Lemma (4)
Let R be a commutative ring. Then the intersection of all prime ideals of R is {a ∈ R|aⁿ = 0, n > 0} = A
Proof
Let p be a prime ideal of R and a be a nilpotent element of R. Let n be the least positive integer where aⁿ ∈ p. Which implies that x * xⁿ⁻¹ ∈ p for n > 1, where by the definition of prime ideal either xⁿ⁻¹ ∈ p or x ∈ p, this is a contradiction thus A ⊆ ∩p=primeideal p
Suppose a is not nilpotent. we want to show that there exists some prime ideal q where a ∉ q and therefore not in the intersection of all prime ideals of R. Let S be the set of all ideals for which aⁿ ∉ S for all positive integers n. Because a is not nilpotent aⁿ ≠ 0 and aⁿ ∉ (0), therefore (0) ∈ S, making S non-empty. Thus by Zorn’s Theorem, S has a maximal element m, where aⁿ ∉ m for all n.
Thus we must show that m is a prime ideal. Let x, y ∈ R, where x, y ∉ m, we must show that xy ∉ m. Because x ∉ m, the ideal generated by m and x, (m, x) is greater than the ideal m, m ⊂ (m, x). However since m is maximal in S, (m, x) ∉ S, therefore aⁿ ∈ (m, x). Thus aʳ ∈ (m, y) for some positive integer r. Thus we see that aⁿ⁺ʳ ∈ (m, xy), therefore (m, xy) ∉ S and m ⊂ (m, xy). We see then that xy ∉ m and that m is prime, which means that since a ∉ m, a prime ideal, then we can conclude that a is not in the intersection of all primes ideals of R.