Integral = Reduced and Irreducible

A scheme $X$ is integral if for every $U \subset X$ open we have $O_X(U)$ is integral.

Proposition. A scheme $X$ is integral iff it is reduced and irreducible.

Proof.
Suppose $X$ is integral. Then $O_X(U)$ is integral and hence reduced for all $U$ open in $X$.  It suffices to show that every two nonempty open subsets of $X$ intersect.

Indeed, suppose $U$ and $V$ are nonempty open subsets of $X$ that do not intersect. By shrinking $U$ and $V$, we can assume that they are affine $U = \text{Spec} A$ and $V = \text{Spec} B$. Then $O_X(U \cup V) = O_X(U) \times O_X(V) = A \times B$ is not integral.

Conversely, suppose $X$ is reduced and irreducible. Let $U =\text{Spec} A$ be an affine open of $X$. Then $U$ is reduced and irreducible so $A$ is a reduced ring i.e. the nilradical of $A$ is $0$ and $U = \mathbb{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p}$, i.e. the nilradical of $A$ is prime. Thus $0$ is a prime ideal of $A$ so $A$ is integral domain.

Now let $U$ be an open subset of $X$. Let $V$ be an affine open contained in $U$. Then the restriction map $O_X(U) \to O_X(V)$ is an injection so $O_X(U)$ must be integral.


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Corollary. An affine scheme $X =\text{Spec} A$ is integral iff $A$ is an integral domain.

Proof. Suppose $A$ is an integral domain. As reduced-ness is an affine-local condition,  $X$ is reduced. On the other hand, $\text{Spec} A = \mathbb{V}(0)$ and $0$ is a prime ideal so $X$ is irreducible.
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Note: Integrality is not a stalk-local condition, as the disjoint union of integral scheme is not integral.
($\text{Spec} A \sqcup \text{Spec} B = \text{Spec} (A \times B)$). However it is almost stalk-local.