Localisation of $\mathbb{Z}/(p^k)$.

Question

I was looking at the wiki that explains localization. It says that the only way to localize $\mathbb{Z}/(p^k)$ is $\{0\}$. The argument is that the elements of $\mathbb{Z}/(p^k)$ are either units or nilpotents elements. So if $x \in S$ (multiplicatively closed subset) is a nilpotent (and $x^n=0$), than clearly $0 \in S$ so the only localization is $0$. But if $u \in S$ is a unit, does this imply that $0 \in S$ in general? Why?

Why if $S=\{(1,0),(1,1)\}$ is the localization $\mathbb{Z}/a \mathbb{Z}$?

Answer 1

Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.

So you are right, $u\in S$ does not implies $ 0\in S$ so $\{0\} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. $\{0\} $ and the ring it self.

Answer 2

If you localize a ring R by units, you actually do nothing, so you get R itself. If you localize by a multiplicative subset which contains nilpotents,you are identifying 0=1,so you get {1=0} .