I have some problems in understanding the relation between domains.
As I understood it I know that we have the following implications PID$\Rightarrow$UFD$\Rightarrow$integrally closed domain$\Rightarrow$integral domain.
Now my confusion arises if I consider the following example. Let us take $R:=\Bbb{Z}/6\Bbb{Z}$. This is not an integral domain since for example $\bar 2\cdot \bar 3=\bar 6=\bar 0$ but $\bar 2, \bar 3\neq \bar 0$. But then using the above implications I would get that $R$ is not a PID. As I understood it, by definition this means that there are ideals which are not generated by a single element of $R$. But on the other side I have seen the remark that ideals in $\Bbb{Z}/n\Bbb{Z}$ are of the form $(\bar m)$ for $\bar m \in \Bbb{Z}/n\Bbb{Z}$.
Where is my mistake?
Thanks for your help.
The negation of "is a domain and all its ideals are principal" is not "it is not a domain AND it has nonprincipal ideals."
The negation would be "It is not a domain OR it has a nonprincipal ideal (possibly both.)"
You have an example of a principal ideal ring that isn't a domain.
Also, $\mathbb Z[x]$ is a domain with nonprincipal ideals.
Finally, $\mathbb Z[x]\times\mathbb Z$ is not a domain and also has a nonprincipal ideal.
All three are examples of rings that aren't PIDs.