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If I have a finitely generated module $M$ over a local noetherian ring $(A,m)$, where $m$ denotes the maximal ideal, and $\operatorname{supp}{(M)}=m$, then there exists a surjection

$$M\rightarrow A/m$$

and an injection

$$A/m\rightarrow M$$

How can I get these?

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1 Answer

up vote 5 down vote accepted

Here are some hints. Let me know if I should say more.

For the first map, we know that $M$ maps onto $M/\mathfrak{m}M$, which is a vector space over the residue field $A/\mathfrak{m}$, so it should be enough to show that $M \neq \mathfrak{m}M$. Do you know techniques for doing this in the local case?

For the second, you need to find an $x \in M$ such that $\operatorname{ann}(x) = \mathfrak{m}$. One way of doing this is to consider the family of ideals $$ \{\operatorname{ann}(y) : y \in M,\, y \neq 0\}. $$ Show that a maximal element — with respect to inclusion — of this family is prime. Why does a maximal element exist? How does this help us?

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Ok, thanks, that's what I needed. The first is Nakayama. For the annihilator use noetherian and an easy maximality argument to show that there is a prime P with P=ann(x) for an element x of M. Then use that the intersection of the associated primes is the intersection of the primes in the support of M and you get that P must be $m$. The rest is clear, you send the class of an $a$ to $ax$. –  Descartes Aug 16 '11 at 22:17
    
Interesting. I wasn't even thinking about primary decomposition (I haven't been able to internalize the theory yet). My thought was that if $\mathfrak p$ is the annihilator of $x \neq 0$, then $\mathfrak p$ is in the support of $M$ because $x$ isn't killed off by anything in $A \setminus \mathfrak p$. –  Dylan Moreland Aug 16 '11 at 22:33
    
well, that is obiously the less complicated way thinking about this ;) –  Descartes Aug 17 '11 at 7:09
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