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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.

George
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    This is not true. $a$ and $b$ could be distinct ideals contained within the annihilator of $M$. – user1090793 Apr 18 '23 at 14:15
  • But what about when a and b contain annihilator? and I modified the question. – George Apr 18 '23 at 14:20
  • @atssit Do not add hypotheses after part of your question gets answered. In this case, your addition did not change the situation and the solution given is still a counterexample. But in general you should not move the goalposts on existing solutions. – rschwieb Apr 18 '23 at 17:27

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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.)

rschwieb
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  • Thanks for your exmaple but i can't come up with any idea. Could you give me any advice. And I modified the question. – George Apr 18 '23 at 16:56
  • This solution still gives a counterexample for the modified question. Don't move the goalposts by editing. If you want to ask a fundamentally different question ask in another post. – rschwieb Apr 18 '23 at 17:27
  • I apologize for editing the question. If M is finitely generated and M=IM, then there exsits s in I s.t 1+s is in ann(M). So when I contain ann(M), then 1 is in I, that is I=A. Your answer seems to contradict this. – George Apr 18 '23 at 17:58
  • @atssit I don't believe that formulation of Nakayama's lemma holds for rings that aren't commutative. Have you secretly been assuming commutativity this whole time? – rschwieb Apr 18 '23 at 18:59
  • I'm ashamed that I was unaware that the examples you gave are noncommutative! I also implicitly assumed commutativity this whole time. – George Apr 19 '23 at 01:03
  • I find counter example in a related question. https://math.stackexchange.com/questions/1310230/does-ij-ik-implies-j-k-always-hold-for-integral-domain-and-finitely-generated?rq=1 – George Apr 19 '23 at 01:21