# Construnction of concrete example about the normal scheme

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.

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 $$\frac{a}{b} \in \mathbb{Z}_\mathfrak{p}$$.