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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.

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If $q\neq 1$ then this is false, as the ring is not even an integral domain. –  Tobias Kildetoft Oct 22 '12 at 7:38
    
why it isn't an Integral Domain? –  Kits89 Oct 22 '12 at 7:44
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Please think a bit longer on this. Finding zero-divisors in the given ring is a good exercise that will really help you understand that ring. –  Tobias Kildetoft Oct 22 '12 at 7:50
    
okay, I understand. Thanks –  Kits89 Oct 22 '12 at 7:52
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@TobiasKildetoft Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 20 '13 at 7:18
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1 Answer 1

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).

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