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Timeline for Closed immersion of closed fiber?

Current License: CC BY-SA 3.0

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Jun 3, 2014 at 3:30 comment added bananastack this is of course because $y \in Y$ was a closed point to begin with.
Jun 2, 2014 at 15:11 vote accept user51197
Jun 2, 2014 at 15:10 comment added user51197 Thank you, I see it now: the associated morphism Spec $ k(y) \rightarrow Y$ is indeed a closed immersion, and that property is stable under base change.
Jun 2, 2014 at 15:04 history answered bananastack CC BY-SA 3.0