I would like to know whether or not the going-up property holds for some classes of finite filtered extensions of non-commutative rings.

Let $S \subseteq R$ be rings. The pair $(S,R)$ has the going-up property if for every pair of prime ideals $\mathfrak{p} \subseteq \mathfrak{q} \subseteq S$ and for every prime ideal $P$ of $R$ such that $P \cap S = \mathfrak{p}$ there exists a prime $Q \subseteq R$ such that $P \subseteq Q$ and $Q \cap S = \mathfrak{q}$.

When $R$ is a commutative, integral extension of $S$ the pair $(S, R)$ is well-known to have the going-up property, and there are versions of this theorem in the case where $R$ is not necessarily commutative. For example, when $S$ is a subring of the centre of $R$ and $R$ is a finite $S$-module the pair $(S,R)$ has the going-up property.

Question: Does the going-up property hold in the case where neither $S$ nor $R$ are assumed to be commutative, but $R$ has a non-negative filtration $R = \bigcup_{i \geq 0} R_i$ such that the associated graded ring is commutative, both $R$ and $S$ are Noetherian and $R$ is a finite, free as a left and right $S$-module?

Any references or counterexamples would be greatly appreciated.


1 Answer 1


I think this is dealt with in McConnell and Robson's book "Noncommutative noetherian rings". See for example Theorem 10.2.4 (there are many results under various assumptions, but that's a good place to start).


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