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
Tags: , , ,

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


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