Annihilator of an ideal in an R-module From Dummit Foot, 10.1 ex.11b)
Let M be the abelian group $\mathbb{Z}/\mathbb{24Z}\times \mathbb{Z}/\mathbb{15Z}\times\mathbb{Z}/\mathbb{50Z}$. Let $I=2\mathbb{Z}$ and describe the annihilator I in M as a direct product of cyclic groups.
For reference, If I is a right ideal of R, the annihilator of I in M is $\{m\in M| am=0 ~\forall a\in I \}$ 
Thanks.
 A: The elements of $2\mathbb{Z}$ are even integers i.e. of the form $2m$ for $m\in\mathbb{Z}$. Write an arbitrary $x\in\mathrm{Ann}(I)$ in component form as $(a,b,c)\in M$. Since $2x=0$ implies $(2m)x=0$ for all $m\in\mathbb{Z}$, it suffices to check that $2$ annihilates $x$ in $M$. This can be written as the following congruence system:
$$2a\equiv0\quad (24)$$
$$2b\equiv 0\quad (15)$$
$$2c\equiv0\quad (50)$$
In order, this implies $12|a$, $15|b$ and $25|c$ - this is elementary number theory.
A: The annihilator of $I$ in $M$ is $\{(x,0,y)\in M:12|x, 25|y\}\cong \mathbb{Z}/2\oplus\mathbb{Z}/2$. Try acting on such an element by an element of $I$ (for example, multiplying each coordinate by 2), and do the same with an element not in the annihilator, and it should become clear how I came up with the answer.
A: If you first prove some basic facts about the notion of "the annihilator of an ideal in a module", it becomes a simple computation:
$$\begin{align*} & \mathrm{Ann}_{2\mathbb{Z}}(\mathbb{Z}/\mathbb{24Z}\times \mathbb{Z}/\mathbb{15Z}\times\mathbb{Z}/\mathbb{50Z}) \\
&= \mathrm{Ann}_{2\mathbb{Z}}(\mathbb{Z}/\mathbb{24Z})\times \mathrm{Ann}_{2\mathbb{Z}}(\mathbb{Z}/\mathbb{15Z})\times\mathrm{Ann}_{2\mathbb{Z}}(\mathbb{Z}/\mathbb{50Z}) \\
&= (\{0,12\} \subseteq \mathbb{Z}/\mathbb{24Z}) \times (\{0\} \subseteq \mathbb{Z}/\mathbb{15Z}) \times (\{0,25\} \subseteq \mathbb{Z}/\mathbb{50Z}) \\
& \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \end{align*}$$
The idea is to look at the above computation, convince yourself each step makes sense, then formalize the theorem used in each step as a precise statement of mathematics and prove it.
