I have this exercise, where I'm a little stuck. Any help would be nice.

1) Find $\displaystyle Ann_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z})$

2) For ideals $\displaystyle I_1$ and $\displaystyle I_2$ in a ring R find $\displaystyle Ann_R(R/I_{1}\oplus R/I_{2})$

3) Define $\displaystyle S=\{2^n3^m | n,m \in \mathbb{N}_0\}$. Is S multiplicatively closed?

Prove that $\displaystyle \overline{5/1}=\overline{0}$ as an element in $\displaystyle S^{-1}(\mathbb{Z}/10\mathbb{Z})$ and that $\displaystyle \overline{2/1}\neq \overline{0}$ as an element in $\displaystyle S^{-1}(\mathbb{Z}/10\mathbb{Z})$.

4) With S as above find $\displaystyle S^{-1}M$ for each of following modules;

$\displaystyle \mathbb{Z}/2\mathbb{Z}$

$\displaystyle \mathbb{Z}/12\mathbb{Z}$

$\displaystyle \mathbb{Z}/5\mathbb{Z}$ and

$\displaystyle \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/12\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}$

Well, regarding (1) I know that the sum module $\displaystyle \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}$ is just rows $\displaystyle \left(

\begin{array}{c}

x \\

y \\

\end{array}\right)$, right? So I would say that the Annulator would be $\displaystyle \left(

\begin{array}{c}

0 \\

0 \\

\end{array}\right)$

And (3) I would say yes because $\displaystyle 2^{n_1}3^{m_1}*2^{n_2}3^{m_2}=2^{n_1+n_2}3^{m_1+m_ 2}=2^{n}3^{m}\in S$?