0
$\begingroup$

I know that morphisms between affine schemes $f\colon\operatorname{Spec}(S)\to\operatorname{Spec}(R)$ are in bijection with morphisms of rings $\varphi\colon R\to S$. So if I ask that $f$ is an affine morphisms, what does that mean for the ring map $\varphi$? What is the corresponding ring morphism condition for affineness?

$\endgroup$
1
  • $\begingroup$ By affine morphism do you mean your $f$ or actually an affine morphism between schemes? If you mean the former, the map $\phi$ is the map between global sections. $\endgroup$
    – Qi Zhu
    Apr 8, 2021 at 11:23

1 Answer 1

2
$\begingroup$

Any map of affine schemes $\operatorname{Spec} B \to \operatorname{Spec} A$ is already affine with no conditions/restrictions on $\varphi$. Indeed, the definition of an affine morphism is one so that the preimage of any affine open subscheme of the target is again an affine open subscheme of the source. So letting $\operatorname{Spec} A_0\subset\operatorname{Spec} A$ be an affine open subscheme, we have that it's preimage is $\operatorname{Spec} B\otimes_A A_0$ which is affine and automatically an open subscheme of $\operatorname{Spec} B$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .