Let be a commutative ring with and its Jacobson radical. Let be the set of non-unit elements of and let

**Problem**. Suppose that is non-empty and finite. Then is finite and for some finite local rings and

**Solution***.*

for all and

*Proof*. Clearly Proving is by contradiction: suppose that i.e. for some But then because That means , and hence , is a unit. Contradiction!

is finite.

*Proof*. Let and define by Clearly is one-to-one. Thus , and therefore , is finite.

, where each is a local ring.

*Proof*. Since every proper ideal of is contained in , our ring has only a finite number of ideals and so it is artinian. Hence , for some positive integer Let , be the maximal ideals of and It is clear that each is a local ring and, by the Chinese remainder theorem,

and , and so each , is finite.

*Proof*. Since , is not local and thus Now suppose, to the contrary, that is infinite. Then at least one of is infinite and so the set

is infinite. But which is a contradiction.

**Remark**. Every is a zero divisor.

*Proof*. We have and so, since is finite, there exist some such that and is minimal. So and thus we need to prove that Well, if and or then would be a unit, which is absurd. Otherwise, we’re done by the minimality of