Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.37
Solution: First we prove a lemma.
Lemma: Let $R$ be a ring, and let $I_1, I_2, J \subseteq R$ be ideals such that $J \subseteq I_1, I_2$. Then $(I_1/J)(I_2/J) = I_1I_2/J$.
Proof: ($\subseteq$) Let $$\alpha = \sum (x_i + J)(y_i + J) \in (I_1/J)(I_2/J).$$ Then $$\alpha = \sum (x_iy_i + J) = (\sum x_iy_i) + J \in (I_1I_2)/J.$$ Now let $\alpha = (\sum x_iy_i) + J \in (I_1I_2)/J.$ Then $$\alpha = \sum (x_i + J)(y_i + J) \in (I_1/J)(I_2/J). \square$$
From this lemma and the lemma to Exercise 7.3.36, it follows by an easy induction that $$(p\mathbb{Z}/p^m\mathbb{Z})^m = (p\mathbb{Z})^m/p^m\mathbb{Z} = p^m\mathbb{Z}/p^m\mathbb{Z} \cong 0.$$ Thus $p\mathbb{Z}/p^m\mathbb{Z}$ is nilpotent in $\mathbb{Z}/p^m\mathbb{Z}$.
