Thread: An exercise in Algebra by Thomas W. Hungerford

A commutive ring wih unity has a minmal prime ideal contains all zero divisor and all nonunits are zero divisor
prove
all nonunits are nilpotent

this is an exercise in Algebra by Thomas W. Hungford page 148
about local ring

Originally Posted by chipai

A commutive ring wih unity has a minmal prime ideal which contains all zero divisors and all nonunits are zero divisor. prove that all nonunits are nilpotent.

this is an exercise in Algebra by Thomas W. Hungford page 148 about local ring.

let be the ring and the minimal prime. so is exactly the set of units of hence any proper ideal of is contained in since is a minimal prime, it cannot contain any prime ideal

properly. thus is the unique prime ideal of now suppose is a non-unit. so suppose is not nilpotent. let then because we assumed that is

not nilpotent. let which is a non-empty set because apply Zorn's lemma to find a maximal element of we claim that is prime: suppose but

then therefore by maximality of so there exists such that but then:

because thus which is contradiction! so is prime and hence thus which is false because



Remark 1: in general, in a (not even necessarily commutative) ring given a multiplicatively closed set with we can always find a prime ideal which is contained in

Remark 2: if you already know about the nilradical of a commutative ring, then your problem is trivial: since is the only prime ideal of the nilradical of is but we know that the nilradical

is exactly the set of nilpotent elements of

Remark 3: the information given in the problem about zero-divisors is only used to conclude that is exactly the set of units of