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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.
96
votes
16
answers
18k
views
Why is it a good idea to study a ring by studying its modules?
This is related to another question of mine. Suppose you met someone who was well-acquainted with the basic properties of rings, but who had never heard of a module. You tell him that modules genera …
42
votes
Accepted
Why are ring actions much harder to find than group actions?
First, I am not really sure what you mean by "it is hard to come across a general theory of ring actions." This is precisely module theory! If $f : R \to S$ is any ring homomorphism whatsoever, then c …
30
votes
intuition for hochschild homology
Slogan: Hochschild homology is a (derived) categorification of the trace.
This means the identity at the end of John Pardon's answer is a categorification of the identity $\text{tr}(AB) = \text{ …
26
votes
Accepted
What's an example of a transcendental power series?
If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any pur …
26
votes
Integer matrices which are not a power
I actually even struggle to find examples of primitives matrices in these groups.
Here is a relatively easy sufficient condition. If $M \in SL_n(\mathbb{Z})$ is the $k^{th}$ power of some other matr …
25
votes
8
answers
6k
views
What is the "right" definition of a ring?
This is somewhat related to Greg's question about groups and abelian groups. Suppose you met someone who was well-acquainted with groups, but who was unwilling to accept rings as a meaningful object …
22
votes
What are Homotopy rings good for?
The rationalization of this ring can be understood in a very nice way, as follows. Suppose for simplicity that $X$ is simply connected. Then we can define its rational homotopy groups
$$\pi_n(X, \math …
19
votes
Accepted
Purely noncommutative algebra-Morita equivalence
An algebra is Morita equivalent to a commutative algebra iff it's Morita equivalent to its center, since the center is Morita invariant. So any representative of a nontrivial class in the Brauer group …
18
votes
Are automorphisms of matrix algebras necessarily determinant preservers?
Here is a positive result. Every finite-dimensional algebra $A$ over a field $K$ has an intrinsic determinant, and in fact an intrinsic characteristic polynomial, which is preserved by all automorphis …
17
votes
What is an example of a ring with two (or more) multiplicative right-identities?
Take the semigroup ring of a semigroup with two or more multiplicative right identities. For example, the semigroup
$$S = \langle a, b | ab = aa = a, ba = bb = b \rangle$$
works (it is the universal e …
17
votes
Accepted
Is there any "fundamental" distinction between min-plus, max-plus, min-product, and max-prod...
Consider the following four semirings, listed in the order underlying set, addition, additive identity, multiplication, multiplicative identity:
$A = (\mathbb{R} \cup \{ \infty \}, \text{min}, \inft …
16
votes
2
answers
1k
views
Which commutative groups are the group of units of some field?
Inspired by a recent question on the multiplicative group of fields. Necessary conditions include that there are at most $n$ solutions to $x^n = 1$ in such a group and that any finite subgroup is cyc …
16
votes
How many Lie and associative algebras over a finite field are there?
Bjorn Poonen addresses this question for commutative (associative, unital) algebras in The moduli space of commutative algebras of finite rank; asymptotically we have
$$q^{\frac{2}{27} n^3 + O(n^{8/3} …
15
votes
Accepted
Left-Module Structure on the Tensor Product ofTwo Left Modules
Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product …
15
votes
Dual of a bimodule
As explained in more detail in this blog post linked by Jakob in the comments, every $(A, B)$-bimodule $M$ has two natural duals:
If $M$ is finitely generated projective as a left $A$-module, it has …