(a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module.
Prove that the module $M$ has a nonzero annihilator.
In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$.
Here $r$ does not depend on $m$.

(b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal.

Proof.

(a) Prove that the module $M$ has a nonzero annihilator.

Since $M$ is a finitely generated $R$-module, there is a finite set
\[A:=\{a_1, a_2, \dots, a_n\} \subset M\] such that $M=RA$

As $M$ is a torsion $R$-module, for each $a_i\in A\subset M$ there is a nonzero element $r_i\in R$ such that
\[r_ia_i=0.\] Let us put $r\in R$ to be the product of these $r_i$:
\[r:=r_1 r_2 \cdots r_n.\] Note that $r$ is a nonzero element of $R$ since each $r_i$ is nonzero and $R$ is an integral domain.

We claim that the element $r$ annihilates the module $M$.
Let $m$ be an arbitrary element in $M$. Since $M$ is generated by the set $A$, we can write
\[m=s_1a_1+s_2a_2+\cdots +s_n a_n\] for some elements $s_1, s_2, \dots, s_n\in R$.

Note that since $R$ is an integral domain, it is commutative by definition.
Hence we can change the order of the product in $r$ freely. Thus for each $i$ we can write
\[r=p_ir_i,\] where $p_i$ is the product of all $r_j$ except $r_i$.

Then it follows that we have
\begin{align*}
ra_i&=p_ir_ia_i=p_i0=0 \tag{*}
\end{align*}
for each $i$.
Using this, we obtain
\begin{align*}
rm&=r(s_1a_1+s_2a_2+\cdots +s_n a_n)\\
&=rs_1a_1+rs_2a_2+\cdots +rs_n a_n\\
&=s_1ra_1+s_2ra_2+\cdots +s_n ra_n && \text{as $R$ is commutative}\\
&=s_10+s_20+\cdots +s_n 0 && \text{by (*)}\\
&=0.
\end{align*}
Therefore, for any element $m\in M$ we have proved that $rm=0$.
Thus the nonzero element $r$ annihilates the module $M$.

(b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal.

Let $R=\Z$ be the ring of integers. Then $R=\Z$ is an integral domain.
Consider the $\Z$-module
\[M=\oplus_{i=1}^{\infty}\Zmod{2^i}.\]

Then each element $a\in M$ can be written as
\[a=(a_1+\Zmod{2}, a_2+\Zmod{2^2}, \dots, a_k+\Zmod{2^k}, 0, 0, \dots)\] for some $a_1, a_2, \dots, a_k\in \Z$.
(Here $k$ depends on $a$.)

It follows that we have
\[2^ka=0,\] and thus $M$ is a torsion $\Z$-module.

We now prove that any annihilator of $M$ must be the zero element of $R=\Z$.
Let $r\in \Z$ be an annihilator of $M$.
Choose an integer $k$ so that $r < 2^k$. Consider the element \[a=(0, 0, \dots, 1+\Zmod{2^k}, 0, 0, \dots)\] in $M$. The only nonzero entry of $a$ is at the $k$-th place. Since $r$ is an annihilator, we have \begin{align*} 0=ra=(0, 0, \dots, r+\Zmod{2^k}, 0, 0, \dots) \end{align*} and this implies that $r=0$ because $r < 2^k$. We conclude that the annihilator is the zero ideal.

The post Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator first appeared on Problems in Mathematics.

Let $R$ be a ring with $1$ and let $M$ be a left $R$-module.
Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be
\[\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.\] (If $rx=0, r\in R, x\in S$, then we say $r$ annihilates $x$.)

Suppose that $N$ is a submodule of $M$. Then prove that the annihilator
\[\Ann_R(N)=\{ r\in R\mid rn=0 \text{ for all } n\in N\}\] of $M$ in $R$ is a $2$-sided ideal of $R$.

Proof.

To prove $\Ann_R(N)$ is a $2$-sided ideal of $R$, it suffices to prove the following conditions:

1. For any $r, s \in \Ann_R(N)$, we have $r+s\in \Ann_R(N)$.
2. For any $r\in R$ and $s\in \Ann_R(N)$, we have $rs\in\Ann_R(N)$.
3. For any $r\in R$ and $s\in \Ann_R(N)$, we have $sr\in \Ann_R(N)$.

Let $r, s \in \Ann_R(N)$. Then for any $n\in N$, we have
\[rn=0 \text{ and } sn=0.\] It follows from these identities that
\begin{align*}
(r+s)n=rn+sn=0+0=0
\end{align*}
for any $n\in N$.
Hence $r+s\in \Ann_R(N)$, and condition 1 is met.

To prove condition 2, let $r\in R$ and $s\in \Ann_R(N)$.
For any $n\in N$, we have
\begin{align*}
&(rs)n=r(sn)\\
&=r(0) && (\text{since } s \in \Ann_R(N))\\
&=0.
\end{align*}
Thus, the element $rs$ annihilates all elements $n$ in $N$.
So $rs\in \Ann_R(N)$ and condition 2 is satisfied.
(At this point, we have proved that $\Ann_R(N)$ is a left ideal of $R$.)

To check condition $3$, let $r\in R$ and $s\in \Ann_R(N)$.
We need to prove that $(sr)n=0$ for any $n\in N$.
Since $N$ is a submodule of $M$, the element $rn$ is in $N$.
Since $s\in \Ann_R(N)$, we have $s(rn)=0$.
It follows from the associativity that
\begin{align*}
(sr)n=s(rn)=0
\end{align*}
for any $n\in N$.
Hence $sr\in \Ann_R(N)$ and condition 3 is proved.

Therefore, $\Ann_R(N)$ is a $2$-sided ideal of $R$.
This completes the proof.

Remark.

Note that the proof of condition 1 and 2 shows that $\Ann_R(S)$ is a left ideal of $R$ for any subset $S$ of $M$.
We needed the assumption that $N$ is a submodule of $M$ when we proved condition $3$.

The post Annihilator of a Submodule is a 2-Sided Ideal of a Ring first appeared on Problems in Mathematics.