Let be a commutative ring with identity. Let be the set of all non-unit elements of . If ( is the jacobson radical of ) is finite , then prove that for some finite local rings and every non-unit is a zero-divisor .
we also need the condition fix an element
claim 1. for all
Proof. clearly proving is by contradiction: suppose that so for some but then because that means , and hence , is a unit and that's a contradiction. Q.E.D.
claim 2. is finite.
Proof. define by clearly is one-to-one and thus , and therefore is finite.
claim 3. 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 is artinian. hence for some positive integer let be the maximal ideals of and put it is clear that each is a local ring and by the chinese remainder theorem for rings we have Q.E.D.
claim 4. each is finite.
Proof. first note that because now suppose, to the contrary, that is infinite for some then is infinite and which is impossible because is finite. so each , and hence itself, is finite.
claim 5. every is a zero divisor.
Proof. and so, since is finite, there exist such that and is minimal. so if we prove that then we've proved that is a zero divisor. well, if and or then would be a unit, which is absurd. if then, since we must have by the minimality of Q.E.D.