MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(I removed my motivation because it may be misleading :) )

Let $A$ be a noetherian commutative ring and let $M \neq 0$ be a finitely generated zero-dimensional (i.e. $\mathrm{dim} \ \mathrm{Supp}(M) = 0$) $A$-module. Then the submodule $0 < M$ has primary decomposition $0 = \bigcap_{\mathfrak{p} \in \mathrm{Supp}(M)} M(\mathfrak{p})$, where $M(\mathfrak{p})$ is the $\mathfrak{p}$-primary component of $M$, i.e. the kernel of the canonical morphism $M \rightarrow M_{\mathfrak{p}}$. I have proven (it's hopefully correct) that the canonical morphism $\mathbf{j}:M \rightarrow \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$ ist an isomorphism and that $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) \leq M/M(\mathfrak{q}) \leq \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$, where $M \lbrack \mathfrak{q}^\infty \rbrack$ is the $\mathfrak{q}$-torsion part of $M$ ($0$-th local cohomology with support $\mathfrak{q}$). Now, my question is if in general $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) = M/M(\mathfrak{q})$ so that $M = \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M \lbrack \mathfrak{p}^\infty \rbrack$, or do I need to assume that $A$ is a Dedekind ring (and perhaps also that $M$ is a torsion module) and if so, how can I prove this?

share|cite|improve this question
Primary decomposition exists for finitely generated modules over noetherian rings. Serge Lang has an excellent proof of the decomposition in chapter 10 of algebra. I believe Bourbaki and Matsumura also have proofs of this (though surprisingly not atiyah-macdonald.) The standard approach uses the associated primes rather than the support. – Harry Gindi Feb 14 '10 at 19:55
Well, of, course. My question was about primary decomposition in a very special case. – user717 Feb 14 '10 at 19:58
I say this only because the method that you're using looks really much more difficult than necessary. You can get unique factorization into prime ideals immediately by assuming the ring is Dedekind. Then if you assume further that the ring is a principal ideal domain, the result follows from unique ideal factorization (since all ideals are principal.) – Harry Gindi Feb 14 '10 at 19:59
You use this idea of isolated primes. In a dedekind ring, all of the primes are isolated, which gives you the result. – Harry Gindi Feb 14 '10 at 20:00
Right, but I wanted to see in what generality this works! I knew that it at least works over Dedekind domains and if I did not make a mistake, it also works for zero-dimensional modules (only one part of the structure theorem of course!). I need this somewhere else in this generality. – user717 Feb 14 '10 at 20:03
up vote 6 down vote accepted

If $M$ is finitely generated and has $0$-dimensional support, then $M$ is in fact supported at finitely many maximal ideals (a $0$-dimensional closed subset of the Spec of a Noetherian ring is just a finite union of maximal ideals), and one has the isomorphism $M = \oplus_{\mathfrak p} M_{\mathfrak p} = \oplus_{\mathfrak p} M[\mathfrak p^{\infty}].$

The first equality just specifies the fact that, since the quasi-coherent sheaf on Spec $A$ attached to $M$ is supported at finitely many closed points, it is a sky-scraper sheaf at these points, and its global sections are just the sum of its stalks at those finitely many points. (In particular, at all ${\mathfrak p}$ not in the support of $M$, the localization $M_{\mathfrak p}$ vanishes, and so does not contribute to the direct sum, so really the direct sum is just over the finitely many points in the support.) To see the second equality (which is the crux of the question as far as I can tell), note that we reduce to the local case: $M_{\mathfrak p}$ is finitely generated over $A_{\mathfrak p}$, and has support equal to the closed point ${\mathfrak p}$ of Spec $A_{\mathfrak p}$. A consideration of the very definition of support will now show that each element of $M_{\mathfrak p}$ is annihilated by some power of ${\mathfrak p}$, and hence that $M_{\mathfrak p} = M[\mathfrak p^{\infty}]$.

From this decomposition everything else is easy to work out: for example, $M(\mathfrak p)$ (which if I understand correctly is defined to be the kernel of the map to $M_{\mathfrak p}$) is precisely $\oplus_{\mathfrak q \neq \mathfrak p} M_{\mathfrak q} = \oplus_{\mathfrak q \neq \mathfrak p} M[\mathfrak q^{\infty}].$

In particular, if we want to isolate $M[\mathfrak q^{\infty}],$ we just intersect the $M(\mathfrak p)$ for all $\mathfrak p \neq \mathfrak q$. This explains the formula in Arminius's answer.

In short: rather than having to memorize or quote technical results from Eisenbud (or elsewhere), one can use geometric reasoning on Spec $A$ to answer such questions. (All the technical reasoning has been embedded in one step: the proof of the correspondence between $A$-modules and quasi-coherent sheaves on Spec $A$.)

share|cite|improve this answer
Thanks a lot for the geometric explanation! – user717 Feb 14 '10 at 20:53

I could answer this now myself (so, sorry for asking!): From a general result (presented e.g. in Eisenbud's commutative algebra book) it follows that $M \lbrack \mathfrak{q}^\infty \rbrack = \bigcap_{\mathfrak{p} \in \mathrm{Supp}(M), \mathfrak{p} \neq \mathfrak{q}} M(\mathfrak{p})$ and then it follows that $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) = M/M(\mathfrak{q})$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.