Soory for my bad English.
Let $X, Y$ be Noetherian separated scheme, and $f:X\to Y$ be morphism of scheme.
Let $U \subseteq Y$ be affine open subscheme.
Then, Is $f^{-1}(U)=X\times_Y U$ affine?
Tell me proof or counter example ,thanks.
algebraic-geometry
Soory for my bad English.
Let $X, Y$ be Noetherian separated scheme, and $f:X\to Y$ be morphism of scheme.
Let $U \subseteq Y$ be affine open subscheme.
Then, Is $f^{-1}(U)=X\times_Y U$ affine?
Tell me proof or counter example ,thanks.
Best Answer
No, morphisms with that property are called affine, see e.g. stacks project. For a counterexample, just pick a non-affine Noetherian separated scheme over a field $k$ and consider the structure morphism to $\text{Spec}(k)$.