# A question related to local global principle

commutative-algebramodules

This question was asked in my quiz on Commutative algebra and I couldn't solve it. I tried it again today and I am struck.

Let V be an A-module over a commutative ring A and let $$A' \subseteq A$$ be an ideal. Suppose that $$V_M=0$$ for all $$M\in Spm A$$ with $$A' \subseteq M$$. Then $$V=A'V$$.

I have recently done a question in which I proved that if V is an A-module over a commutative ring A then V =0 is equivalent to $$V_M =0$$ for all $$M\in Spm A$$. I was thinking if I could prove that in A/A' -module V/A'V , $$(V/A'V)_M'=0$$, for all $$M'\in Spec A$$ where M' are maximal ideals in $$V/A'V$$.

Let $$W=V/A’V$$. Let $$M$$ be a maximal ideal of $$A$$ with $$M \supset A’$$. Then $$W_M=V_M/A’V_M=0$$ by assumption.
Assume that $$M \not\supset A’$$ is a maximal ideal of $$A$$. Then there is some $$a \in A’$$ such that $$a \notin M$$. Then $$a$$ acts as an automorphism on $$V_M$$ so $$aV_M=V_M$$ so $$A’V_M=V_M$$ so $$W_M=0$$.
So for any maximal ideal $$M$$ of $$A$$, $$W_M=0$$.
Now suppose for the sake of contradiction that there exists a nonzero $$z \in W$$. Let $$I$$ be the annihilating ideal of $$z$$, $$I \neq A$$ so $$I$$ contained in some maximal ideal $$M$$. Since $$z$$ vanishes in $$W_M$$, there is some $$t \in A \backslash M$$ such that $$tz=0$$, so $$t \in I \backslash M$$, a contradiction. .