Annihilators - Solution

Suppose $R$ is a ring and $M$ is a left $R$ -module.

For each submodule $N$ on $M$ , the annihilator of $N$ in $R$ is defined to be the set of elements $r \in R$ such that $r n = 0$ for all $n \in N$ . Prove that the annihilator of $N$ in $R$ is a 2-sided ideal of $R$ .
For each right ideal $I$ of $R$ , the annihilator of $I$ in $M$ is defined to be the set of all elements $m \in M$ such that $i m = 0$ for all $i \in I$ . Prove that the annihilator of $I$ in $M$ is a submodule of $M$ .
Consider the $Z$ -module $M = Z_{24} \times Z_{15} \times Z_{50}$ and ideal $I = 2 Z$ . Determine the annihilator of $M$ in $Z$ and the annihilator of $I$ in $M$ .

Solution

We first verify the annihilator of $N$ in $R$ is an abelian subgroup. First recall that $0_{R} \cdot n = 0_{M}$ for every $n\in $N$ . Take any $r_{1}, r_{2}$ in the annihilator of $N$ and observe that for all $n \in N$ we have $(r_{1} - r_{2}) n = r_{1} n - r_{2} n = 0_{N} - 0_{N} = 0_{N}$ , and so $r_{1} - r_{2}$ is in the annihilator of $N$ . By the Subgroup Criterion, we've thus proven the annihilator is a subgroup of $R$ under addition. Now fix any element $r$ of the annihilator and let $s \in R$ be an arbitrary element. Then for every $n \in N$ observe that $(s r) n = s (r n) = s \cdot 0_{M} = 0_{M}$ , where for the last equality we once again used the fact that zero acts trivially. This proves $s r$ is in the annihilator for every $s \in R$ and hence the annihilator is a left ideal of $R$ . Similarly, we have $(r s) n = r (s n) = r \cdot 0_{M} = 0_{M}$ , where for the second equality we used the fact that $N$ is a submodule so $s n \in N$ and hence it is annihilated by $r$ . This proves $r s$ is in the annihilator for every $s \in R$ , and hence the annihilator is a right ideal of $R$ .
This will prove very similar to the argument above. First note once again that $i \cdot 0_{M} = 0_{M}$ for every $i \in I$ , and so $0_{M}$ is the annihilator of $I$ in $M$ . Next suppose $m_{1}, m_{2}$ are in the annihilator of $I$ in $M$ . Then for every $i \in I$ we have $i (m_{1} - m_{2}) = i m_{1} - i m_{2} = 0_{M} - 0_{M} = 0_{M}$ , and so $m_{1} - m_{2}$ is in the annihilator of $I$ in $M$ . We've thus proven the annihilator of $I$ in $M$ is a subgroup of $M$ . Finally, take any $r \in R$ and $m$ in the annihilator of $I$ in $M$ . Then $i (r m) = (i r) m = i^{'} m = 0_{M}$ , since $i^{'} \in I$ (as $I$ is a right ideal) and $m$ annihilates everything in $I$ . Thus $r m$ is in the annihilator of $I$ in $M$ , and hence that annihilator is a left $R$ -submodule.
By the definition, the annihilator of $M$ in $Z$ is the collection of all integers $n$ such that $n m = 0$ for every $m \in M$ . The module $M$ is a Cartesian product, which means that its operation is component-wise (addition) and that its zero element is the triple $(0, 0, 0)$ . If we denote the general element in $M$ as $(m_{1}, m_{2}, m_{3})$ then $n \cdot (m_{1}, m_{2}, m_{3}) = (n m_{1}, n m_{2}, n m_{3})$ , and so $n$ annihilates that element exactly when $n m_{1} = 0$ in $Z_{24}$ , $n m_{2} = 0$ in $Z_{15}$ , and $n m_{3} = 0$ in $Z_{50}$ . In other words, the desired annihilator is $A_{1} \cap A_{2} \cap A_{3}$ , where each $A_{i}$ is the annihilator in $Z$ of the corresponding factor group $M_{i}$ of $M$ . By our basic knowledge of the cyclic groups $Z_{k}$ , we know these annihilators are the ideals $A_{1} = 24 Z$ , $A_{2} = 15 Z$ , and $A_{3} = 50 Z$ . The intersection of those three ideals in $Z$ is the ideal generated by their three generators, which (by definition!) is the least common multiple $lcm (24, 15, 50) = 600$ . So in summary, the annihilator of $M$ in $Z$ is the ideal $600 Z$ .

Turning things around, the annihilator of the ideal $I = 2 Z$ in the given module $M$ is the collection of all elements $(m_{1}, m_{2}, m_{3})$ such that $(i m_{1}, i m_{2}, i m_{3}) = (0, 0, 0)$ for every $i \in I$ . Given the description of the ideal $I$ , the annihilator are those triples such that for every integer $k$ we have $2 k m_{1} = 0$ in $Z_{24}$ , $2 k m_{2} = 0$ in $Z_{15}$ , and $2 k m_{3} = 0$ in $Z_{50}$ . These conditions are satisfied exactly by $m_{1} \in 12 Z_{24}$ , $m_{2} = 0$ in $Z_{15}$ , and $m_{3} \in 25 Z_{50}$ . The annihilator of $I$ in $M$ is therefore the submodule $(12 Z_{24}) \times 0 \times (25 Z_{50})$ , which is isomorphic as an abelian group to $Z_{2} \times Z_{2}$ .