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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.26


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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