All Questions
9 questions
9
votes
5
answers
2k
views
Proof a Weyl Algebra isn't isomorphic to a matrix ring over a division ring
Can anyone prove that a Weyl Algebra is not isomorphic to a matrix ring over a division ring?
7
votes
2
answers
489
views
How big can a commutative subalgebra of Weyl algebra be?
Consider the smallest Weyl algebra $A_1=\{q,p; qp-pq=1\}$. It is known that there exist pairs of commuting elements, say $L$ and $M$, that obey various polynomial relations, e.g. elliptic curves. I ...
7
votes
1
answer
567
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, ...
5
votes
2
answers
340
views
Gelfand-Kirillov dimension of generalized Weyl algebras
I believe that the Gelfand-Kirillov (GK) dimension for a generalized Weyl algebra $D(\sigma,a)$ is just the GKdim$(D) + 1$.
Does anyone have a reference for this?
I can find partial results, and I ...
4
votes
1
answer
242
views
Locally nilpotent operators of the Weyl algebra
$\newcommand{\ad}{\operatorname{ad}}$As my recent post (here) did not receive any answers yet, I thought I would ask a similar question in which I'm also interested.
Let $A=$ $^{k \langle x,y\rangle }...
4
votes
0
answers
152
views
Nilpotent operator of the Weyl algebra
For a research project I'm currently working on, I came across the following problem:
Let $A=$ $^{k <x,y> }\Big/_{(yx-xy-1)}$ be the Weyl Algebra over a field $k$ of characteristic $p$, where $...
3
votes
1
answer
197
views
Gelfand-Kirillov dimension of the first Weyl algebra
How can we compute the Gelfand-Kirillov dimension (GK for short) of the first Weyl algebra?
As we know we can look at the Weyl algebra as a generalized Weyl algebra in the following way:
Let $A=\...
2
votes
1
answer
341
views
Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra
In afew threads I've read that the Weyl algebra $A_{n+1}(k)$ is isomorphic to the $k$-tensor product of $A_n(k)$ with $A_1(k)$, why is this true?
0
votes
1
answer
63
views
What are the fixed points of $\alpha^n-\mu_j$ for a fixed $j$?
Let us consider the polynomial ring $\Bbb C[x_1,...,x_s]$ and $\alpha(x_i)= x_i + \mu_i$ where $\mu_i \in \Bbb C$ are not all zero. Then $\alpha \in \mathrm{Aut}(\Bbb C[x_1,...,x_s])$.
What are ...