Show the stalk of $Spec( \mathbb{Z}/60)$ at $(2)$ is $\mathbb{Z}/4$

Question

Show the stalk of $Spec( \mathbb{Z}/60)$ at $(2)$ is $\mathbb{Z}/4$

It seems that this question has appeared at this site, but I want to use the following proposition to solve it. https://stacks.math.columbia.edu/tag/00CT

Applying this we can have the stalk is $\mathbb{Z}_{(2)}/(60\mathbb{Z})_{(2)}$. But I have no idea how to show that it’s $\mathbb{Z}/4 \mathbb{Z}$...

Thanks for any help..

Answer 1

$3$ and $5$ are invertible in the localization $\mathbb Z_{(2)}$, so that $\mathbb Z_{(2)} / (60\mathbb Z)_{(2)} = \mathbb Z_{(2)} / (4\mathbb Z)_{(2)}$.

There is a natural isomorphism $\mathbb Z_{(2)} / (4\mathbb Z)_{(2)} \cong (\mathbb Z / 4\mathbb Z)_{(2)}$. Because $\mathbb Z / 4\mathbb Z$ is a local ring and $(2)$ is its maximal ideal, localization at $(2)$ doesn't do anything: $(\mathbb Z / 4\mathbb Z)_{(2)} \cong \mathbb Z / 4 \mathbb Z$.