Affinization of a normal variety By affinization of $X$ I mean $\text{Aff}(X) := \text{Spec}(\Gamma(X, \mathcal{O}_X))$.  First, I claim that if $X$ is reduced, then $\text{Aff}(X)$ is reduced.  The argument goes: if $\Gamma(X, O_X)$ has a nilpotent element, i.e. a nilpotent global function, it is nonzero and nilpotent on some open affine of $X$, contradicting that $X$ is reduced.
Does this argument sound convincing?
If it does, then I have a follow up question: if $X$ is normal, does that mean $\text{Aff}(X)$ is normal?
 A: Fact: Let $X$ be a scheme. Consider the following assertions:
$(i)$ Every connected component of $X$ is irreducible.
$(ii)$ $X$ is the disjoint union of its irreducible components.
$(iii)$ For all $x \in X,$ the nilradical of the local ring $\mathcal O_{X,x}$ is a prime ideal.
Then we have the following implications: $(i) \Rightarrow (ii) \Rightarrow (iii).$ In addition if the scheme $X$ is locally Noetherian then all the three are equivalent.
As a corollary of the above fact we get that, if $X$ is connected and locally Noetherian then $X$ is integral if and only if for all $x \in X,$ the local ring $\mathcal O_{X,x}$ is a domain.
Claim: Let $X$ be a locally Noetherian normal scheme, and let $U \subseteq X$ be a connected open subset. Then $\Gamma(U, \mathcal O_X)$ is a normal domain.
We can assume $X=U,$ i.e. $X$ is connected. Then $X$ is integral (from the above discussion). Let $K(X)$ be the function field of $X$ and let $t \in K(X)$ be integral over $A=\Gamma(X, \mathcal O_X).$ Now $X$ is normal and hence $t \in \mathcal O_{X,x}, \forall x \in X.$ This shows that $t \in A.$
