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 ...

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 ...

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)...

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 ...

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

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 $...

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

### 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 ...

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. ...

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 $a$ ...

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, ...

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 ...

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 ...

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 ...

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 ...

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 (...

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 ...

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 ...

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

### 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?...

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 ...

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-...

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 ...

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 $G$ ...

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 ...

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 $...

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)....

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 ...

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 ...

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 ...

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