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5 votes
2 answers
397 views

Ring with vanishing $K_0$

Suppose we have a ring $R$ such that the Grothendieck group $K_{0}(R)=0$. Question 1: Does it follow that there exists two positive natural numbers $n\neq m$ such that $R^{m}$ is isomorphic to $ R^{n}$...
cellular's user avatar
  • 855
1 vote
0 answers
123 views

Quiver representations and the standard matrix decompositions

Many matrix decompositions - like the Jordan Normal Form, the SVD, the spectral theorem, the Takagi decomposition - have the property that they express a matrix $M$ as the form: $$M = A D B$$ where $D$...
wlad's user avatar
  • 4,943
12 votes
1 answer
312 views

Group ring with infinite stable rank

In searching for a counterexample in homological stability, I came across the following question: Is there a known example of a finitely presented group $G$, so that the group ring $\mathbb{Z}[G]$ ...
user124543's user avatar
5 votes
1 answer
273 views

Examples of noetherian local rings $R$ such that $K'_0(R)$ is not isomorphic to $\mathbb Z$

Does there exist a simple example of a commutative noetherian local ring $R$ such that $K'_0(R) = K_0(\mbox{Mod-}R)$ (by $\mbox{Mod-}R$ I mean the abelian category of finitely generated $R$-modules) ...
Mykola Pochekai's user avatar
5 votes
2 answers
470 views

Exact subcategory with trivial Grothendieck group: what are the consequences and examples

Let $C$ be (a full) exact subcategory of the category of $R$-modules. We suppose that $C$ is essentially small. If the Grothendieck group $K_{0}(C)=0$, what can be said about the higher groups $K_{n}(...
GSM's user avatar
  • 223
2 votes
1 answer
268 views

Does $A\oplus M_n(R)\cong B\oplus M_n(R)$ imply $A\cong B$? $R$ Dedekind domain

Let $R$ be a Dedekind domain and $A, B$ be finitely generated projective $M_n(R)$-modules. Is it true that $A\oplus M_n(R)\cong B\oplus M_n(R)\:\:\Rightarrow\:\:A\cong B$? Here, the isomorphism is ...
Sam Williams's user avatar
3 votes
1 answer
236 views

For $T$ the $2\times 2$ triangular matrices over $R$, can we write $GL_2(T)=U(T)E_2(T)$?

Let $R$ be a commutative ring with identity, and let $T = T_2(R)$ be the ring of $2\times 2$ upper triangular matrices over $R$. Is it true that the following identity holds? $$GL_2(T)=U(T)E_2(T)$$ ...
Sam Williams's user avatar
7 votes
1 answer
212 views

What does $K_1(R)$ tell us about $GL_n(R)/E_n(R)$?

Let $D$ be a division ring, and $R=D[t_1,\ldots,t_n]$. If $GL_m(R)$ is the usual group of invertible matrices over $R$, then by $E_m(R)$ I mean the subgroup of $GL_m(R)$ generated by the elementary ...
Sam Williams's user avatar
7 votes
1 answer
518 views

When is $GL_m(R)$ generated by elementary and diagonal matrices?

Let $D$ be a division ring and $R=D[t_1,\ldots,t_n]$ the polynomial ring in $n$ variables. Now let $GL_m(R),\,E_m(R)$ be the usual general linear group and its subgroup generated by the elementary ...
Sam Williams's user avatar
5 votes
1 answer
342 views

Can the trivial module be stably free for a monoid ring?

Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ ...
Benjamin Steinberg's user avatar
4 votes
1 answer
283 views

Geometric contractibility of noetherian rings

Let $A$ be a noetherian ring. Let us define $A$ to be $n$-contractible if All locally free sheaves of rank $\le n$ over $\text{Spec} A$ is trivial. There exist a non-trivial locally free sheaf of ...
Shubhodip Mondal's user avatar
6 votes
1 answer
302 views

Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?

Let $A$ be a finite dimensional semisimple $F$-algebra and $K_1(A)$ the Whitehead group of $A$. By splitting $A$ into its Wedderburn components, the reduced norm map $\operatorname{nr}_A:K_1(A)\to Z(...
eddie's user avatar
  • 255
8 votes
0 answers
307 views

How bad can $SK_1$ of a commutative ring be?

For a commutative ring $R$ define $\mathrm{SK}_1(n, R)=\mathrm{SL}(n, R)/\mathrm{E}(n, R)$, the quotient of the special linear group by its subgroup generated by the elementary matrices. When $n\...
Andrei Smolensky's user avatar
7 votes
1 answer
1k views

The saturation of Murray von Neumann relation

Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version: Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows: We say $a M b$ iff $...
Ali Taghavi's user avatar
12 votes
1 answer
458 views

Algebraic K-theory of a ring

I started to learn some algebraic $K$-theory and its relation to geometric topology problems. My question is: What is the list of rings such that all their algebraic $K$-theory groups are known? I ...
sphere's user avatar
  • 433
19 votes
2 answers
703 views

Besides $F_q$, for which rings $R$ is $K_i(R)$ completely known?

With the exception of finite fields and "trivial examples", which rings $R$ are such that Quillen's algebraic $K$ groups $K_i(R)$ are completely known for all $i\geq 0$? Here, by "trivial examples" ...
user avatar
6 votes
1 answer
550 views

A generalized K- theory via generalized idempotents

Edit After the answer by Neil Strickland, I add the word "a ring" in this new version. In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach ...
Ali Taghavi's user avatar
18 votes
4 answers
2k views

For which rings $R$ is $\mathrm{SL}_n(R)$ generated by transvections?

Let $R$ be a commutative ring $R$ with $1$, and $n \geq 2$ an integer. Under which conditions is the group $\operatorname{SL}_n(R)$ generated by transvections? (A transvection is a matrix with $1$ ...
Tom De Medts's user avatar
  • 6,614
7 votes
3 answers
1k views

Group rings of infinite products of groups

Given a infinite family of groups $(G_i)$ for $i\in I$. Is there a ring theoretic construction, that produces $R[\prod_{i\in I} G_i]$ using only the rings $(R[G_i])_{i\in I}$ ? For the case of a ...
HenrikRüping's user avatar
15 votes
2 answers
2k views

Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left modules instead of right modules?

I am under the impression that in the definition of the Grothendieck group $K_0(R)$ of a (non-commutative) ring it doesn't matter whether we apply the usual $K_0$ construction to the exact category of ...
Beren Sanders's user avatar