Let $(X, \mathcal O_X)$ be a Noetherian scheme. If $X$ is an integral scheme (i.e. $X$ is irreducible and reduced scheme), then is it true that for every open subset $U$ of $X$, the open subscheme $(U, \mathcal O_U)$ is integral ? (Here $\mathcal O_U=\mathcal O_X|_U$)
I think this is true and here are my thoughts: $U$ being an open subset of $X$, is irreducible. Moreover, for every $u\in U$, we have $\mathcal O_{U,u}\cong \mathcal O_{X,u}$ is reduced , hence $U$ is reduced. Thus $U$ is integral.
Am I correct ?
Definition: Let $(X, \mathcal{O}_X)$ be a scheme. We say $(X, \mathcal{O}_X)$ is "integral" iff for every open set $U\subseteq X$ it follows $\mathcal{O}_X(U)$ is an integral domain.
Example: Consider $(V,\mathcal{O}_V)$ with $V\subseteq X$ an open subscheme. Let $U \subseteq V$ be an open set. It follows $\mathcal{O}_V(U)\cong \mathcal{O}_X(U)$ is an integral domain since $X$ is integral, hence $(V, \mathcal{O}_V)$ is an integral scheme.