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Results tagged with ra.rings-and-algebras
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user 4008
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.
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 …
- 22.6k
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 …
- 22.6k
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. …
- 22.6k
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 …
- 22.6k
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 …
- 22.6k
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 …
- 22.6k
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 …
- 22.6k