As we know every normal Noetherian domain $R$ can be written as $$R=\bigcap\_{\mathsf{ht}(\mathfrak p)=1}R\_\mathfrak p.$$ I'm asking myself the following question:


**Question:** If the normalization of $\widetilde{R}$ is given by $$\widetilde{R}=\bigcap\_{\mathsf{ht}(\mathfrak p)=1}R\_\mathfrak p,$$ then is $R$ normal? 


Every reference or counterexample is well accepted.



$\widetilde{R}=\bigcap_{\mathsf{ht}(\mathfrak p)=1}R_\mathfrak p$
I am just posting my comments as an answer. First, in the positive direction, for a Noetherian, integral domain $R$ with fraction field $K$, if $R$ satisfies Serre's condition $S\_2$, then the natural inclusion of subrings of $K$, $$R\to \bigcap\_{\text{ht}(\mathfrak{p}) = 1} R\_{\mathfrak{p}}, $$ is an isomorphism. Thus, for such a ring, $R$ is normal if and only if the normalization $\widetilde{R}$ equals the intersection above as an $R$-subalgebra of $K$.


However, there are many examples of non-normal, Noetherian, integral domains that fail Serre's condition $S\_2$, for which $\widetilde{R}$ equals the intersection. For instance, for every field $k$ of characteristic different from $2$, the following $k$-subalgebra $R$ of the polynomial ring $k[x,y]$ is a non-normal, Noetherian, integral domain that fails Serre's condition $S\_2$, and for which the normalization $\widetilde{R}$ equals the intersection above, $$R = k[x,xy,y^2-1,y^3-y]\subset k[x,y].$$


