**Remark.** If is a left primitive ring, then, by the density theorem and remark 4 in this post, there exists a division ring such that either for some positive integer or for any positive integer there exists a subring of and an onto ring homomorphism

**Example 1**. A finite ring is left primitive if and only if for some integer and some finite field

*Proof*. The “if” part is trivial because is a simple ring and hence left primitive. Now suppose that is a finite left primitive ring. By the above remark we’ll have two possibilities.

*Case 1*. There exists a division ring such that Then is finite and so a field.

*Case 2*. There exists a division ring such that for any positive integer there exists a subring and an onto ring homomorphism But this case is not possible because is finite for all but the cardinality of goes to infinity as goes to infinity.

**Example 2**. A ring has the property if for every there exists an integer depending on such that A left primitive ring which has the property is a division ring.

*Proof*. Again, the same as example 1, we’ll consider two cases.

*Case 1*. for some integer and division ring But for the ring doesn’t have the property because for all So and thus

*Case 2*. Suppose that there exists an onto ring homomorphism from a subring of to Then for some ideal of Then, since has the property the ring must have the property But we showed in Case 1 that does not satisfy So this case is impossible.

**Example 3**. Let be a left primitive ring and suppose that for all Then is a division ring.

*Proof*. The proof is the same as example 2. The only thing we’ll need is to note that for the ring doesn’t satisfy the above property because, for example, if we let then but

**Example 4**. Let be a left primitive ring such that is a unit for all Then is a division ring.

*Proof*. Let be a faithful simple left -module and So, by the above remark, we only need to prove that the dimension of as a vector space over is Suppose that the dimension is at least and let be a -linearly independent set. By the density theorem, there exists such that and But then and thus which gives us the contradiction because is a unit.

**Example 5**. (Wedderburn-Artin) Let be a left primitive ring. If is left Artinian, then for some division ring and positive integer

*Proof*. Let be a faithful simple left -module and If is finite dimensional as a vector space over we’re done by the above remark. Otherwise, choose an infinite -linearly independent set For any positive integer let Clearly for all Now, by the density theorem, there exists such that and Thus but which means So we have which is a contradiction because is Artinian.