## New answers tagged algebraic-k-theory

4
votes

Accepted

### Is there an ∞-categorical interpretation of the Quillen S⁻¹S construction?

A published reference for the claim (right after the question in boldface) is the proof given by Thomason on pages 1657-1658 of "First quadrant spectral sequences in algebraic K-theory via ...

2
votes

### Is there an ∞-categorical interpretation of the Quillen S⁻¹S construction?

Consider an $E_n$-monoid X. We can deloop $X$ to an $\infty$-category $\mathbf{B}X$. There's a natural functor $X^\circlearrowleft : \mathbf{B}X \rightarrow \text{Spc}$ given by the left action of $X$ ...

4
votes

Accepted

### Homotopy coherence datum for composition of Becker-Gottlieb transfers

$\newcommand{\S}{\mathcal S}
\newcommand{\Top}{\mathrm{Top}}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Span}{\mathrm{Span}}
\newcommand{\Fin}{\mathrm{Fin}}
\newcommand{\Fun}{\mathrm{Fun}}$
This is a bit ...

0
votes

### f.g. modules vs. f.g. projective modules

Let $R = \mathbb Z_4$ (by which I mean integers modulo $4$). Then the abelian group $K_0(R)$ is freely generated by $[\mathbb Z_4]$ and $G(R)$ is freely generated by $[\mathbb Z_2]$, and we have $[\...

4
votes

### Algebraic K-theory of a ring

The comments point out that this question is very similar to another question, but an attempt to close as a duplicate failed. So, rather than leave this question on the unanswered queue, I will try to ...

5
votes

Accepted

### Grothendieck group and an almost localization

Not in general, no. Let $S$ be any triangulated category, and consiser the inclusion of the left (say) summand $S \to S \times S$.
You can choose the functor $F$ to be the direct sum functor $\oplus : ...

5
votes

Accepted

### The action of the Grothendieck group on higher K-theory groups

No this is not true in general. Let $R$ be a flat $k$-algebra with the two maps $K_1(R) \to K_1(R\otimes_k R)$ being different. For example, $R = k[t^{\pm 1}]$ for some nonzero ring $k$ and then the ...

3
votes

Accepted

### Example of triangulated category with vanishing $K_0$

The answer is no in general: let $R= \mathbb Z$ and $E= \mathbb Z/p$. Then $[E] = 0$ because of the co/fiber sequence $\mathbb Z\to \mathbb Z\to \mathbb Z/p$.
However, by the theorem of the heart, $...

9
votes

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

In the ten years since this question was asked, there has been a lot of progress in algebraic $K$-theory. For example, Achim Krause, Ben Antieau, and Thomas Nikolaus came up with an algorithm to ...

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