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Exercise 5.6 b) of Chapter II of Hartshorne's Algebraic Geometry asks to prove that if $A$ is a Noetherian ring and $M$ a finitely generated $A$-module then $Supp(\tilde{M})=V(Ann(M))$. Where $\tilde{M}$ is the sheaf on the scheme $Spec(A)$ associated to the module $M$.

I have "proved" the statement without using the Noetherianity condition on $A$ (or without noticing I'm using it!). So I guess I'm missing something. I would like some help in understanding where my error lies!

Here follows my "proof" of the statement:

First inclusion $V(Ann(M))\subseteq Supp(\tilde{M})$

Let $p\in V(Ann(M))$, so $Ann(M)\subseteq p$. By contradictions I suppose $M_p=0$, in specific for each $m \in M$ there exists $a\in A - p$ such that $am=0$. Let $\{m_i\}$ be a finite set of generators for $M$ and $\{a_i\} \subseteq A$ such that $a_im_i=0$ for each $i$. On one hand $\prod a_i\in A-p$ because $p$ is a prime ideal, but on the other hand $(\prod a_i)m_j=0$ for each $j$, so $\prod a_i\in \bigcap Ann(m_i) = Ann(M) \subseteq p$. Contradiction.

Second inclusion $Supp(\tilde{M}) \subseteq V(Ann(M))$

Let $p \in Supp(\tilde{M})$, then $M_p\neq 0$ and there exists $m\in M$ such that it does not exist $a \in A - p$ such that $am=0$. Then $Ann(M)\bigcap (A-p)=\emptyset$ so $Ann(M)\subseteq p$, in specific $p\in V(Ann(M))$.

So, what's wrong? Thank you for your help!

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1 Answer 1

up vote 4 down vote accepted

I am happy to report that your statement is perfectly correct: if $M$ is a finitely generated $A$-module, then $Supp (M)=V(Ann(M))$, and the fact that $A$ is or isn't noetherian is irrelevant.

NB Many people use noetherianity as a blanket assumption (for example Eisenbud in his book Commutative Algebra) but that doesn't mean that their theorems become false without that assumption.
In contrast, Bourbaki and Matsumura tend to use noetherianity only when it is really needed.

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I'm happy to hear it! Thank you again! –  Giovanni De Gaetano Feb 22 '12 at 10:01

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