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?

From wikipedia:
"Integral domain" is defined almost universally as above, but there is some variation. This article follows the convention that rings have a 1, but some authors who do not follow this also do not require integral domains to have a 1

So authors who ask rings to have a $1$ ask their domains to have a $1$, but authors who don't ask for their rings to have a $1$ don't ask for their domains to have a $1$ and don't ask it of  their PID's either.
I personally would suggest that you ask rings to have an identity and learn the theory there, I have heard from my proffesors that you can later study the subject without asking for rings to have $1$ without much difficulty (Authors who ask that their rings have a $1$ call the rings without $1$ Rngs since they lack the $i$ for identity)
