I am reading Ravi Vakil's foundation of algebraic geometry, 4.4.11. He views the Chinese Remainder theorem $$\mathbb Z/60 \cong \mathbb Z/4 \times \mathbb Z/3 \times \mathbb Z/5$$ by using scheme theory.
We look at the structure sheaf of $\operatorname{Spec}(\mathbb Z/60)=\{(2),(3),(5)\}$(easy to see it has the discrete topology). He claims that
the stalk of the structure sheaf at those three points are $\mathbb Z/4, \mathbb Z/3, \mathbb Z/5$ respectively.
I have trouble verifying this. By his exercise 4.3.F, we know these stalks should be $(\mathbb Z/60)_{(2)}, (\mathbb Z/60)_{(3)}, (\mathbb Z/60)_{(5)}$ (localizations at the primes ideals) respectively. But how to show for example
$$(\mathbb Z/60)_{(2)}\cong \mathbb Z/4$$
Please let me know if I interpret his meaning wrongly.
Let $n$ be a natural number, $p$ be a prime dividing $n$ and $m$ be the exponent of $p$ in $n$, i.e. $p^m$ divides $n$ but $p^{m+1}$ does not. Then $(\mathbb{Z}/n)_{(p)}=\mathbb{Z}/(p^m)$.
The proof goes like this : There is a canonical quotient map $\varphi:\mathbb{Z}/(n)\rightarrow\mathbb{Z}/(p^m)$. Note that if $x\in\mathbb{Z}/(n)-(p)$ then $x$ is mapped to $x+(p^m)$, which is a unit in $\mathbb{Z}/(p^m)$. In other words, $\varphi$ maps $\mathbb{Z}/(n)-(p)$ to the units of $\mathbb{Z}/(p^m)$. Using the universal property of the localization, this induces a map $\psi:(\mathbb{Z}/n)_{(p)}\rightarrow\mathbb{Z}/(p^m)$. Can you show that $\psi$ is an isomorphism?