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Let $A$ be a Noetherian domain. I'm interested on the relations of the following properties of $A$:

1. being regular in codimension $1$ (localizations at primes with height $1$ are regular),

2. having $A=\bigcap\limits_{\substack{\mathfrak p\in \mathrm{Spec}A\\\mathrm{ht}(\mathfrak p)=1}}A_{\mathfrak p}$ and

3. $A$ being normal.

More specifically, if $A$ is normal, it's straightforward to see that $A$ is regular in codimension $1$ (localizations are normal, and normal local Noetherian domains of dimension $1$ are DVRs, hence regular), and the intersection property is a well-known result (e.g. Matsumura, p.132). That is, 3.$\implies$1. and 2..

There also are examples of domains regular in codimension $1$ that are not normal (e.g. here). That is, 1.$\;\not\!\!\!\implies$ 3..

My interest is how the properties of being regular in codimension $1$ and having the intersection of localizations property relate to being normal. I want a proof or counterexamples for both 2.$\implies$3. and (1. and 2.)$\implies$ 3..

Note: this question is motivated by a step in the proof that the class group of Cartier divisors in an integral locally Noetherian scheme $X$ embbeds in the class group of Weil divisors. More specifically, a crucial step is that, in the affine case, if $f\in\mathrm{Frac}(A)$ satisfies $v_{\mathfrak{p}}(f)=0$, it then implies that $f$ is in the intersection of all localizations at prime divisors, so $f\in A$. I assumed normality, but assuming 1. and 2. are sufficient. I'm interested to see whether assuming 1. and 2. is weaker than assuming 3., which gives a proof to a more general case.

Lucas Henrique
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    Statements (1) and (2) together imply (3) and I think it is in fact an equivalent condition for Noetherian rings. For starters, see https://en.wikipedia.org/wiki/Krull_ring. – cqfd Feb 15 '25 at 16:58
  • @cqfd: It's good to know this. I'll accept this as an answer if you're willing to explore the details. I might try to find a counterexample for the other cases to complete the answer later. – Lucas Henrique Feb 17 '25 at 19:34
  • See the first comment here for a counterexample to (2) implies (3): https://mathoverflow.net/q/216145. I might write an answer later, but feel free to self-answer. – cqfd Feb 18 '25 at 06:24

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It is easy to see that the conditions (1) and (2) together implies (3). This boils down to showing the intersection of normal rings is normal.

For a counterexample to (2) implies (3), take any one dimensional non-regular local domain $(A,\mathfrak m)$, as mentioned here. Then we have $A=A_\mathfrak m=\displaystyle\bigcap_{\operatorname{ht} P=1}A_P$, but $A$ is not normal.

cqfd
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  • Thank you for your answer! For future readers: the intersection of normal domains is normal as any monic polynomial with coefficients on the intersection is a monic polynomial with coefficients in each ring, then we apply normality for each one. For the counterexample, take for example the local ring of $y^2=x^3$ at $p=(0,0)$, i.e., the localization of $A=k[x,y]/(y^2-x^3)$ at the ideal $(x,y)$. I don't know how to prove the normal case for general normal rings (i.e. $R$ not necessarily a domain, but $R_{\mathfrak p}$ is a normal domain for every maximal $\mathfrak p$). – Lucas Henrique Feb 19 '25 at 18:07