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

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- Oct 17th 2008, 11:26 PMchipaiAn 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 - Oct 18th 2008, 01:43 AMNonCommAlg
let $\displaystyle R$ be the ring and $\displaystyle P$ the minimal prime. so $\displaystyle R - P$ is exactly the set of units of $\displaystyle R.$ hence any proper ideal of $\displaystyle R$ is contained in $\displaystyle P.$ since $\displaystyle P$ is a minimal prime, it cannot contain any prime ideal

properly. thus $\displaystyle P$ is the unique prime ideal of $\displaystyle R.$ now suppose $\displaystyle x \in R$ is a non-unit. so $\displaystyle x \in P.$ suppose $\displaystyle x$ is not nilpotent. let $\displaystyle \mathcal{S}=\{x^n : \ n = 1, 2, \cdots \}.$ then $\displaystyle 0 \notin \mathcal{S},$ because we assumed that $\displaystyle x$ is

not nilpotent. let $\displaystyle \mathcal{C}=\{I \lhd R: \ I \cap \mathcal{S} = \emptyset \},$ which is a non-empty set because $\displaystyle <0> \in \mathcal{C}.$ apply Zorn's lemma to find $\displaystyle Q,$ a maximal element of $\displaystyle \mathcal{C}.$ we claim that $\displaystyle Q$ is prime: suppose $\displaystyle ab \in Q$ but

$\displaystyle a \notin Q, \ b \notin Q.$ then $\displaystyle Q \subset Q + Ra, \ Q \subset Q+Rb.$ therefore $\displaystyle (Q+Ra) \cap \mathcal{S} \neq \emptyset, \ (Q+Rb) \cap \mathcal{S} \neq \emptyset,$ by maximality of $\displaystyle Q.$ so there exists $\displaystyle i,j$ such that $\displaystyle x^i \in Q+Ra, \ x^j \in Q+Rb.$ but then:

$\displaystyle x^{i+j} \in (Q+Ra)(Q+Rb) \subseteq Q+Rab =Q,$ because $\displaystyle ab \in Q.$ thus $\displaystyle Q \cap \mathcal{S} \neq \emptyset,$ which is contradiction! so $\displaystyle Q$ is prime and hence $\displaystyle Q=P.$ thus $\displaystyle P \cap \mathcal{S} = \emptyset,$ which is false because $\displaystyle x \in P \cap \mathcal{S}. \ \ \ \Box$

__Remark 1__: in general, in a (not even necessarily commutative) ring $\displaystyle R,$ given a multiplicatively closed set $\displaystyle S \subset R$ with $\displaystyle 0 \notin S,$ we can always find a prime ideal $\displaystyle Q$ which is contained in $\displaystyle R - S.$

__Remark 2__: if you already know about the nilradical of a commutative ring, then your problem is trivial: since $\displaystyle P$ is the only prime ideal of $\displaystyle R,$ the nilradical of $\displaystyle R$ is $\displaystyle P.$ but we know that the nilradical

is exactly the set of nilpotent elements of $\displaystyle R.$

__Remark 3__: the information given in the problem about zero-divisors is only used to conclude that $\displaystyle R-P$ is exactly the set of units of $\displaystyle R.$