Exercise 3.13

Proof that $T(S^{-1}M)=S^{-1}(\mathit{TM})$. Suppose $0\notin S$, for this would make everything equal $0$, making the claim trivial. Let $m∕s\in T(S^{-1}M)$; then, there exists non-zero $a∕s^{\prime }\in S^{-1}A$ such that $(m∕s)(a∕s^{\prime })=\mathit{am}∕(s{s}^{\prime })=0$, i.e., there exists $s^{\mathit{\prime \prime \prime }}\in S$ such that $s^{\mathit{\prime \prime \prime }}\mathit{am}=0$. But since $A$ is an integral domain, we have $s^{\mathit{\prime \prime }}a\ne 0$ since $s^{\mathit{\prime \prime }},a\ne 0$. Thus, $m\in T(M)$, and so $m∕s\in S^{-1}(\mathit{TM})$.

Now let $m\in T(M)$, i.e., there exists non-zero $a\in A$ such that $\mathit{am}=0$. Then, $a∕1\ne 0∕1$ since $0\notin S$, and therefore, $(m∕s)(a∕1)=0$ for arbitrary $s$, i.e., $S^{-1}(\mathit{TM})\subseteq T(S^{-1}M)$. □

Proof of equivalence. $i)\Rightarrow \mathit{ii})$. By the previous proof, $T(M_{𝔭})={(\mathit{TM})}_{𝔭}=0_{𝔭}=0$.

$\mathit{ii})\Rightarrow \mathit{iii})$. True since maximal $⟹$ prime.

$\mathit{iii})\Rightarrow i)$. Suppose $T(M)\ne \varnothing$. Then, there is some $x\in T(M)$ and so $\mathit{ax}=0$ for some non-zero $a\in A$. Let $𝔪$ be the maximal ideal that contains $\mathrm{Ann}<!--\; FUNCTION\; APPLICATION\; -->(x)$, which exists via AM Corollary $1.4$. Since $A$ is an integral domain, $a∕1\ne 0$ in $M_{𝔪}$. Then, $a(x∕1)=0$ in $M_{𝔪}$ which is torsion-free, and so $x∕1=0$ in $M_{𝔪}$. This implies $a^{\prime }x=0$ for some $a^{\prime }\in A\setminus 𝔪$, which contradicts our choice of $𝔪\supseteq \mathrm{Ann}<!--\; FUNCTION\; APPLICATION\; -->(x)$. □