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In a previous question I asked about the fiber $M(P)=M_P / PM_P$ where $M$ is an $A$-module and $P$ a prime ideal of $A$. Later I introduced the rank function $$rk_M : \text{Spec} A \to \mathbb{N} \cup \{ \infty \}$$ given by $$P \mapsto \dim_{A(P)} M(P).$$ I am supposed to prove that if $M$ is finitely generated then $rk_M(P)$ is finite and $rk_M(P) > 0$ iff $P \in \text{Supp}(M)$. I managed to prove everything expect the last implication $P \in \text{Supp}(M) \implies rk_M(P) > 0$, I am stuck on this one although I feel it should not be hard, any hints?

Community
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2 Answers2

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Its easy: if $P \in Supp(M)$, then $M(P)$ is a non-zero $A(P)$-vector space. For it were the zero space then we would have $M_P = P M_P$ and Nakayama's Lemma would imply $M_P=0$. But this contradicts the assumption that $P \in Supp(M)$. Thus $M(P)$ must have a positive dimension.

Manos
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  • Can I ask a question related to this one quick? I dont want to start a new thread, I want to know if a statement I made is true. –  Jan 05 '15 at 19:26
  • Give an example of a ring $A$ and an $A$-module $M$ with $rk_M(P)=0$ but $P \in Supp(M)$. I chose $A= \mathbb{Z}$ and $M= \mathbb{R}$, and the prime ideal $(2) \in Supp(\mathbb{Z})$. Is this example ok? I was unsure about one step I took in proving it. –  Jan 06 '15 at 11:16
  • @Sodan: This should be a new post. Please start one and add what you did, where are you stuck and why. – Manos Jan 06 '15 at 15:20
  • Hmm ok, do you see something wrong with the example right away? –  Jan 06 '15 at 17:05
  • Its unclear: What is $rk_M(P)$? – Manos Jan 07 '15 at 18:34
  • In our case i chose $P=(2)$ so that $rk_M(P)= dim_{A(P)}M(P)$, however $M(P)=M_P/PM_P$ and since $M_P=PM_P$ we have that $M(P)$ only consist of the zero element, hence its dimension over $A(P)$ must be zero, that is, $rk_M(P)=0$- –  Jan 07 '15 at 18:56
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It results from Nakayama's lemma: if $A$ is a local ring with maximal ideal $\mathfrak m$ and $M$ is a finitely generated $A$-module, then $$M/\mathfrak m M=0 \Rightarrow M=0.$$

Bernard
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