Annihilator of an Element in a R-Module

[Sudharaka]

Hi everyone, :)

I was reading the book A Course in Ring Theory by Passman and in it is the following lemma;

and after this lemma there's a example which I don't quite understand;

I have several things to discuss. Let me start with this;

I want to confirm my understanding about how to show that that the annihilator of $a$ is equal to $bR$. Tell me whether this is correct: By the definition of the right annihilator $ann(a)=\{r=k_1+k_2a\in R\mid a(k_1+k_2a)=0\}$. Since $a^2=0$ we have, $ak_1=0$. Since $a$ is any arbitrary element(the indeterminate) this implies $k_1=0$. Hence, $ann(a)=k_2 a=ak_2=bk_2$ (since $a=b$) and since $k_2\in K$ is arbitrary, $ann(a)=bR$. Am I correct?

 

[jvicens]

Yes, your understanding is correct. To show that the annihilator of $a$ is equal to $bR$, we need to show that every element in $bR$ is also in the annihilator of $a$, and vice versa.

First, let $r \in bR$, then $r=kb$ for some $k \in K$. Since $a^2=0$, we have $ar=0$, which means $a(kb)=0$. Since $a$ is arbitrary, this implies $kb=0$. But since $a=b$, we have $kb=ka=0$. Therefore, $r \in ann(a)$.

Conversely, let $r \in ann(a)$. Then by definition, $ar=0$. Since $a$ is any arbitrary element, this implies $br=0$. But since $a=b$, we have $br=ba=0$. Therefore, $r \in bR$.

Hence, we have shown that $ann(a)=bR$. Great job on understanding the lemma and its application in the example! Keep up the good work.