Given the following localization of ring: $\mathbb{Z}_{(p)}=\{\frac{m}{n} \in \mathbb{Q}:m,n \in \mathbb{Z} ; p \nmid n\} \subseteq \mathbb{Q}$. We look at an Ideal $I\subset \mathbb{Z}_{(p)}$ and take its restriction $I'=I\cap \mathbb{Z}$. Why can we say that this is an ideal in $\mathbb{Z}$?
I know that $\mathbb{Z}$ is a principal ideal domain and that there exist a homomorphism between $\mathbb{Z}$ and $\mathbb{Z}_{(p)}$. But still I don't know how to justify this.
You can very easily check the definition directly; if $I$ is closed under addition and under multiplication by $\Bbb{Z}_{(p)}$, then $I\cap\Bbb{Z}$ is closed under addition and under multiplication by $\Bbb{Z}$.
Alternatively, you could note that $I\cap\Bbb{Z}$ is the kernel of the restriction of the quotient map $\Bbb{Z}_{(p)}\ \longrightarrow\ \Bbb{Z}_{(p)}/I$ to the subring $\Bbb{Z}\subset\Bbb{Z}_{(p)}$.