For nilpotent elements of the commutative ring, choose positive integers with . The binomial theorem giveseach term has either or . Also , , and for every . Thus is an additive subgroup stable under multiplication by arbitrary ring elements:Commutativity is essential to the simple binomial argument; an arbitrary sum of nilpotent elements in a noncommutative ring need not be nilpotent.
Consider all ideals containing and disjoint from the nonempty multiplicative subset . This collection is nonempty because it contains . The ascending chain condition for the Noetherian ring gives a maximal member : otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since is nonempty and misses it, .
Suppose but . Maximality implies that and both meet , so write and with . Thenwhile closure under products gives , a contradiction. HenceThis proves the prime ideal avoiding a multiplicative subset result without needing . The assumptions themselves rule out , since every ideal contains zero.
Conversely, if , none of its positive powers is zero. The multiplicative subset is disjoint from . Part (b) supplies a prime ideal disjoint from , hence not containing . Thus is outside the intersection, proving the nilradical as the intersection of prime ideals identity:In the zero ring there are no prime ideals; the empty intersection is interpreted as the whole ring, which is also its nilradical. For nonzero rings this argument proves existence of a prime ideal whenever a nonnilpotent element is available.
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