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.



A Version of Nullstellensatz for Rings of Dİfferential Operators
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). 


