Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]

(a) Prove that if $R$ is an integral domain, then $\Tor(M)$ is a submodule of $M$.
(Remark: an integral domain is a commutative ring by definition.) In this case the submodule $\Tor(M)$ is called torsion submodule of $M$.

(b) Find an example of a ring $R$ and an $R$-module $M$ such that $\Tor(M)$ is not a submodule.

(c) If $R$ has nonzero zero divisors, then show that every nonzero $R$-module has nonzero torsion element.

 

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Proof.

(a) Prove that if $R$ is an integral domain, then $\Tor(M)$ is a submodule of $M$.

To prove $\Tor(M)$ is a submodule of $M$, we check the following submodule criteria:

1. $\Tor(M)$ is not empty.
2. For any $m, n\in \Tor(M)$ and $t\in R$, we have $m+tn\in M$.

It is clear that the zero element $0$ in $M$ is in $\Tor(M)$, hence condition 1 is met.

To prove condition 2, let $m, n \in \Tor(M)$ and $t\in R$.
Since $m, n$ are torsion elements, there exist nonzero elements $r, s\in R$ such that $rm=0, sn=0$.

Since $R$ is an integral domain, the product $rs$ of nonzero elements is nonzero.
We have
\begin{align*}
rs(m+tn)&=rsm+rstn\\
&=s(rm)+rt(sn) &&\text{($R$ is commutative)}\\
&=s0+rt0=0.
\end{align*}
This yields that $m+tn$ is a torsion elements, hence $m+tn\in \Tor(M)$.
This condition 2 is met as well, and we conclude that $\Tor(M)$ is a submodule of $M$.

(b) Find an example of a ring $R$ and an $R$-module $M$ such that $\Tor(M)$ is not a submodule.

Let us consider $R=\Zmod{6}$ and let $M$ be the $R$-module $R$.
We just simply write $n$ for the element $n+6\Z$ in $R=\Zmod{6}$.
Then we have
\[3\cdot 2=0 \text{ and } 2\cdot 3=0.\] This implies that $2$ and $3$ are torsion elements of the module $M$.

However, the sum $5=2+3$ is not a torsion element in $M$ since if $r\cdot 5=0$ in $\Zmod{6}$, then $r=0$.
Thus, $\Tor(M)$ is not closed under addition. Hence it is not a submodule of $M$.

(c) If $R$ has nonzero zero divisors, then show that every nonzero $R$-module has nonzero torsion element.

Let $r$ be nonzero zero divisors of $R$. That is, there exists a nonzero element $s\in R$ such that $rs=0$. Let $M$ be a nonzero $R$-module and let $m$ be a nonzero element in $M$.
Put $n=sm$.

If $n=0$, then this implies $m$ is a nonzero torsion element of $M$, and we are done.
If $n\neq 0$, then we have
\begin{align*}
rn=r(sm)=(rs)m=0m=0.
\end{align*}
This yields that $n$ is a nonzero torsion element of $M$.

Hence, in either case, we obtain a nonzero torsion element of $M$. This completes the proof.

The post Torsion Submodule, Integral Domain, and Zero Divisors first appeared on Problems in Mathematics.

Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.

(a) Prove that $T(A)$ is a subgroup of $A$.

(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion elements.)

(b) Prove that the quotient group $G=A/T(A)$ is a torsion-free abelian group. That is, the only element of $G$ that has finite order is the identity element.

Proof.

(a) $T(A)$ is a subgroup of $A$

We write the group operation multiplicatively.
Let $x, y\in T(A)$. Then $x, y$ have finite order, hence there exists positive integers $m, n$ such that $x^m=e, y^n=e$, where $e$ is the identity element of $A$. Then we have
\begin{align*}
(xy)^{mn}&=x^{mn}y^{mn} \qquad \text{ (since $A$ is abelian)}\\
&=(x^m)^n(y^m)^n=e^me^n=e.
\end{align*}
Therefore the element $xy$ has also finite order, hence $xy \in T(A)$.

Also, we have
\begin{align*}
(x^{-1})^m=(x^m)^{-1}=e^{-1}=e.
\end{align*}
Hence the inverse $x^{-1}$ of $x$ has finite order, hence $x^{-1}\in T(A)$.

Therefore, the subset $T(A)$ is closed under group operation and inverse, hence $T(A)$ is a subgroup of $A$.

(b) $A/T(A)$ is a torsion-free abelian group

Since $A$ is an abelian group, the quotient $G=A/T(A)$ is also an abelian group.
For $a\in A$, let $\bar{a}=aT(A)$ be an element of $G=A/T(A)$. Suppose that $\bar{a}$ has finite order in $G$. We want to prove that $\bar{a}=\bar{e}$ the identity element of $G$.

Since $\bar{a}$ has finite order, there exists a positive integer $n$ such that
\[\bar{a}^n=\bar{e}.\] This implies that
\[a^nT(A)=T(A)\] and thus $a^n\in T(A)$.

Since each element of $T(A)$ has finite order by definition, there exists a positive integer $m$ such that $(a^n)^m=e$.
It follows from $a^{nm}=e$ that $a$ has finite order, and thus $a\in T(A)$.
Therefore we have
\[\bar{a}=aT(A)=T(A)=\bar{e}.\]

We have proved that any element of $G=A/T(A)$ that has finite order is the identity, hence $G$ is the torsion-free abelian subgroup of $G$.

The post Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group first appeared on Problems in Mathematics.