**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.30**

Solution:

First, we certainly have $I \subseteq \mathsf{rad}(I)$ since for all $a \in I$, $a^1 \in I$. In particular, $\mathsf{rad}(I)$ is nonempty. Now let $a,b \in \mathsf{rad}(I)$ with $a^n$,$b^m \in I$. Now $$(a+b)^{m+n} = \sum_{k=0}^{m+n} \binom{m+n}{k} a^k b^{m+n-k}$$ by Exercise 7.3.25. Note that if $k \geq n$, then $a^k \in I$, and if $k < n$, then $b^{m+n-k} \in I$, since $I$ is an ideal. Thus every term in the expansion of $(a+b)^{n+m}$ is in $I$, and thus $(a+b)^{m+n} \in I$. We also have, for all $r \in R$, $(ra)^n = r^na^n \in I$, and in particular $(-a)^n \in R$. Thus $\mathsf{rad}(I)$ is an ideal of $R$.

Now note the following. $x + I \in \mathsf{rad}(I)/I$ if and only if $x \in \mathsf{rad}(I)$, if and only if $x^n \in I$ for some $n \geq 1$, if and only if $x^n + I = 0$ in $R/I$ for some $n \geq 1$, if and only if $(x+I)^n = 0$ in $R/I$ for some $n \geq 1$, if and only if $x+I \in \mathfrak{N}(R/I)$. Thus $\mathsf{rad}(I)/I = \mathfrak{N}(R/I)$.