Exercise 3.12

Question

Let $A$ be an integral domain and $M$ an $A$-module. An element $x \in M$ is a \emph{torsion element} of $M$ if $\operatorname{Ann}(x) 
e 0$, that is if $x$ is killed by some non-zero element of $A$. Show that the torsion elements of $M$ form a submodule of $M$. This submodule is called the \emph{torsion submodule} of $M$ and is denoted by $T(M)$. If $T(M) = 0$, the module $M$ is said to be torsion-free. Show that
  \begin{enumerate}[label=oman*)]
    \item If $M$ is any $A$-module, then $M/T(M)$ is torsion-free.
    \item If $f : M 	o N$ is a module homomorphism, then $f(T(M)) \subseteq T(N)$.
    \item If $0 	o M^{\prime} 	o M 	o M^{\prime\prime}$ is an exact sequence, then the sequence $0 	o T(M^{\prime}) 	o T(M) 	o T(M^{\prime\prime})$ is exact.
    \item If $M$ is any $A$-module, then $T(M)$ is the kernel of the mapping $x \mapsto 1 \otimes x$ of $M$ into $K \otimes_A M$, where $K$ is the field of fractions of $A$.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of first claim]
  Let $x,y \in T(M)$ be distinct; we have $ax = 0, a^{\prime}y = 0$ for some non-zero $a,a^{\prime} \in \operatorname{Ann}(x)$. Then, $aa^{\prime}(x-y) = aa^{\prime}x - aa^{\prime}y = 0$ where $aa^{\prime} 
e 0$ since $A$ is an integral domain. Thus, $T(M)$ is an abelian group.
  \par $x \in T(M) \implies ax = 0$ for some non-zero $a \in \operatorname{Ann}(x)$. Then, $\operatorname{Ann}(\alpha x) 
i a$ for all $\alpha \in A$ since $\alpha xy = 0$. Thus, $T(M)$ is closed under external multiplication, and $T(M) \subseteq M$ is a submodule.
\end{proof}
\begin{proof}[Proof of $i)$]
  Suppose $\overline{x} \in T(M/T(M))$. Then, there is non-zero $a \in A$ such that $0 = a\overline{x} = \overline{ax}$, and so $ax \in T(M)$. There exists non-zero $a^{\prime} \in A$ such that $a^{\prime}ax = 0$, but $a^{\prime}a 
e 0$ since $A$ is an integral domain, and so $x \in T(M)$, i.e., $\overline{x} = 0$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  Suppose $x \in T(M)$, then there exists non-zero $a \in A$ such that $ax = 0$. We then see $0 = f(ax) = af(x)$, and so $a \in \operatorname{Ann}(f(x)) \implies f(x) \in T(N)$.
\end{proof}
\begin{proof}[Proof of $iii)$]
  We have the exact sequence
  \begin{equation*}
    0 	o M^{\prime} \xrightarrow{f} M \xrightarrow{g} M^{\prime\prime}.
  \end{equation*}
  We claim that
  \begin{equation*}
    0 	o T(M^{\prime}) \xrightarrow{\overline{f}} T(M) \xrightarrow{\overline{g}} T(M^{\prime\prime})
  \end{equation*}
  is exact, where $\overline{f}$ is the restriction of $f$ on $T(M^{\prime})$, and likewise for $\overline{g}$. Note that the image of each map is contained in the next module of the sequence by $ii)$. At $T(M^{\prime})$, we see that $f$ is injective and so $\overline{f}$ is injective since it is simply a restriction, and so the sequence is exact at $T(M^{\prime})$. At $T(M)$, we claim $\operatorname{im}(\overline{f}) = \ker(\overline{g})$. Suppose $m \in \ker(\overline{g}) \subseteq \ker(g) = \operatorname{im}(f)$. Then, choose $m^{\prime} \in M^{\prime}$ such that $f(m^{\prime}) = m$; since $m \in T(M)$ there exists non-zero $a \in A$ such that $am = 0$. Then, $0 = am = af(m^{\prime}) = f(am^{\prime})$. By the injectivity of $f$, we see $am^{\prime} = 0$, and so $m^{\prime} \in T(M^{\prime})$, i.e., $\ker({g}) \subseteq \operatorname{im}(\overline{f})$. Next we see $m \in \operatorname{im}(\overline{f}) \subseteq \operatorname{im}(f) = \ker(g)$, and so $\overline{g}(m) = 0$. Thus, $\operatorname{im}(\overline{f}) \subseteq \ker(\overline{g})$, and our sequence is exact at $T(M)$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  Let $\varphi: x \mapsto 1 \otimes x$. We see that $T(M) \subseteq \ker(\varphi)$ since $x \in T(M) \implies \exists a 
e 0 \in A$ such that $ax = 0 \implies x \mapsto 1 \otimes x = a/a \otimes x = 1/a \otimes ax = 1/a \otimes 0 = 0$.
  \par To show $\ker(\varphi) \subseteq T(M)$, we first show that
  \begin{equation*}
    K = \varinjlim_{\mathclap{a \in A \setminus \{0\}}} A[1/a].
  \end{equation*}
  We see that for $A[1/a], A[1/a^{\prime}]$, $A[1/a] + A[1/a^{\prime}] \subseteq A[1/(aa^{\prime})]$, and so by AM Exercise $2.17$, we have that $K$ is the direct limit we claimed. Tensoring with $M$ gives
  \begin{equation*}
    K \otimes_A M = \varinjlim_{\mathclap{a \in A \setminus \{0\}}} (A[1/a] \otimes_A M),
  \end{equation*}
  by AM Exercise $2.20$. 
  \par Now if $1 \otimes x = 0$ in $K \otimes_A M$, we see that $1 \otimes x = 0$ in some $A[1/a] \otimes_A M$ for some $a \in A$ by AM Exercise $2.15$. Since $1 = a/a$, we have $1 \otimes x = (a/a) \otimes x = (1/a) \otimes ax = 0$. Moreover, $A[1/a] \otimes_A M \hookrightarrow M$ is injective by having $b/a \otimes x \mapsto bx$, which only maps to zero if $b = 0$ by the fact that $A$ is an integral domain. Thus, $ax = 0$, and so $x \in T(M)$.
\end{proof}

Exercise 3.12

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Exercise 3.12

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