Let $R$ be a ring with unity and $A$ a right $R$-module. The singular submodule $Z(A)$ of $A$ is defined to be $Z(A)=\lbrace a\in A: \text{ann}(a) \subseteq^{\text{ess}} R_R \rbrace$. It can be shown that \begin{align} Z(A)= \lbrace a\in A: aE=0 \text{ for some essential right ideal $E \subseteq R$} \rbrace. \end{align} The second singular submodule $Z_2(A)$ is defined by the equality \begin{align} Z(A/Z(A)) = Z_2(A) /Z(A). \end{align} It is shown that \begin{align} Z_2(A)= \lbrace a\in A: aE \subseteq Z(A) \text{ for some essential right ideal $E \subseteq R$} \rbrace. \end{align} I need to show that $A/Z_2(A)$ is nonsingular, i.e., $Z(A/Z_2(A))=0$.
Attempt: Let $a+Z_2(A) \in Z(A/Z_2(A))$ for some $a\in A$ and show that $a\in Z_2(A)$. Hence, $\text{ann}(a+Z_2(A))=\lbrace r\in R: ar\in Z_2(A) \rbrace \subseteq^{\text{ess}} R_R$.
How can I proceed and show that $a\in Z_2(A)$ ?!
Thanks in advance.
This is a tough one to keep straight. In fact, Lam refers to it as "a theorem of Goldie" in his book Lectures on modules and rings in section 7. Lam gives a different argument there, but here I give the elementary one I stumbled across.
Let's abbreviate $Z_i=Z_i(A)$ (even for $Z(A)=Z_1(A)$.)
Lemma: $Z_1\subseteq_e Z_2$. Proof: it suffices to show that if $x\in Z_2\setminus Z_1$, there exists an $r\in R$ such that $xr\in Z_1\setminus\{0\}$. Suppose $x\in Z_2\setminus Z_1$. Then since $ann(x)$ is not essential in $R$, there exists a nonzero $a$ such that $ann(x)\cap aR=\{0\}$. Since $x\in Z_2$, $ann(x+Z_1)\subseteq_e R$, and there must exist a nonzero $ab\in ann(x+Z_1)\cap aR$. Therefore $xab\in Z_1$, and furthermore $xab\neq 0$ since $ann(x)\cap aR=\{0\}$. Therefore $Z_1\subseteq_e Z_2$.
For the main proposition, suppose $x\in Z_3\setminus Z_2$, and try the same line of thought.
Because $x\notin Z_2$, there is some nonzero $c\in R$ such that $ann(x+Z_1)\cap cR=\{0\}$ (that annihilator is not essential.) A key observation here is that $ann(x)\cap cR=\{0\}$ too because $ann(x)\subseteq ann(x+Z_1)$. To recapitulate, no nonzero multiple of $c$ annihilates $x$.
Since $x\in Z_3$, we have $ann(x+Z_2)\cap cR\neq\{0\}$, there is a nonzero multiple of $c$ that annihilates $x+Z_2$. If it is written $cd$, then $xcd\in Z_2\setminus \{0\}$ (nonzero because no nonzero multiple of $c$ annihilates $x$.)
By the lemma, there exists an $e$ such that $xcde\in Z_1\setminus \{0\}$. But this contradicts what we first established that $ann(x+Z_1)\cap cR=\{0\}$.
Therefore we have to conclude that $Z_3=Z_2$, and unpacking that it means that $\{0\}=Z_3/Z_2=Z(A/Z_2)$, which was to be proven.