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Let $M$ be a finitely generated module over the commutative noetherian ring $R$. Let ${\cal C}(M)$ be the set of all primes $P$ in $R$ such that $R/P$ appears as a quotient in every composition series for $M$. Clearly ${\cal C}(M)$ includes all the primes associated to $M$. But it can contain other primes as well. For example, let $R$ be the local ring of a singular point on an irreducible curve and let $M$ be the maximal ideal of $R$. Then $R/M$ appears as a quotient in every composition series for $M$, but $M$ (as an ideal) is not associated to $M$ (as a module).

So: Is there a nice characterization of the set ${\cal C}(M)$?

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Hi Steven, nice question! I am just curious, could you share the motivation with us? –  Hailong Dao Dec 7 '10 at 6:06
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Hailong: Given two modules with the same support, one would ideally like to believe that they represent the same class in $K$-theory up to classes of modules from some well-behaved subcategory. (This is true in the smooth case, with the well-behaved subcategory being modules of higher codimension --- but in the singular case one often wants a more restricted subcategory, e.g. modules that not only have higher codimension but also meet the singular locus in higher codimension.) Primes in ${\cal C}(M)$ are potential obstructions to that program, so one wants to understand those primes. –  Steven Landsburg Dec 7 '10 at 6:33
    
Hailong: The problem is interesting even in the smooth case; for understanding K-theory relative to a collection of subschemes, for example, where the "well behaved subcategory" should consist of modules that meet all those subschemes in higher codimensions. –  Steven Landsburg Dec 7 '10 at 6:37
    
I see, in fact I have a K-theoretic answer in mind! I will post it, but may be you can ask your original question, it looks very interesting! –  Hailong Dao Dec 7 '10 at 6:57
    
Hi Steven, a silly question: if $P\cong Q$ as $R$-modules, do you consider them the same? I asked because for example if $R$ is a domain, then $(f) \cong (g)$ for any 2 elements $f,g$ but $R/(f)$ and $R/(g)$ are not isomorphic unless f=g*unit. –  Hailong Dao Dec 8 '10 at 3:17

1 Answer 1

up vote 8 down vote accepted

I think this is a subtle question. The best result I am aware of is the following paper "Filtrations of Modules, the Chow Group, and the Grothendieck Group", by Jean Chan. She proved the following: Let $\mathcal F$ be any composition series of $M$. Let $c_i(\mathcal F)$ be the formal sum of primes of height $i$. Then $c_0(\mathcal F), c_1(\mathcal F)$, as elements in the Chow group of $R$, does not depend on the composition $\mathcal F$, so we can talk about $c_0(M), c_1(M)$.

So, for example, any such series will always contain the same minimal primes. As for height one primes, we only know that the sum of them is constant up to rational equivalences. However, if there is only one height one prime (your example of local ring of curves), then you can deduce its presence in any series by showing $c_1(M) \neq 0$ in the Chow group of $R$.

I will also note that in Eisenbud's "Commutative Algebra..." after Proposition 3.7, the author remarks that modules which always have a filtration consisting only the associated primes are called clean. Clearly for clean modules one has $C(M) = Ass(M)$. There is no clean(!) criterion for cleanliness, as far as I know, but see this paper for some partial results!

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