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I have some confusion on Atiyah Commutative algebra

On Page No $20$ it is written that

An $A-$Module is faithful if $Ann(M)=0$.If $Ann(M)=\alpha$ ,then $M$ is faithful as an $A/{\alpha}$ module

My confusion : Why is $M$ faithful as an $A/{\alpha}$ module ?

My thinking : $Ann_{A/{\alpha}}(M)=\{r+\alpha | r \in A \ \text{and} \ rM={0}\}=\{ r+\alpha |r \in Ann(M)\}$

$\implies Ann_{A/{\alpha}}(M) = \frac{Ann(M)}{\alpha}=\frac{\alpha}{\alpha}=\frac{0}{0} \neq 0$

Therefore $M$ is not faithful as an $A/{\alpha}$ module

wasiu
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    What you write is correct up to the point $\alpha/\alpha$. This is simply the zero module. I cannot tell where your thinking is going wrong at the end exactly. Not sure if it's just notation, but I advise you to recall that quotients of modules are not literal fractions. – Thorgott Aug 21 '22 at 21:33
  • @Thorgott I thought that $0/0 $ undefine – wasiu Aug 21 '22 at 21:36
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    If $0$ means the natural number $0$, then the fraction $0/0$ is undefined. If $0$ means the trivial abelian group, then the quotient $0/0$ is defined and just $0$ again. You have to always keep in mind what the notation you use actually means. Also, while $\alpha/\alpha=0/0$ is accurate, it would probably help if you indicate how you arrive at this step. "Cancelling" quotients does not work quite the same way as cancelling fractions. – Thorgott Aug 22 '22 at 10:53

1 Answers1

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The action of $A/\alpha$ on $M$ is given by $(r+\alpha)x=rx$ for $x \in M$. If $(r+\alpha)M=rM=0$, then by assumption $r \in \alpha$ so $r+\alpha=\alpha$ which is $0$ in $A/\alpha$.

metalspringpro
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