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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 $a \in \mbox{Ann}(M/P)$.

edited May 9 2010 at 18:31

Qiaochu Yuan
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asked May 9 2010 at 16:57

ashpool
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	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 2010 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 2010 at 0:25

zamanjan · Accepted Answer · 2010-08-17 06:00:55Z UTC

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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 $n>>0$).

answered Aug 17 2010 at 6:00

zamanjan
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Dimension of module

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