7
votes
1answer
159 views

Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$

Consider the semiring $$\mathbb{N}[H,H^{-1}]/(H^p+H^q = H^{p+q}+1)_{p,q \in \mathbb{Z}}.$$ Is it finitely presentable? Is there any simplification of the relations (except for $p \geq q \geq 0$)? ...
3
votes
1answer
310 views

What sort of ring-theoretic properties does the representation ring of a compact Lie group possess?

Recall the definition of the representation ring $R(G)$ of a compact Lie group $G$. I'd like a reference that gives me basic ring-theoretic properties that $R(G)$ always has, or enough info that I can ...
7
votes
0answers
159 views

Is the generation of rings by their units a question in K-theory?

Susan's question When can number rings be spanned (as $\mathbb{Z}$-modules) by units? smells like an algebraic K-theory question in disguise. I'll reformulate the question first: Given an integral ...
4
votes
0answers
73 views

Homology of special linear group over local field

I am trying to compute the group $H_1(Sl_2(\mathbb{Z}_2),M)$, where $\mathbb{Z}_2$ are $2$-adic integers and M is a module $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. I suppose that the group acts on $M$ by ...
0
votes
1answer
162 views

About regular local rings and Socles

Let R be a regular local ring with $ \text{dim} R = d $. If $ 0\rightarrow R\rightarrow I_0\rightarrow ...\rightarrow I_d\rightarrow 0 $. Then why for $ 0\leq i\leq d-1 $, the socle of $ I_i $ is ...
0
votes
1answer
228 views

Length of a module

Let R be a commutative ring, M an R-module of finite length and let N be an Injective R-module with zero socle. Then why $ \text{Hom}_R(M, N) $ is zero?
1
vote
1answer
115 views

Does the global dimension gldim R equal the projective dimension of R as bimodule over its enveloping algebra?

I know that generally the answer is no, for example the weyl algebra。 But is this true for commutative algebra? or we may restrict to affine commutative algebras。 Maybe ,it is a classical result. ...
15
votes
2answers
1k views

Who named it the Snake Lemma?

What is the history behind the colorful name of this result? Cartan-Eilenberg states it without any particular fanfare.
12
votes
3answers
643 views

Injective dimension of graded-injective modules.

In "Existence theorems..." Van den Bergh proposes the following "pleasant excercise in homological algebra": Let $A$ be a connected graded noetherian $k$-algebra (that is, $\mathbb N$-graded with ...
8
votes
2answers
1k views

State of the art for Gersten's conjecture for K-theory?

Does anyone know (of a reference to) under what restrictions on the regular scheme $X$ it is known that we have an exact sequence $$0 \to \mathcal{K}_n(X) \to \bigoplus_{x \in X^{(0)}} K_n(k(x)) \to ...
6
votes
1answer
378 views

Projective modules over free groups

Consider the ring of Laurent polynomials $R := \mathbb{Z}[s,s^{-1}]$ with integer coefficients. Are all projective $R$-modules free? (Let's say left modules by convention.) More generally, let $G$ ...
4
votes
0answers
625 views

$Ext$ functor, filtered complexes and spectral sequences

Let $\mathcal{A}$ an abelian category. Take $M$ an object of $\mathcal{A}$, and $K_*$ a bounded complex in $\mathcal{A}$ equipped with a bounded increasing filtration $F$. By using homological and ...
6
votes
2answers
1k views

Zero divisor conjecture and idempotent conjecture

Let $G$ be a torsion-free group and $C$ the ring of complex numbers. The zero divisor (idempotent, resp.) conjecture is that there is no nontrivial zerodivisor (idempotent, resp.) in $CG$. The wiki ...
2
votes
1answer
778 views

Automorphism theorem

Help me please to find reference for the proof of the following theorem: Theorem. Let $\theta$ be a Leibniz cocycle on the Leibniz algebra L with values in $V,$ and assume $\theta^{\bot}\cap ...
8
votes
2answers
390 views

Maps between K-groups induced by rings homomorphism

Let $f: R\to S$ be a map between two commutative Noetherian rings. Let $G_0(R)=K_0(mod R)$ be the Grothendieck group of finite generated modules over $R$. It means $G_0(R)$ is the quotient of the free ...