I know that if we have a surjection $f:B\rightarrow A$, this induces an injection on the spectra $f^* \colon \operatorname{Spec} A\hookrightarrow \operatorname{Spec} B.$

What about the opposite? Does an injection in the spectra of affine schemes induce a surjection in rings?

I would assume if such a nice property held I would have found it somewhere online, so can you provide a counterexample? Also, is there a natural set of conditions such that if $f^*$ satisfies them, then we can conclude that $f$ is a surjection?


closed as off-topic by Pace Nielsen, Qiaochu Yuan, abx, Sean Lawton, user44191 Mar 15 at 1:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Pace Nielsen, Qiaochu Yuan, abx, Sean Lawton, user44191
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Is $A\rightarrow S^{-1}A$ surjective?. $\endgroup$ – Einfacher Schreiberling Mar 14 at 17:49
  • 1
    $\begingroup$ Nope. Think of a localization $A\to A[f^{-1}]$ $\endgroup$ – Qfwfq Mar 14 at 17:49
  • 1
    $\begingroup$ Ok.. almost at the same time! :D $\endgroup$ – Qfwfq Mar 14 at 17:49
  • $\begingroup$ Oh of course you are right! Thank you both very much. $\endgroup$ – Ioannis Zolas Mar 14 at 18:16
  • 1
    $\begingroup$ Or think about $\mathbb{Q} \subseteq \mathbb{R}$. But you do want to think about closed embeddings of (affine) schemes, those come from surjective ring maps. See also II, Exercise 3.11 in Hartshorne as well. $\endgroup$ – Karl Schwede Mar 14 at 18:30