**Definition. **A partially ordered set is called a **directed set** if for every there exists some such that and

**Example 1**. Let be a collection of subsets of a set, where is a directed set and iff Let be the direct system defined in Example 1 in this post. Then

*Proof*. Let For every we define to be the inclusion map.

1) If and then Thus

2) Suppose that there exists a set and the maps such that for all Let’s see how we have to define Suppose that is a map with for all Let Then for some and hence Now, if is also in then choosing with and we will have and Thus So we must define whenever We just proved that is well-defined. Clearly for all

**Remark**. Clearly the result in Example 1 holds if were groups, rings, modules, etc.

**Example 2**. Let be a multiplicatively closed subset of a commutative ring with and Let be the direct system defined in Example 2 in this post. Then

*Proof*. Let and, for every define to be the inclusion map.

1) If then for some and hence for all and That means

2) Suppose that there exists a ring and the ring homomorphisms such that for all Let be a ring homomorphism such that for all Let’s see how we have to define Let Then and thus Also if then choosing we’ll have Thus

and So we must define We just proved is well-defined and clearly is a ring homomorphism and for all

**Example 3**. Let be a commutative ring with 1 and let be the polynomial ring in the indeterminate Let the ideal of generated by For every positive integer let Let be the inverse system in Example 3 in this post. Then

*Proof.* Let and define the map by for all and

1) If and then Thus

2) Suppose first that there exists a ring and the ring homomorphisms such that for all Suppose also that there exists a ring homomorphism such that for all Then for any positive integer and all we will have So here is how we have to define : for every positive integer we have

where Now we define

Clearly is well-defined and for all So, to prove that is a ring homomorphism we must show that for all To prove this, we only need to show that for every integere the coefficient of in both and are equal. To do so, we begin with the fact that for all which gives us

for all Now, the coefficient of in is

On the other hand, the coefficient of in is But, since we will have

It is clear from (1), (2) and (3) that the coefficients of in both and are equal.

Finally, using (1), we have

and thus