Must PID contain 1? Must PID contain 1? My concern arises when i consider the gcd of say a and b in the PID. Since it is a PID, it is generated by one element say k. k obviously $\in (k)$. However, if PID does not contain 1, then i can't write $k = k \cdot 1$, can 1? So is it true that then k must be equal to $k = k \cdot q$ for some $q \in$ PID. I find this kind of weird. Am I misunderstanding anything?

May has or not. For Z: integers, we can say 2Z is a PID but it does not have 1.
PID means: every ideal of it is generated by only one element. And it does not have zero divisor. Domain part says that.
R is a domain <=> every a,b element of R, if ab=0 => a=0 or b=0
