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Non-commutative rings and algebras, non-associative algebras. Can be used in combination with ra.rings-and-algebras

26 votes
1 answer
990 views

Idempotents in Rings of Differential Operators

Differential Operators on General Commutative Rings Let k be an algebraically closed field of characteristic zero, and let R be a commutative k-algebra. Then a (Grothendieck) differential operator on …
Greg Muller's user avatar
23 votes
Accepted

How much theory works out for "almost commutative" rings?

Don't get too excited about the theory of algebraic geometry for almost commutative algebras. A ring can be almost commutative and still have some very weird behavior. The Weyl algebras (the differe …
Greg Muller's user avatar
10 votes

Proof a Weyl Algebra isn't isomorphic to a matrix ring over a division ring

A different proof would be to show that a Weyl algebra is not semisimple, that is, that it is not a direct sum of simple submodules as a left module over itself. However, note that there is an infini …
Greg Muller's user avatar
7 votes
1 answer
566 views

Depth Zero Ideals in the Homogenized Weyl Algebra

Let $\mathcal{D}$ be the $n$th Weyl algebra $ \mathcal{D} :=k[x_1,...,x_n,\partial_1,...,\partial_n] $, where $\partial_ix_i-x_i\partial_i=1$. Let $\widetilde{\mathcal{D}}$ be its Rees algebra, whi …
Greg Muller's user avatar
6 votes
1 answer
334 views

A potential resolution of $R/r$

The DGA For $k$ some field, let $R$ be a $k$-algebra, and let $r\in R$. Define a differential graded algebra $\mathbf{R}_r$ as follows. As a graded algebra, it is isomorphic to $R\langle t\rangle$, t …
Greg Muller's user avatar
5 votes

Idempotents in Rings of Differential Operators

Here's a cute partial result, classifying idempotents of order 1. Let $\delta$ be an idempotent differential operator of order 1. Then there is a unique decomposition $R\simeq A\oplus M$, with $A$ a …
Greg Muller's user avatar
3 votes
0 answers
163 views

Pulling out factors in a Noetherian Domain

Let $R$ be a Noetherian domain (not-necessarily commutative), and let $S$ be a Noetherian subring of $R$. An element $r\in R$ is left $S$-irreducible if, for any $s\in S$ and $r' \in R$ with $sr'=r$, …
Greg Muller's user avatar
3 votes

When does the converse to Schur's Lemma hold?

Let $p$ be a prime, and let $R(p)$ be the residue field at $p$. If $R \to R(p)$ is not a surjection, then then $R(p)$ is an $R$ module whose endomorphism ring is $R(p)$, but such that the image of $R …
Greg Muller's user avatar