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According to the Hartshorne textbook, $X=\operatorname{Spec}A $ is normal if $A$ is UFD. (such context is proof of why the divisor class of the group of $X$ is zero if $A$ is UFD) I have a curiosity why such a fact holds. However, I want to make a concrete example rather than to prove the given fact.

For example, take $A=\mathbb{Z}$ (It is undoubtly UFD!). Then, the rational number, $\mathbb{Q}$, is a ring extension of $\mathbb{Z}$.(or, such extension is viewed as a field of fraction). $X=\operatorname{Spec}A = \left \{ (0) \right \}U \left \{ (p) | ~p:prime~number \right \} $. (I deonte $(p)$ by $\mathfrak{p}$). Next, when considering the defintion of normal scheme, all of local rings are integrally closed domain, our claim is

Claim : The local ring $\mathcal{O}_{X,\mathfrak{p}}$ is a integrally closed domain. i.e (1) $\mathcal{O}_{X,\mathfrak{p}}$ is an integral domain (2) its integral closure in its field of fractions is $\mathcal{O}_{X,\mathfrak{p}}$ for all $\mathfrak{p} \in \operatorname{Spec}A$.(*Now, I do not want to foucs on the generic point.. )

When obseriving $\mathcal{O}_{X,\mathfrak{p}}$, $\mathcal{O}_{X,\mathfrak{p}}$ is isomorphic to $\mathbb{Z}_{\mathfrak{p}}$ by prop2.2 in Hartshorne. And $\mathbb{Z}_{\mathfrak{p}}=\left \{ \frac{a}{b}| a \in \mathbb{Z}, p \not | b \right \} \subset \mathbb{Q} $ (Here is local ring at prime ideal $\mathfrak{p}$. And it is clearly integral domain. However, (2) is not easy for me ; I do not know how to collect integral closure $I$. Clearly, $\mathbb{Z} \subset \mathbb{Z}_{\mathfrak{p}} \subset \mathbb{Q} $ holds. Then is $I\overset{??}=\left \{ q \in \mathbb{Q} : q ~ is~integrable~over~ \mathbb{Z}_{\mathfrak{p}} \right \}$ the collection of integral closure? I do not see which elements are integral closure. Anyway, the proof will finish just by showing $I= \mathbb{Z}_{\mathfrak{p}} $ , but I do not imagine the collection of integral closure.

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Assuming $\frac{a}{b} \in \mathbb{Q}$ is integral over $\mathbb{Z}_\mathfrak{p}$, it means that there exists $c_1,\ldots,c_n \in \mathbb{Z}_\mathfrak{p}$ satisfied $$(\frac{a}{b})^n+c_1(\frac{a}{b})^{n-1}+\ldots+c_n=0.$$ Since $c_i=\frac{x_i}{y_i}$ where $y_i \nmid p$, after multiplying $b^ny_1y_2\ldots y_n$ one gets $$a^n(y_1\ldots y_n)+x_1ba^{n-1}(y_2\ldots y_n)+\ldots+x_nb^n(y_1\ldots y_{n-1})=0.$$ If $\frac{a}{b} \notin \mathbb{Z}_\mathfrak{p}$, one can assume $p \mid b$ and $p \nmid a$. But it's clearly wrong since the first term is not divisible by $p$ but the others are not. It means $\frac{a}{b} \in \mathbb{Z}_\mathfrak{p}$.

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