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Jun 10, 2020 at 6:56 comment added Lennart Meier @k.j. The citation is indeed wrong, but the statement that an injective morphism of rings induces a dominant morphism on their spectra is still true. See for example math.stackexchange.com/questions/389036/… for a proof.
Jun 10, 2020 at 2:48 comment added k.j. In the proof of lemma 2.5.(3) of your note, using Stack Project Tag0CC1, you conclude that $\operatorname{Spec}R \to \operatorname{Spec}R_0$ has the dense image. But this proposition assumes that $R$ is an integral domain. Why do we use this for general $R$?
Mar 18, 2019 at 10:59 vote accept k.j.
Aug 28, 2018 at 5:46 history answered Lennart Meier CC BY-SA 4.0