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 $V/A'V$.
Can you please help me with this?
Best Answer
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. .