The nilradical of a commutative ring is contained in every prime ideal

Let R be a commutative ring with 1 \neq 0. Prove that if P \subseteq R is prime, then \mathfrak{N}(R) \subseteq P, where \mathfrak{N}(R) denotes the nilradical of R. Deduce that \mathfrak{N}(R) is contained in the intersection of all prime ideals of R.


Let P \subseteq R be a prime ideal, and let x \in R be nilpotent with x^n = 0. Let 1 \leq m \leq n be minimal such that x^m \in P. Note that x^m + P = 0 in R/P, which is an integral domain. If m \geq 2, we have (x + P)(x^{m-1} + P) = 0, so that x+P is a zero divisor in R/P, a contradiction. Thus we have m = 1, and x \in P. Thus \mathfrak{N}(R) \subseteq P, and moreover if \mathfrak{P}(R) denotes the collection of all prime ideals of R, we have \mathfrak{N}(R) \subseteq \bigcap \mathfrak{P}(R).

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