original semiring unital?

The answer is no. Let $S$ be a finite meet semilattice without maximum. For concreteness, take $S$ to be the proper subsets of $\{1,2\}$ under intersection. Let $\mathbb NS$ be the semigroup ...
Benjamin Steinberg's user avatar
8 votes
Accepted

Morita equivalences and centers of some algebras

The answer is that in your matrix $\left(\begin{smallmatrix} 0&x_0\\0&0\end{smallmatrix}\right)$, the $x_0$ denotes the isomorphism of modules given by left multiplication by $x_0$, so it ...
Dave Benson's user avatar
  • 11.2k
8 votes
Accepted

Kaplansky inverse element theorem on group C-star algebra

This is not accurate as far as discrete groups are concerned. First of all it is an open question, called Kaplansky's direct finiteness conjecture, whether every group ring over a field has the ...
Benjamin Steinberg's user avatar
7 votes

Does $R$ is Dedekind-finite imply $\mathbb{M}_n(R)$ is Dedekind-finite

The answer is no, even for $n=2$. One thng to note is that over the decades, different groups of authors have used different terminology. The property you call Dedekind-finite has been called ...
Yemon Choi's user avatar
  • 25.4k
7 votes

Idempotent Laurent polynomials (in noncommuting variables)

No, it can not. $R$ is a group ring of the free group with $n$ generators. This group is locally indicable (any non-trivial subgroup has a homomorphism onto $\mathbb{Z}$), thus by result of Higman (...
Fedor Petrov's user avatar
7 votes
Accepted

$H^1(X, GL(n, \mathcal{O}_X))$ and Vector Bundles

Yes, this is called non-abelian sheaf cohomology. If $X$ is a topological space and $\mathcal{G}$ is a sheaf of groups, then $H^0(X, \mathcal{G})$ is the global sections of $\mathcal{G}$, and there is ...
David E Speyer's user avatar
7 votes
Accepted

Infinite linearly independent set in finitely generated module

In a commutative ring $R$ this does not exist. Better for any $n\ge 0$, if $M$ is an $R$-module generated by $n$ elements, then $R^{n+1}$ doesn't embed into $M$. Indeed, lifting if necessary, we can ...
YCor's user avatar
  • 60k
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 ...
Neil Strickland's user avatar
6 votes
Accepted

Quotient Ring number

The definition of $N_i(R,I)$ depends only on $R/I$, which is a ring isomorphic to $M_n(q)$ as HeinrichD has already observed. Thus, the question is: what is the sum $$ S = \sum_{A\in M_n(q)} A^i \quad?...
Keith Kearnes's user avatar
6 votes

the relation between projective and quasi-projective modules

Every simple module is trivially quasi-projective, and if every simple $R$-module is projective then $R$ is semisimple. So semisimple rings are the only rings for which quasi-projective implies ...
Jeremy Rickard's user avatar
6 votes
Accepted

number of indecomposable summands of an extension of two modules

The answers to both questions are no in general, with a counter-example being given by the path algebra of a quiver of type $\mathsf{D}_4$—the category of left modules over this algebra is a Hom-...
Matthew Pressland's user avatar
6 votes
Accepted

invertibility of matrix over free associative algebra

You are asking whether the following holds. Claim. Let $F$ be a field. Then the free associative algebra $F\langle x,y\rangle$ is a $GE_2$-ring in the sense of P. M. Cohn [2]. The answer ...
Luc Guyot's user avatar
  • 7,343
6 votes
Accepted

Is every (left) graded-Noetherian graded ring (left) Noetherian?

The answer is yes, by Corollary 2.2 in C. Nastasescu, F. Van Oystaeyen, Graded rings with finiteness conditions II, Comm. Algebra 13 (1985), 605-618. More generally, we have the following. Let $G$ ...
Fred Rohrer's user avatar
  • 6,650
6 votes
Accepted

Curious anti-commutative ring

I noticed this now, and I want to remark that the underlying abelian group can in fact be described very precisely. To do that, note that: (1) the defining relations easily imply that the abelian ...
Vladimir Dotsenko's user avatar
6 votes
Accepted

Do you know which is the minimal local ring that is not isomorphic to its opposite?

I learned this example from MO-user Johannes Hahn: The algebra is $A=K<x,y>/(x^3,y*x,y^2,x^2*y)$ over a field $K$ with 2 elements. Then $A$ as an $A$-module as 20 submodules, but $A^{op}$ as an $...
Mare's user avatar
  • 25.5k
6 votes

Idempotent Laurent polynomials (in noncommuting variables)

Here's a self-contained proof (which is certainly Higman's proof), following Fedor Petrov's answer. Let $G$ be a locally indicable group (= every nontrivial f.g. subgroup has $\mathbf{Z}$ as quotient)....
YCor's user avatar
  • 60k
6 votes
Accepted

Abelian groups such that $A \cong \mathrm{End}(A)$ and "complete rings"

The rings, you call ``complete'' are known as $E$-rings (as Ulrich Pennig mentioned in the comments). Some comments on your questions There are too many results on the $E$-rings to list them here and ...
Adam Przeździecki's user avatar
6 votes

Ideals of an ordered ring

Here's an example showing that the answer is negative. Consider the monoid algebra $\mathbf{Z}[\mathbf{R}_{\ge 0}]$. It thus consists of finitely supported sums $q=\sum_{t\ge 0}q_tX^t$. Say that such ...
YCor's user avatar
  • 60k
6 votes
Accepted

Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?

$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need ...
Jeremy Rickard's user avatar

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