## 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 $\mathfrak{N}(R) \subseteq \bigcap \mathfrak{P}(R)$.