Question

Here is one of the classical versions of the nullstellensatz: Let $K$ be a field and let $\mathfrak{m}$ be a maximal ideal of the polynomial ring $K[T_1,\ldots,T_n]$. Then $K[T_1,\ldots,T_n]/\mathfrak{m}$ is a finite extension of $K$.

I am interested in a noncommutative version of this theorem. To be more precise: Let $K$ be a field and let $\delta_1,\ldots,\delta_n$ be commuting derivations on $K$. Suppose $\mathfrak{m}$ is a maximal left ideal of $K[\delta_1,\ldots,\delta_n]$. Is $K[\delta_1,\ldots,\delta_n]/\mathfrak{m}$ finite dimensional as a vector space over $K$? If not, is there a simple counterexample?

This seems like a very natural question to ask but I was unable to find anything relevant in the literature. I also tried to lift the proof of the classical theorem to this setting using the technology of associated graded rings/modules but I could not make it work.

Answer 1

I don't think it's true. Namely, Stafford showed ("Non-holonomic modules over Weyl algebras and enveloping algebras," Inventiones Mathematicae, 1985, Volume 79, Number 3, Pages 619-638) that if you choose $\lambda_2, \dots, \lambda_n$ linearly independent over $\mathbb{Q}$ then the element $x_1 + (-\partial_1)\big(\sum_2^{n}\lambda_i x_i(-\partial_i)\big) + \sum_2^{n}(x_i+\partial_i)$ in the $n$th Weyl algebra generates a maximal ideal (this was later vastly generalized by Bernstein-Lunts). If you localize to $K=\mathbb{C}(x_1,\dots, x_n)$, you should still get a maximal cyclic ideal and using induced filtrations on the ideal and the quotient you should be able to see that the quotient has GK dimension $n-1$ (thought of as a module over the localized algebra as a $K$-algebra).