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Let $R$ be a noetherian commutative ring, and let $\mathfrak{m}$ be a maximal ideal of $R$. Let $M$ be a finitely-generated torsion $R_\mathfrak{m}$-module, considered as an $R$-module. Is it possible that $M_\mathfrak{p} \ne 0$ for some prime ideal $\mathfrak{p}$ not contained in $\mathfrak{m}$?

Intuitively/geometrically, this doesn't seem possible. What we have here is a coherent sheaf on $\operatorname{Spec} R_\mathfrak{m}$ and its direct image on $\operatorname{Spec} R$. What I want to argue is that for any point $\mathfrak{p}$ of $\operatorname{Spec} R$, $\mathfrak{p} \ne \mathfrak{m}$, there is an open neighbourhood $U$ of $\mathfrak{p}$ not containing $\mathfrak{m}$, and so therefore sections of $M$ over $U$ must be $0$... but this isn't quite true, since the image of $\operatorname{Spec} R_\mathfrak{m}$ is not just $\{ \mathfrak{m} \}$.

We know that $\mathfrak{p}$ is in the support of $M$ if and only if $\mathfrak{p}$ contains the annihilator of $M$. To show that $\mathfrak{p}$ is not in the support of $M$, we must exhibit an element of $R \setminus \mathfrak{p}$ that annihilates $M$. But why does such an element exist?


In case the answer to my very first question is "yes", let's add more hypotheses: suppose $R$ is a Dedekind domain. Then $\operatorname{Spec} R_\mathfrak{m}$ consists of an open point and a closed point, and the sections/stalk over the open point must be $0$ because $M$ is a torsion $R_\mathfrak{m}$-module. I feel convinced that this proves that $M_\mathfrak{p} = 0$ for all $\mathfrak{p} \ne \mathfrak{m}$... but how does this translate into algebra?

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Note that if $f \in \mathfrak{m} \backslash \mathfrak{p}$, then the open set $D(f)$ contains $[\mathfrak{p}]$ and not $[\mathfrak{m}]$. –  Aaron Mazel-Gee May 21 '12 at 21:49
    
@AaronMazel-Gee: Yes, that's obvious. But the fact that sections of $M$ over $D(f)$ are elements of $M[1/f]$ is non-trivial – I'm looking for an argument that doesn't go through this machinery. –  Zhen Lin May 21 '12 at 21:54
    
Dear Zhen, By torsion $R_{\mathfrak m}$-module, do you mean one in which every element is annihilated by a power of $\mathfrak m$. If so, this is stronger than what others would usually mean, and so perhaps you would say so. My own view is that torsion is somewhat problematic as a phrase without clarification if $R_{\mathfrak m}$ is non-reduced, or even not a domain, but when it is a domain, it would just mean that every element of the module is annihilated by some non-zero element of $R_{\mathfrak m}$. And such a module need not be supported just at the closed point $\mathfrak m$; –  Matt E May 23 '12 at 11:04
    
... it might be supported, e.g., along a hypersurface passing through $\mathfrak m$. Regards, –  Matt E May 23 '12 at 11:05

3 Answers 3

up vote 2 down vote accepted

The answer to your first question is yes. Let $R$ be such that there exist two distinct maximal ideals $m_1, m_2$ containing a same (non-associted) prime ideal $ \mathfrak q$. Let $M=(A/\mathfrak q)_{m_1}=A_{m_1}/\mathfrak q A_{m_1}$. It is clearly finitely generated and torsion over $A_{m_1}$.

I claim that $M_{m_2}\ne 0$. Suppose the contrary. Let $e$ be the class of $1$ in $M$. There exists $s_2\in A\setminus m_2$ such that $s_2.e=0$ in $M$. So there exists $s_1\in A\setminus m_1$ such that $s_1s_2=0$ in $A/\mathfrak q$. Therefore $s_1s_2\in \mathfrak q$. So one of the $s_i$'s belongs to $\mathfrak q\subset m_i$. Contradiction.

When Spec($R$) is irreducible and has dimension $\le 1$, the answer of Makoto Kato shows that such examples don't exist.

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I found this question somewhat confusing, because of the OP's expectations, and the possible interpretation of the term torsion. For this reason I am giving an answer. It may be more geometric than the OP wants, but my own view is that thinking geometrically makes the makes the answers to these sorts of questions clear, while thinking algbraically can be confusing.

Let's suppose that $R$ (and hence $R_{\mathfrak m}$) are integral domains. A module over an integral domain is called torsion if every element of the module is annihilated by a non-zero element of the domain. If the module is finitely generated, then choosing such a non-zero element of the domain for each generator and multiplying them together, we find a single non-zero element of the domain which annihilates the entire module.

Now if $M$ is a f.g. module over a $A$, and $I$ is its annihilator, then $M_{\mathfrak p}$ is non-zero for a point $\mathfrak p \in $ Spec $A$ if and only if $\mathfrak p \supset I$, and so the support of $M$ is equal to the closd subset $V(I)$ of Spec $A$.

Combining the two paragraphs, we are considering a f.g. module over $R_{\mathfrak m}$ whose support is contained in $V(f)$ for some non-zero $f \in R_{\mathfrak m}$, and asking if its support as an $R$-module consists of the single point $\mathfrak m$. We may as well make the support in Spec $R_{\mathfrak m}$ as large as possible, to try and make the support in Spec $R$ as large as possible, and so let's assume the support actaully is $V(f)$ (e.g. by taking $M = R/\mathfrak m$). Clearing denominators, we may also assume that $f \in R$. In fact, we should assume $f \in \mathfrak m$, so that it does not become a unit in $R_{\mathfrak m}$. (Otherwise $M$ would be annihilated by a unit of $R_{\mathfrak m}$, and hence would equal $0$.)

Our question then comes down to asking, for a non-zero $f \in \mathfrak m$, whether or not the image of $V(f) \subset$ Spec $R_{\mathfrak m}$ necessarily equals the single point $\mathfrak m$. The answer is no in general, which means that the support of $M$, thought of as an $R$-module, will typically be larger than just $\mathfrak m$.

The reason is that $V(f)$ (in Spec $R_{\mathfrak m}$) is the germ at $\mathfrak m$ of the hypersurface in Spec $R$ cut out by $f$ (i.e. it is the germ at $\mathfrak m$ of the vanishing locus of $f$ in Spec $R$), so its Zariski closure in Spec $R$ will be the irreducible component of that hypersurface passing through $\mathfrak m$. (Let's assume that $R$ is Noetherian, so that irreducible components exist and are well behaved.)

So the only way that the image of $V(f)$ could be just $\mathfrak m$ is if $\mathfrak m$ itself were a hypersurface in Spec $R$. This will be possible if $R$ has dimension one, but only in that case.

So if $R$ is one-dimensional, then the image of $V(f)$ will be just $\mathfrak m$, but not otherwise.


On the other hand, if we assume that $M$ is torsion in a very strong sense, namely that every element is annihilated by a power of $\mathfrak m$, then its support will simply be $\mathfrak m$, and the image of in Spec $R$ will again be $\mathfrak m$. So in this case $M$ will be supported at $\mathfrak m$ as an $R$-module.


The point is that looking at $R_{\mathfrak m}$ is not at all the same as looking at $R/\mathfrak m$, or $R/\mathfrak m^n$. The latter two rings are supported at $\mathfrak m$, but $R_{\mathfrak m}$ sees the whole germ of Spec $R$ near $\mathfrak m$.

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Very clearly explained: thanks! –  Zhen Lin May 23 '12 at 13:55

In case the answer to my very first question is "yes", let's add more hypotheses: suppose $R$ is a Dedekind domain. Then $\operatorname{Spec} R_\mathfrak{m}$ consists of an open point and a closed point, and the sections/stalk over the open point must be $0$ because $M$ is a torsion $R_\mathfrak{m}$-module. I feel convinced that this proves that $M_\mathfrak{p} = 0$ for all $\mathfrak{p} \ne \mathfrak{m}$... but how does this translate into algebra?

Proof Let $R$ be a Dedekind domain. Let $P$ be a maximal ideal of $R$ . Let M be a finitely generated torsion $R_P$-module. Since $R_P$ is a principal ideal domain, $M$ is a finite product of cyclic torsion $R_P$-modules. Hence we can assume that $M$ is a finite product of modules of type $R_P/(P^n)R_P$. Since $R_P/(P^n)R_P$ is isomorphic to $R/P^n$, the support of $M$ as an $R$-module is {$P$}.

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