Quasi-finite morphism into affine scheme is affine?

Question

Let $f: X\rightarrow\operatorname{Spec}A$ be a quasi-finite morphism of schemes, then is $f$ an affine morphism? If this is wrong, what if $A$ is a field?

Thanks for any answers.

Answer 1

No. Consider the punctured plane mapping in to the plane: $\Bbb A^2_k\setminus \{(0,0)\}\to \Bbb A^2_k$. This is quasi-finite (every fiber is one or zero points) but isn't affine because an affine morphism with affine target must have affine source.

Answer 2

This is false in general, even if you assume that $f$ is of finite presentation. For example, take $f$ to be an open embedding of a non-affine scheme into an affine scheme.

If $A$ is a field then things are better. Namely, by [1, Tag06RT] if you assume that $f$ is of finite type, then $X$ is finite and discrete as a topological space. It’s then easy to see that if $X=\{x_1,\ldots,x_n\}$ then $X=\mathrm{Spec}\left(\prod\mathcal{O}(\{x_i\})\right)$.

[1] https://stacks.math.columbia.edu/