[proofplan] The implications $(1) \implies (2) \implies (3)$ are immediate: localization of the zero module is zero, and every maximal ideal is prime. The content is $(3) \implies (1)$. We prove the contrapositive: if $M \neq 0$, then there exists a maximal ideal $\mathfrak{m}$ with $M_\mathfrak{m} \neq 0$. Given a nonzero element $m \in M$, its annihilator $\operatorname{Ann}_R(m)$ is a proper ideal, hence contained in some maximal ideal $\mathfrak{m}$. The localization $M_\mathfrak{m}$ then contains the nonzero element $\frac{m}{1}$, because the condition $\frac{m}{1} = 0$ in $M_\mathfrak{m}$ would require an element of $R \setminus \mathfrak{m}$ to annihilate $m$, contradicting $\operatorname{Ann}_R(m) \subseteq \mathfrak{m}$. [/proofplan] [step:Verify $(1) \implies (2) \implies (3)$] $(1) \implies (2)$: If $M = 0$, then for every $\mathfrak{p} \in \operatorname{Spec}(R)$, the localization $M_\mathfrak{p} = S^{-1}M$ (where $S = R \setminus \mathfrak{p}$) consists only of the element $\frac{0}{1}$, since every element of $M$ is $0$. So $M_\mathfrak{p} = 0$. $(2) \implies (3)$: Every maximal ideal of $R$ is prime, so the set of maximal ideals is a subset of $\operatorname{Spec}(R)$. If $M_\mathfrak{p} = 0$ for all primes, then in particular $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$. [/step] [step:Prove $(3) \implies (1)$ by contrapositive: a nonzero element forces some $M_\mathfrak{m} \neq 0$] We prove the contrapositive: assume $M \neq 0$ and produce a maximal ideal $\mathfrak{m}$ with $M_\mathfrak{m} \neq 0$. Choose a nonzero element $m \in M$. Define the annihilator $\operatorname{Ann}_R(m) = \{r \in R : rm = 0\}$. This is an ideal of $R$. Since $m \neq 0$, we have $1 \notin \operatorname{Ann}_R(m)$ (as $1 \cdot m = m \neq 0$), so $\operatorname{Ann}_R(m)$ is a proper ideal of $R$. Every proper ideal of $R$ is contained in a maximal ideal (by Zorn's lemma). Let $\mathfrak{m}$ be a maximal ideal of $R$ with $\operatorname{Ann}_R(m) \subseteq \mathfrak{m}$. We claim $\frac{m}{1} \neq 0$ in $M_\mathfrak{m}$. By definition, $\frac{m}{1} = 0$ in $M_\mathfrak{m}$ if and only if there exists $u \in R \setminus \mathfrak{m}$ with $um = 0$. Such a $u$ would satisfy $u \in \operatorname{Ann}_R(m) \subseteq \mathfrak{m}$, contradicting $u \in R \setminus \mathfrak{m}$. Therefore $\frac{m}{1} \neq 0$, so $M_\mathfrak{m} \neq 0$. [guided] We prove the contrapositive of $(3) \implies (1)$: suppose $M \neq 0$ and find a maximal ideal $\mathfrak{m}$ where $M_\mathfrak{m} \neq 0$. Pick any nonzero $m \in M$. The annihilator $\operatorname{Ann}_R(m) = \{r \in R : rm = 0\}$ is an ideal of $R$. It is proper because $1 \cdot m = m \neq 0$, so $1 \notin \operatorname{Ann}_R(m)$. By Zorn's lemma, every proper ideal is contained in some maximal ideal, so choose $\mathfrak{m} \supseteq \operatorname{Ann}_R(m)$. Now consider $\frac{m}{1} \in M_\mathfrak{m}$, where $M_\mathfrak{m} = (R \setminus \mathfrak{m})^{-1}M$. The element $\frac{m}{1}$ is zero in $M_\mathfrak{m}$ if and only if there exists $u \in R \setminus \mathfrak{m}$ with $um = 0$ in $M$, i.e., $u \in \operatorname{Ann}_R(m)$. But $\operatorname{Ann}_R(m) \subseteq \mathfrak{m}$, so any $u \in \operatorname{Ann}_R(m)$ satisfies $u \in \mathfrak{m}$, contradicting $u \in R \setminus \mathfrak{m}$. Therefore no such $u$ exists, $\frac{m}{1} \neq 0$ in $M_\mathfrak{m}$, and $M_\mathfrak{m} \neq 0$. The key mechanism is that the annihilator of a nonzero element is a proper ideal, so it is "trapped" inside some maximal ideal. At that maximal ideal, the localization cannot kill the element because the only elements that act as denominators (those in $R \setminus \mathfrak{m}$) lie outside the annihilator. [/guided] [/step]