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Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.

23 votes

Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left mo...

Here is an alternative to Andreas proof (which if you unfold it is not so different): We have a functor $M\mapsto \mathrm{Hom}_R(M,R)=:M^\ast$ which gives both a contravariant functor from left $R$-mo …
Torsten Ekedahl's user avatar
19 votes
Accepted

Analogue of Smith normal form for matrices over $\mathbb Z[t]$

For Q1 the problem is that one invariant of the matrix is the (isomorpism class of the) cokernel and any $\mathbb Z[t]$-module generated by $n$ elements and $n$-relations appears as such an invariant. …
Torsten Ekedahl's user avatar
14 votes

What is a "block" in an abelian category?

It seems clear to me that blocks should have something to do with the decomposition of the category as a direct product of subcategories. A decomposition into a product of two factors corresponds exac …
Torsten Ekedahl's user avatar
12 votes
Accepted

Is the functor of divided powers a weakly monoidal functor?

This is a well-known result and, apart from terminology, should be found in Roby, Norbert Lois polynômes multiplicatives universelles. (French. English summary) C. R. Acad. Sci. Paris Sér. A-B 290 (19 …
Torsten Ekedahl's user avatar
7 votes

Reflexive modules over a 2-dimensional regular local ring

If you accept the fact that a $2$-dimensional (local) ring has global dimension $2$, the following is a (somewhat) alternative proof. Choose a free f.g. presentation $F_1 \to F_0 \to M^\ast \to 0$ and …
Torsten Ekedahl's user avatar
5 votes

Comparing lower central series and augmentation ideal completions

As Simon points out, the answer is no in a simple case and if you think about his argument the answer should probably be no as soon as $G^p$ is infinite. However, there is a statement that is very clo …
Torsten Ekedahl's user avatar
3 votes

Invertible elements in monoid rings of unital monoids without non-trivial invertible elements

Let $R$ be a finite dimensional algebra over $\mathbb Z/2$. Then $\{1\}\neq R^\times$ unless $R=(\mathbb Z/2)^n$. Indeed, if $N$ is the radical of $R$, then $1+N\subseteq R^\times$ so we may assume $R …
Torsten Ekedahl's user avatar