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 .

Printable View

- June 26th 2010, 12:08 AMxixidirect product of finite local rings
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 .

- June 26th 2010, 10:42 PMNonCommAlg
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.