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
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
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 $\displaystyle R.$