mulitiplication of module and ideal

Question

Is this right?

Let $M$ finitely generated a $A$ module and $a$,$b$ ideals of $A$ s.t $a$,$b$ contain $ann(M)$.

$aM=bM$ imply $a=b$.

I got partial result when a=A by Nakayama lemma but i can neither prove or disprove the original statement.

Answer 1

I'll offer an answer for the first one because sometimes examples can be tough to find.

Is this right? (Let $M$ a $A$ module and $a$,$b$ ideals of $A$ s.t $a$,$b$ contain $ann(M)$. $aM=bM$ imply $a=b$.)

It isn't right.

Let $R$ be any primitive ring that isn't simple, like the endomorphism ring of an infinite dimensional vector space, and let $M$ be a faithful simple module for that ring. Then $IM=M$ for every nonzero ideal $I$ of $R$, and all $I$'s contain the annihilator of $M$ (which is zero) and there are many such $I$'s.

If it's wrong, when [does] this hold.

Gonna pass on that. You owe us efforts of your own along these lines.

Let $M$ finitely generated a $A$ module and $a$,$b$ ideals of $A$ s.t $a$,$b$ contain $ann(M)$. $aM=bM$ imply $a=b$.

Still wrong after your revision. The example I gave is finitely generated (it is cyclic.)