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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?

1 Answers 1

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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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    well, that is obiously the less complicated way thinking about this ;)2011-08-17