Throughout is a ring with 1 and all modules are left -modules. In Definition 2 in this post, we defined the singular submodule of a module

**Problem 1**. Let be an -module and suppose that are submodules of Prove that if and only if for all

**Solution**. We only need to solve the problem for If then and because both and contain Conversely, let be a nonzero submodule of Then because and therefore because

**Problem 2**. Prove that if is an -module, then is a submodule of and is a proper two-sided ideal of In particular, if is a simple ring, then

**Solution**. First note that because Now suppose that Then by Problem 1. Therefore and hence Now let and We need to show that Let be a nonzero left ideal of Then is also a left ideal of If then and thus If then because So there exists such that and Hence So and thus is a submodule of Now, considering as a left -module, is a left ideal of by what we have just proved. To see why is a right ideal, let and Then and so i.e. Finally, is proper because and so

**Problem 3**. Prove that if are -modules, then Conclude that if is a semisimple ring, then

**Solution**. The first part is a trivial result of Problem 1 and this fact that if where the sum is direct, then The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem.

**Problem 4. **Suppose that is commutative and let be the nilradical of Prove that

1)

2) it is possible to have

3) if then as -modules or -modules.

**Solution**. 1) Let Then for some integer Now suppose that Then Let be the smallest integer such that Then and hence

2) Let and put For every let and consider It is easy to see that

3) Let Then and thus there exists such that and Hence and so Thus implying that is an essential -submodule of Now, we view as a ring and we want to prove that as an essential ideal of Again, let Then and thus there exists such that and Let Then and thus implying that is an essential ideal of