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If $_{R}M$ is a left $R$-module of a ring $R$, show that $ann(M)=\left \{ r\in R:rM=0 \right \}$ is an ideal in $R$.

I have used the fact that the left annihilator is a left ideal in $R$ to prove that $ann(M)$ is a subring of $R$ and also a left ideal in $R$. But how do I show $ann(M)$ is a right ideal in $R$? Do I use the right annihilator for this? For example, if $r\in R$ and $b\in ann(M)$, then $bM=0$. This implies $rbM=0$ or $rb\in ann(M)$. But how do I show that $brM=0$?


marked as duplicate by rschwieb abstract-algebra Apr 4 at 13:26

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  • $\begingroup$ As Melody shows below: it comes down to $rM\subseteq M$, so $b(rM)=0$. $\endgroup$ – Dave Apr 4 at 2:29

Let $m\in M$, $r\in R,$ and $b\in\text{ann}(M).$ Then $rm\in M,$ so $b(rm)=0.$ Since $m\in M$ was arbitrary we have $brM=0.$

  • $\begingroup$ Excuse me for my beginner knowledge, but may I ask why $rm\in M$? $\endgroup$ – numericalorange Apr 4 at 2:33
  • 1
    $\begingroup$ That's from the definition of left $R$ module. Since $M$ is a left module there is a map $\cdot:R\times M\to M.$ This map is denoted typically by $r\cdot m$ or $rm$ for all $r\in R$ and $m\in M.$ $\endgroup$ – Melody Apr 4 at 2:46

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