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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.

12 votes
Accepted

Intuitive Example of a Jacobson Radical

I think my favourite characterization for rings with identity is that y is in the Jacobson radical of R if and only if 1-yx is right invertible for any x in R - so y is sufficiently "zero-like" that m …
Greg Stevenson's user avatar
28 votes
Accepted

What is the "right" definition of a ring?

Well rings are naturally the objects which act on abelian groups - indeed composition always endows the endomorphisms of an abelian group with the structure of a ring. So if one is interested in the e …
Greg Stevenson's user avatar
7 votes

Are quotients of polynomial rings almost UFDs?

I would say that it is not true that quotients of polynomials rings are "almost UFDs". For starters, being a quotient of $k[x_1,\ldots,x_n]$ for some $n$ just says that the ring is finitely generated …
Greg Stevenson's user avatar
8 votes

What is interesting/useful about big Witt Vectors?

From my point of view (which is I should say basically my, probably incorrect, interpretation of Borger's) part of the interest in the big Witt ring is that, at least in the flat case where there are …
Greg Stevenson's user avatar
15 votes

Why is it a good idea to study a ring by studying its modules?

I think the main reason is the flexibility of working in the category of $R$-modules rather than just with the ring $R$. For instance suppose we stick to rings - we have some ways of building new ring …
Greg Stevenson's user avatar
8 votes
Accepted

Is the tensorproduct of a triangulated category with a ring again triangulated?

I would imagine it is false in general that given a triangulated category $T$ the category $T\otimes R$ is also triangulated. The following is a concrete counterexample. Consider $D^b(\mathbb{Z})$ an …
Greg Stevenson's user avatar