

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Does dimension of a module (say, dimension of its support) have anything to do with the supremum length of chains of prime submodules like rings? Let's restrict to finitely generated modules over Noetherian ring. Prime submodules are defined analogously to primary submodules: a submodule P in M is prime if P$\neq$M and $M/P$ has no zero divisors, i.e. $am\in P$ implies $m\in P$ or $a \in \mbox{Ann}(M/P)$.

share|cite|improve this question
meta: try to indicate which field of mathematics you're talking about as you begin using terms. For me, a module is more likely to be over a von Neumann algebra or over a tensor category than a Noetherian ring. No one has a monopoly on modules anymore! You did explain which sense you meant, of course, but it took until the second sentence, and most of the terminology of the first sentence doesn't even make sense until you've done so. – Scott Morrison May 9 '10 at 19:16
So you are asking for the relation between the Krull dimension and the prime dimension of a module. I think the two dimensions are equal for multiplication modules. Would that be of interest? – Gjergji Zaimi May 10 '10 at 0:25
up vote 2 down vote accepted

Let $R$ be an integral domain, then for the module $R^n$ its maximal length of chains of prime submodules is much larger than its dimension (for $n>>0$).

share|cite|improve this answer

Your Answer


