How can I prove that $\mathbb{Z}/p^q$, where $p$ is prime, is a UFD?
Well, I know that if $q=1$, it is because then it is a field.
How can I prove that $\mathbb{Z}/p^q$, where $p$ is prime, is a UFD?
Well, I know that if $q=1$, it is because then it is a field.
The ring $\mathbb{Z}/n\mathbb{Z}$ is an integral domain if and only if $n$ is a prime (or $0$). In fact, this is one of the ways to define what it means for something to be a prime element of an arbitrary ring.
To be more specific, if $n = ab$ with both $a$ and $b$ strictly greater than $1$, then neither $a$ nor $b$ will be $0$ in $\mathbb{Z}/n\mathbb{Z}$, but $ab = n = 0$ in $\mathbb{Z}/n\mathbb{Z}$.
By a lucky coincidence, whenever $\mathbb{Z}/n\mathbb{Z}$ is an integral domain, it is also a UFD (in fact a PID). This is because in that case, either $n$ is a prime, in which case the ring is finite, and any finite integral domain is a field (and clearly any field is a PID), or $n = 0$ in which case we are dealing with the ring $\mathbb{Z}$, which is an Euclidean domain (which implies that it is a PID).