Let the group $G=\mathbb Z^n$ ($n>1$) be ordered lexicographically. By a technique of Krull's, a valuation domain $R$ can be constructed having $G$ as its divisibility group. It is known that the Krull dimension of this ring is the rank of $G$ (which we chose as $n$.)
Now on one hand, $R$ cannot satisfy the ACC on principal ideals (ACCP). This is because that, as a valuation domain, its finitely generated ideals are principal, and the ACCP would then imply the ACC on f.g. ideals, which is known to be equivalent to being Noetherian. That would make it a PID, but then its Krull dimension would have to be $1$, which it isn't.
It is also known (see here for example) that a Krull domain satisfies the ACCP, so we have that $R$ is not a Krull domain.
All of the above seems pretty ironclad to me.
But someone offered me the following reasoning a while ago to argue it is a Krull domain. I'd like to ask in what ways the argument doesn't work.
Any localization of a valuation ring is again a valuation ring and the value group of the localization is a quotient of the original value group. The Krull dimension of a valuation ring is the rank of the value group. The only rank $1$ quotient group of $\Bbb Z^n$ is $\Bbb Z$, which furnishes the proof.
Surely $R_P$ where $P$ is the unique minimal nonzero prime has Krull dimension $1$. The proposed reasoning above says that the value group of the localized ring has to be $\mathbb Z$, and therefore $R_P$ is a discrete valuation ring.
Apparently something is off in one of the claims:
- Localization of valuation domain is valuation
- The value group of $R_P$ is a quotient of the value group of $R$
- The only quotient of $\mathbb Z^n$ of rank $1$ is $\mathbb Z$.
These all seem very plausible to me, and I do not know the details for these particular claims, so I'm unsure what is going wrong here.
It might also be instructive to know what the value group of $R_P$ is, if $R_P$ is indeed not a DVR.
Epilogue
The definition of Krull domain is more involved than I thought! I had interpreted the list as a list of equivalents rather than a list of conditions for a single definition. So nothing is wrong with either argument... it's just that the second argument isn't finished!
The cited argument correctly shows that the localization at a height one prime ideal is a DVR. However, there are more conditions for being a Krull domain. In this case, it is not true that $R$ is the intersection of all localizations at height one prime ideals, because the prime ideals are totally ordered and $R$ itself is not a DVR.