Irreducible closed set, local property?

Question

Let $Y \subset X$ be a connected closed subset of a scheme $(X,\mathcal O_X)$. If $Y \cap U$ is irreducible inside $U$, for every affine $U \subset X$, then is it true that $Y$ is irreducible inside $X$?

Answer 1

Yes, this is true. If a connected set $Y$ has multiple irreducible components, then two of them, say $Y_1,Y_2$, must meet at a point $y$. Take some affine open neighborhood $U\subset X$ of $y$. Then $U\cap Y = (U\cap Y_1)\cup(U\cap Y_2)$ which demonstrates that $U\cap Y$ is reducible. To see that $U\cap Y\neq U\cap Y_i$, note that if this was true, then $U\cap Y_i \supset U\cap Y_j$, and then taking closures $Y_i\supset Y_j$, which is impossible since no irreducible component is contained in any other irreducible component. So your claim is correct.