For the definition and some examples of Dedekind-finite rings see this post. We will assume that is a ring with 1.

**Notation**. For and integers we let

The following shows that if then behave like matrix units!

**Problem 1**. If then for all

**Solution**. Since an easy induction shows that for every integers we have

We also have

Now apply to the terms in brackets on the right hand side of to finish the proof. You will see very quickly that the right hand side of is if or and it is if

**Problem 2**. Prove that a ring which is not Dedekind-finite has infinitely many nilpotent elements.

**Solution**. Since is not Dedekind-finite, there exist such that and Now consider the set By Problem 1, we have So every element of is nilpotent. We now need to show that these elements are pairwise distinct and so is an infinite set of nilpotent elements of So suppose, to the contrary, that for some distinct integers Then by Problem 1,

contradicting

**Example**. A trivial result of Problem 2 is that every finite ring is Dedekind-finite. See also Example 3 in here.