## A singular module is a quotient of a module by an essential submodule

Posted: December 4, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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We will assume that $R$ is a ring (not necessarily commutative) with 1 and all modules are left $R$-modules.

Definition 1. Let $M$ be an $R$-module and $N$ a nonzero submodule of $M.$ We say that $N$ is an essential submodule of $M,$ and we will write $N \subseteq_e M,$ if $N \cap X \neq (0)$ for any nonzero submodule $X$ of $M.$ Clearly, that is equivalent to saying $N \cap Rx \neq (0)$ for any nonzero element $x \in M.$ So, in particular, a nonzero left ideal $I$ of $R$ is an essential left ideal of $R$ if $I \cap J \neq (0)$ for any nonzero left ideal $J$ of $R,$ which is equivalent to the condition $I \cap Rr \neq (0)$ for any nonzero element $r \in R.$

Definition 2. Let $M$ be an $R$-module and $x \in M.$ Recall that the (left) annihilator of $x$ in $R$ is defined by $\text{ann}(x)=\{r \in R: \ rx = 0 \},$ which is obviously a left ideal of $R.$ Now, consider the set $Z(M):=\{x \in M: \ \text{ann}(x) \subseteq_e R \}.$ It is easy to see that $Z(M)$ is a submodule of $M$ (see Problem 2 in this post for the proof!) and we will call it the singular submodule of $M.$ If $Z(M)=M,$ then $M$ is called singular. If $Z(M)=(0),$ then $M$ is called nonsingular. We will not discuss nonsingular modules in this post.

Problem 1. Prove that if $F$ is a free $R$-module, then $Z(F) \neq F,$ i.e. a free module is never singular.

Solution. Let $x \in F$ be any element of an $R$-basis of $F.$ Let $r \in \text{ann}(x).$ Then $rx = 0$ and so $r = 0.$ Thus $\text{ann}(x)=(0)$ and so $x \notin Z(F). \ \Box$

Next problem characterizes singular modules.

Problem 2. Prove that an $R$-module $M$ is singular if and only if $M = A/B$ for some $R$-module $A$ and some submodule $B \subseteq_e A.$

Solution. Suppose first that $M=A/B$ where $A$ is an $R$-module and $B \subseteq_e A.$ Let $x = a + B \in M$ and let $J$ be a nonzero left ideal of $R.$ If $Ja = (0),$ then $Ja \subseteq B$ and so $\text{ann}(x) \cap J = J \neq (0).$ If $Ja \neq (0),$ then $B \cap Ja \neq (0)$ because $B \subseteq_e A.$ So there exists $r \in J$ such that $0 \neq ra \in B.$ That means $0 \neq r \in \text{ann}(x) \cap J.$ So we have proved that $x \in Z(M)$ and hence $Z(M)=M,$ i.e. $M$ is singular. Conversely, suppose that $M$ is singular. We know that every $R$-module is the homomorphic image of some free $R$-module. So there exists a free $R$-module $F$ and a submodule $K$ of $F$ such that $M \cong F/K.$ So we only need to show that $K \subseteq_e F.$ Note that $K \neq (0),$ by Problem 1. Let $\{x_i \}$ be an $R$-basis for $F$ and suppose that $0 \neq x \in F.$ We need to show that there exists $s \in R$ such that $0 \neq sx \in K.$ We can write, after renaming the indices if necessarily,

$x = \sum_{i=1}^n r_ix_i, \ \ \ \ \ \ \ \ \ \ (1)$

where $r_1 \neq 0.$ For any $y \in F,$ let $\overline{y}=y+K \in F/K.$ Now, since $F/K$ is singular, $\text{ann}(\overline{x_1}) \subseteq_e R$ and so $\text{ann}(\overline{x_1}) \cap Rr_1 \neq (0).$ So there exists $s_1 \in R$ such that $s_1r_1 \neq 0$ and $s_1r_1x_1 \in K.$ Hence $(1)$ gives us

$s_1x= s_1r_1x_1 +\sum_{i=2}^n s_1r_ix_i. \ \ \ \ \ \ \ \ \ (2)$

Note that $s_1x \neq 0$ because $s_1r_1 \neq 0.$ Now, if $s_1r_i = 0$ for all $2 \leq i \leq n,$ then $s_1x =s_1r_1x_1 \in K$ and we are done. Otherwise, after renaming the indices in the sum on the right hand side of $(2)$ if necessary, we may assume that $s_1r_2 \neq 0.$ Repeating the above process gives us some $s_2 \in R$ such that $s_2s_1r_2 \neq 0$ and $s_2s_1r_2x_2 \in K.$ Then $(2)$ implies

$s_2s_1x = s_2s_1r_1x_1 + s_2s_1r_2x_2 + \sum_{i=3}^n s_2s_1r_ix_i. \ \ \ \ \ \ \ (3)$

The first two terms on the right hand side of $(3)$ are in $K$ and $s_2s_1x \neq0$ because $s_2s_1r_2 \neq 0.$ If we continue this process, we will eventually have a positive integer $1 \leq m \leq n$ and $s = s_ms_{m-1} \cdots s_1 \in R$ such that $0 \neq sx \in K. \ \Box$