a) We will prove the contrapositive.
Suppose is an ideal that is not irreducible. then there are ideals such that if , and . Since if , there are elements and . Note that . Since and , we know . Similarly, . Thus, we have found elements such that but . Hence, is not prime.
Thus, if is an ideal that is not irreducible then is not prime. Equivalently, if is prime, then is irreducible.
b) Let be a non-zero element. Let be the set of all ideals such that . Note that is not empty since the zero ideal is contained in it. Partially order by inclusion. Then every chain of ideals in has an upper bound in , namely,
because it is an ideal and does not contain . By Zorn's lemma, has a maximal element. Call it . This is an ideal in , so . We claim that is irreducible. Note that proving this claim will prove the statement in question.
Proof of claim: If possible, suppose that there are ideals , neither is contained in the other, such that . Since , one of them, say , must be such that it does not contain . Thus, . Since is maximal in , and , we have . Since and , we conclude . But then , a contradiction. Thus, there are no ideals , such that , and neither is contained in the other. Hence, is irreducible.
c) Let be an element in . Let be an ideal as in b). Since every irreducible ideal is prime, is prime. Thus, there is a prime ideal such that . Since every nilpotent element is contained in the intersection of all prime ideals, we conclude that is not nilpotent. Thus, the only nilpotent element is .
66. Let R be a commutative ring with identity. An ideal I of R is irreducible if it cannot be expressed as the i...
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