In this post we will keep the notation in the previous part. We will assume that is a directed set and is a direct systen of -modules. In Remark 2 in there we proved that

Our goal now is to find a necessary and sufficient condition for an element of to be , i.e. we want to see when we will have . Let’s start with a notation which will simlify our work.

**Notation**. For any with and we let Note that are just the generators of as an -submodule of

Now two easy properties of :

**Lemma 1.** If then

1)

2)

*Proof*. Trivial because and are all -module homomorphisms.

**Lemma 2**. If and then

Proof. This is also a trivial result of this fact that

And the main result of this section:

**Theorem**. if and only if for some with

*Proof.* If for some with then Conversely, supppose that Then by Lemma 1, part 1, we can write

Let be such that for all occuring in (1). Since we will have

by (1). Applying Lemma 2 to (2) we will get

where the sum runs over Note that by Lemma 1, part 2, we may assume that all the indices in (3) are distinct. We can rewrite (3) as

But since every element of is uniquely written as in (4) we must have and hence for all Thus (4) becomes

which gives us

**Corollary**. Let be a direct system of rings. We proved in Remark 3 in this post that is also a ring. If each is a domain, then is also a domain.

*Proof*. Suppose that are two elements of and That means and hence, by the theorem, there exists some with such that

Therefore, since is a domain, we have either or Applying the theorem again, we get either or