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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.
3
votes
Accepted
Alternative algebras in characteristic 2, especially scalar extension
A good samaritan has delivered an answer directly to my email account. It is indeed almost obvious, as long as one ignores the bit about the multiplication operators!
Take for instance the left alte …
6
votes
Which R-algebras are the group ring of some group over a ring R?
Reid Barton's very nice answer to Computing the structure of the group completion of an abelian monoid, how hard can it be? contained a pointer to the wikipedia page for the Eilenberg-Mazur swindle. …
11
votes
Indeterminate "$x$" in algebra/ring Theory
Among infinitely many other places, this issue is discussed in Sections 4.3 and 4.4 of my notes on commutative algebra:
http://alpha.math.uga.edu/~pete/integral.pdf
In Section 4.3 I give Mariano's def …
8
votes
Rings with right inverses
For a memorable explicit example, let $V = \mathbb{R}[x]$ be the real vector space of polynomial functions, and let $R = \operatorname{End}(V)$ be the ring of $\mathbb{R}$-linear endomorphisms (aka li …
3
votes
Accepted
Fractional powers of Dirichlet series?
For your first question: let $R$ be any commutative ring, and let $D(R)$ be the ring of formal Dirichlet series over $R$, i.e., the set of all functions $f: \mathbb{Z}^+ \rightarrow R$ under pointwise …
6
votes
Is there an algebraic proof of the infinitude of primes?
[I don't really know what constitutes an "algebraic proof" of infinitude of prime numbers.]
The proof that BCnrd alludes to above is described in somewhat more length on p. 5 of
http://alpha.math.uga. …
10
votes
Accepted
notation for formal Laurent series
I agree with the mathematician of your acquaintance -- well, okay, I am the mathematician of your acquaintance.
Here are some references for the notation $K((x))$ for the field of formal Laurent seri …
9
votes
What does the semiring of ideals of a ring R tell us about R?
This is sort of a sideways answer, but: in many ways the monoid $\operatorname{Prin}(R)$ of principal ideals carries more information. If $R$ is a domain $\operatorname{Prin}(R)$ is a cancellative mo …
7
votes
$n$-forms representing zero (versus division rings)
You are asking about a venerable and active area in Diophantine equations lying at the border of arithmetic geometry and analytic number theory. Some keywords are forms in many variables and circle m …
18
votes
Accepted
Which commutative groups are the group of units of some field?
The following paper claims an answer to this question:
Dicker, R. M.
A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field.
Proc. London Math. …
49
votes
Why is it a good idea to study a ring by studying its modules?
I want to answer your question twice: first with a "top-down" approach and second with a "bottom-up" approach. Let me limit myself to the first answer here and see how I do.
I claim the following an …
24
votes
Simplest examples of rings that are not isomorphic to their opposites
To amplify on Bugs Bunny's answer: let $D$ be a finite dimensional central division algebra over a field $K$. Then $D \otimes_K D^{\operatorname{op}} \cong \operatorname{End}_K(D)$. From this it fol …
15
votes
transcendental Galois theory
Even in our day of sophisticated search engines, it still seems that the success of a search often turns on knowing exactly the right keyword.
I just followed up on Sylvain Bonnot's comment above. T …
12
votes
What makes a theorem *a* "nullstellensatz."
For a field $k$, by a "Nullstellensatz" over $k$, I mean an explicit description of the Galois connection between subsets of $k^n$ and ideals in the polynomial ring $k[x_1,\ldots,x_n]$. See this MO q …
6
votes
Infinite dimensional central simple algebras
I only know a little bit about this, so for your sake I hope someone more knowledgeable comes along...
As for your first question:
It is not hard to see that the tensor product of any two central simp …