Let $X,Y$ be reduced, irreducible, separated schemes with same function field. Suppose $U_i,V_i$ are affine coverings of $X,Y$ s.t. $O_X(U_i)\cong O_Y(V_i)$. Then $X\cong Y$.
$\textbf{Q:}$ It seems that in the following, I did not even use integrality of affine subschemes of $X,Y$ at all(i.e. reduced, irreducible) in the following idea. Since $X,Y$ are separated, take $U_{ij}=U_i\cap U_j$ and $V_{ij}=V_i\cap V_j$. From separation of $X,Y$, I know $U_{ij},V_{ij}$ are affine. Now $O_X(U_i)\cong O_Y(V_i)$ compatible with restriction map and this gives $O_X(U_{ij})\cong O_Y(V_{ij})$.(i.e. This isomorphism glues to a global sheaf isomorphism of $O_X,O_Y$ and globally I have $X\cong Y$ by identification of points giving rise to topological isomorphism.) What is wrong with the argument?
Ref. Mumford Algebraic Geometry 2, Sec 2.4, Prop 2.4.8(Proof in the book is omitted.)