Non integral connected normal scheme? A connected Noetherian scheme with stalks that are integral domains is an integral scheme.
Hence a connected Noetherian normal scheme is integral.
But is it possible that there is a connected normal scheme that is not integral (hence not irreducible), if the hypothesis on Noetherianity is dropped?
 A: I guess the example in https://stacks.math.columbia.edu/tag/0568 works as a counterexample in this case, too.
With the notation there, let $\mathfrak{p}\in\operatorname{Spec} A$ and we show that $A_{\mathfrak{p}}$ is normal.
Suppose $x/s\in\operatorname{Frac} A_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$. That is, $x\in A_{\mathfrak{p}}$, $s\in A_{\mathfrak{p}}\backslash\{0\}$ and there should exist a positive integer $n$ and $a_0,\ldots,a_{n-1}\in A_{\mathfrak{p}}$ such that $$(x/s)^n+\displaystyle\sum_{i = 0}^{n-1} a_i(x/s)^i = 0 \space \text{in} \space A_{\mathfrak{p}}$$
Let $\mathfrak{p}_k\in\operatorname{Spec}A_k$ be the image of $\mathfrak{p}$ for each nonnegative integer $k$. First, note that $A_{\mathfrak{p}}$ is the colimit of $(A_k)_{\mathfrak{p}_k}$ by the obvious map.
Now, $$x^n+\displaystyle\sum_{i = 0}^{n-1} a_ix^is^{n-i} = 0\in A_{\mathfrak{p}}$$ by the assumption. Take a positive integer $k$ so that $x,s$ and all the $a_i$ come from elements of $(A_{k-1})_{\mathfrak{p}_{k-1}}$, also denoted by the same symbols, and that $$x^n+\displaystyle\sum_{i = 0}^{n-1} a_ix^is^{n-i} = 0\in (A_{k-1})_{\mathfrak{p}_{k-1}}$$ If (not $\mathfrak{p}_{k-1}$ but) $\mathfrak{p}_k\in\operatorname{Spec} A_k$ is a smooth point, then $(A_k)_{\mathfrak{p}_k}$ is a discrete valuation ring and we are done since $x\in s(A_k)_{\mathfrak{p}_k}$ and the same in $A_{\mathfrak{p}}$.
We suppose that $\mathfrak{p}_k\in\operatorname{Spec} A_k$ is a singular point. The local ring $(A_k)_{\mathfrak{p}_k}$ is isomorphic to $(\mathbb{C}[X,Y]/(XY))_{(X,Y)}$ and we identify them in the sequel.
Since every other term of the elements of $A_k$ coming from $A_{k-1}$ is constant starting from the second term, one of $$\mathbb{C}[X]_{(X)}\subseteq (\mathbb{C}[X,Y]/(XY))_{(X,Y)}$$ or $$\mathbb{C}[Y]_{(Y)}\subseteq (\mathbb{C}[X,Y]/(XY))_{(X,Y)}$$ includes $$\operatorname{Im}((A_{k-1})_{\mathfrak{p}_{k-1}}\to (A_k)_{\mathfrak{p}_k}).$$ We may assume this is true for $\mathbb{C}[X]_{(X)}\subseteq (\mathbb{C}[X,Y]/(XY))_{(X,Y)}$.
Since the equation of $x$, $s$ and $a_i$ also holds in the discrete valuation ring $\mathbb{C}[X]_{(X)}$, $x\in s\mathbb{C}[X]_{(X)}$ and the same holds in $A_{\mathfrak{p}}$, completing the proof.
