We will assume that is a ring (not necessarily commutative) with 1 and all modules are left -modules.

**Definition 1**. Let be an -module and a nonzero submodule of We say that is an **essential** submodule of and we will write if for any nonzero submodule of Clearly, that is equivalent to saying for any nonzero element So, in particular, a nonzero left ideal of is an essential left ideal of if for any nonzero left ideal of which is equivalent to the condition for any nonzero element

**Definition 2**. Let be an -module and Recall that the (left) annihilator of in is defined by which is obviously a left ideal of Now, consider the set It is easy to see that is a submodule of (see Problem 2 in this post for the proof!) and we will call it the** singular submodule** of If then is called **singular**. If then is called nonsingular. We will not discuss nonsingular modules in this post.

**Problem 1**. Prove that if is a free -module, then i.e. a free module is never singular.

**Solution**. Let be any element of an -basis of Let Then and so Thus and so

Next problem characterizes singular modules.

**Problem 2**. Prove that an -module is singular if and only if for some -module and some submodule

**Solution**. Suppose first that where is an -module and Let and let be a nonzero left ideal of If then and so If then because So there exists such that That means So we have proved that and hence i.e. is singular. Conversely, suppose that is singular. We know that every -module is the homomorphic image of some free -module. So there exists a free -module and a submodule of such that So we only need to show that Note that by Problem 1. Let be an -basis for and suppose that We need to show that there exists such that We can write, after renaming the indices if necessarily,

where For any let Now, since is singular, and so So there exists such that and Hence gives us

Note that because Now, if for all then and we are done. Otherwise, after renaming the indices in the sum on the right hand side of if necessary, we may assume that Repeating the above process gives us some such that and Then implies

The first two terms on the right hand side of are in and because If we continue this process, we will eventually have a positive integer and such that