Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.32
Solution:
We begin with a lemma.
Lemma: If $a,b \in \mathbb{Z}$ are relatively prime, then $$(a) + (b) = \mathbb{Z}.$$ Proof: we can write $xa + yb = 1$ by Bezout’s identity for some integers $x$ and $y$, so that $1 \in (a)+(b)$. $\square$
(1) By Proposition 11 in the text, $\mathcal{M}_I$ is not empty. Thus by Exercise 7.3.18, $\mathsf{Jac}(I)$ is an ideal of $R$. Since $I \subseteq M$ for all $M \in \mathcal{M}_I$, we have $I \subseteq \mathsf{Jac}(I)$.
(2) Let $x \in \mathsf{rad}(I)$, and let $M \in \mathcal{M}_I$. Now $x^m \in I \subseteq M$ for some $m \geq 1$; let $m$ be minimal with this property. If $m \geq 2$, then $xx^{m-1} \in M$; since $M$ is prime, we have $x \in M$ or $x^{m-1} \in M$, a contradiction. Thus $m = 1$, and we have $x \in M$. Thus $\mathsf{rad}(I) \subseteq M$ for all $M \in \mathcal{M}_I$, and hence $\mathsf{rad}(I) \subseteq \mathsf{Jac}(I)$.
(3) Write the prime factorization of $n$ as $n = \prod p_i^{k_i}$. Recall that the maximal ideals of $\mathbb{Z}$ are precisely those of the form $(p)$ for prime $p$. Now $(n) \subseteq (p)$ precisely when $p$ divides $n$; thus $\mathcal{M}_{(n)}$ consists precisely of the ideals $(p_i)$ for the primes $p_i$ dividing $n$. Thus $$\mathsf{Jac}(n\mathbb{Z}) = \bigcap_{p_i | n} (p_i).$$ Using the lemma (with induction) and Exercise 7.4.12, $\mathsf{Jac}(n\mathbb{Z}) = (\prod p_i)$.
