Exact sequence proof

Question

Show that if $R$ is an integral domain and $0\rightarrow{}A\xrightarrow{f}B\xrightarrow{g}C$ is an exact sequence of left $R$-modules then $0\rightarrow{}T(A)\xrightarrow{f_T}T(B)\xrightarrow{g_T}T(C)$ is an exact sequence as well.

$T(A)$ is set of all torsion elements of $A.$

We know that $g$ is surjective, $f$ is injective and $\operatorname{im}f=\ker g.$ We want to show $g_T$ is surjective, $f_T$ is injective and $\operatorname{im}f_T=\ker g_T.$ Because $f_T, g_T$ are only restrictions of $f,g$ we have only to show $\operatorname{im}f_T=\ker g_T$. Am I correct?

How to do that?

Answer 1

You are right in that we only have to show im$f_T=$ ker$g_T$. The injectivity of $f_T$ is clear. And im$f_T\subseteq$ ker$g_T$ is also clear.

Say $b\in$ ker$g_T$. Then $b\in$ ker$g$ $\cap$ $T(B)$. By exactness of the previous sequence, $b\in$ im$f$. Say $f(a)=b$. WTS $a\in T(A)$. We have $rb=0$ for some $r\neq 0$. Then $rb=rf(a)=f(ra)=0$. The injectivity of $f$ implies $ra=0$. Thus $a\in T(A)$.

That's the idea.

Answer 2

As @Joshua Mundinger mentioned, $g$ need not be surjective.

The statement $f_T$ is injective can be obtained directly by considering $f$ is injective. (If $f_T(x) = 0$, then ...)

About $\textrm{im}f_T = \textrm{ker}g_T$, here's a hint: $$\textrm{ker}g_T = \textrm{ker}g \cap T(B) = \textrm{im}f \cap T(B).$$

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Proof of $\textrm{im}f_T = \textrm{ker}g_T$ (by the hint)

Since $f_T$ is the restriction of $f$, $\textrm{im}f_T \subseteq \textrm{im}f \cap T(B) = \textrm{ker}g_T$.

On the other hand, given $b \in \textrm{ker}g_T = \textrm{im}f \cap T(B)$. We may write $f(a) = b$ for some $a \in A$. Since $b \in T(B)$, there exist non-zero $r \in R$ such that $rb= 0$. Hence $f(ra) = rf(a) = rb= 0$, which implies $ra = 0$ since $\textrm{ker} f = 0$. Thus $a \in T(A)$ and therefore $b= f(a) = f_T(a) \in \textrm{im}f_T$. Our proof is complete.