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This question might be a bit simple for many of those reading this text, but I was wondering why, given the fact that $\mathbb{Z}_8$ is not even an integral domain (and therefore not a PID), all the ideals of this ring are generated by one element. Is there any ideal in this ring generated by two elements? If the answer is negative, doesn't it imply that $\mathbb{Z}_8$ is a PID? Can a ring have only principal ideals but not be a domain?

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    $\begingroup$ You are correct that all the ideals in $\mathbb{Z}_8$ are principal, but that it is not a PID. See the wikipedia page on "principal ideal ring" for more about rings like this. $\endgroup$
    – CoArp
    Commented Jan 2 at 16:44
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    $\begingroup$ The last sentence of your post is 10^6 times better as a title than the random piece of your question you selected, so I took the liberty of making it the title. In the future be sure to make your title as indicative of the key question as possible. $\endgroup$
    – rschwieb
    Commented Jan 2 at 18:58

3 Answers 3

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In the definition of a principal ideal domain you assume the ring is already an integral domain, i.e. a ring $A$ is a principal ideal domain if it is a domain in which every ideal is principal. So yes, it is possible for a ring to have only principal ideals but not to be a domain.
For any $n\in \mathbb{N}_{>0}$ all the ideals of $\mathbb{Z}/n\mathbb{Z}$ are of the form $d\mathbb{Z}/n\mathbb{Z}$ for $d|n$ so they are principal but $\mathbb{Z}/n\mathbb{Z}$ is not a domain if $n$ is not prime.

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  • $\begingroup$ Wouldn't it be "principal ideal domain"? $\endgroup$
    – Emmy N.
    Commented Jan 2 at 17:02
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    $\begingroup$ You are right, thanks! I'll edit the answer $\endgroup$
    – Temoi
    Commented Jan 2 at 17:04
  • $\begingroup$ Besides, is there a ring such that satisfies the same condition as in an UFD (without the domain one), but it is not a domain? $\endgroup$
    – Emmy N.
    Commented Jan 2 at 17:05
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    $\begingroup$ @DanielC. For sure people have floated some versions of that, but the issue is that for nondomains one has to rethink what uniqueness of factorizations means. In an integral domain $aR=bR$ implies $au=b$ for some unit $u$, but this is not necessarily true without the domain condition. It's not a very "popular" topic as far as I know. $\endgroup$
    – rschwieb
    Commented Jan 2 at 18:16
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In the trivial ring, every ideal is principal, but it is not a domain.

What you are looking for is the notion of a principal ideal ring.

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All the ideals of $\mathbb{Z}_{n}$ are $m\mathbb{Z}_{n}$, where $m$ is a divisor of $n$ and $m\mathbb{Z}_{n} = \{mx : x \in \mathbb{Z}_{n}\}$. Thus, the ideals of $\mathbb{Z}_{8}$ are $\mathbb{Z}_{8}$, $2\mathbb{Z}_{8}$, $4\mathbb{Z}_{8}$ and $8\mathbb{Z}_{8} = \{0\}$.

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    $\begingroup$ I guess you mean $\mathbb{Z}/n\mathbb{Z}$, not $\mathbb{Z}_n$ (ring of $n$-adics). $\endgroup$ Commented Jan 2 at 20:19
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    $\begingroup$ Yes, that's a common confusion. But you are correct. It is better to use $\mathbb{Z}/n\mathbb{Z}$ because it is a common practice that ring of $n$ adic integers are denoted by $\mathbb{Z}_{n}$. But here I assume $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_{n}$. $\endgroup$
    – Afntu
    Commented Jan 3 at 5:23
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    $\begingroup$ Why would you assume that $n$-adics are the same as $\mathbb Z/n\mathbb Z$? $\endgroup$ Commented Jan 3 at 16:46
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    $\begingroup$ No, no, I do not assume that. I am saying that $n$ adic integers are denoted as $\mathbb{Z}_{n}$ but in abstract algebra $\mathbb{Z}/n\mathbb{Z} $ also denoted as $\mathbb{Z}_{n}$. $\endgroup$
    – Afntu
    Commented Jan 3 at 17:00

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