Timeline for Question about the statement of the going-down theorem of Cohen-Seidenberg in Mumford
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2020 at 7:07 | comment | added | Johnny T. | @roysmith O right... | |
| Mar 26, 2020 at 1:02 | comment | added | roy smith | That mark * is not a star on the ring but an asterisk referring the reader to a footnote at the bottom of the page. This the one instance where the Springer edition is clearer than the original. | |
| Mar 25, 2020 at 19:28 | comment | added | Johnny T. | Ok, thank you very much! | |
| Mar 25, 2020 at 19:23 | comment | added | Laurent Moret-Bailly | It is always finite if $x$ is a closed point. | |
| Mar 25, 2020 at 18:52 | history | edited | Johnny T. | CC BY-SA 4.0 |
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| Mar 25, 2020 at 18:16 | comment | added | Johnny T. | @LaurentMoret-Bailly I see your point now thank you. And if I understand you correctly, then for example you are considering the morphism $f: \operatorname{Spec} \kappa(x) \to Y$, where the unique point is mapped to $x$ and $\kappa(x)$ is the residue field of $x$. In this case the map of the stalks is $O_{Y, x} \to O_{Y, x}/\mathfrak{m}_x \hookrightarrow \kappa(x)$. This means some of the non-zero elements of $O_{Y,x}$ gets mapped to $0$, so by the $0$-divisor he is probably including $0$.. Is there an easy example where this morphism is finite? | |
| Mar 25, 2020 at 13:40 | comment | added | Laurent Moret-Bailly | Yes, of course: $x_1=x$, $y_0$=any generization of $x$ in $Y$, assuming of course $Y\neq\{x\}$. | |
| Mar 25, 2020 at 12:54 | history | edited | Johnny T. | CC BY-SA 4.0 |
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| Mar 25, 2020 at 10:44 | comment | added | Johnny T. | @LaurentMoret-Bailly I don't quite follows what you mean. To apply the result one needs to specify a pair of points $x_1 \in X$ and $y_0 \in Y$ such that $f(x_1) \in \overline{\{ y_0 \}}$. But with your example one can not get such a pair to begin with? | |
| Mar 25, 2020 at 7:24 | comment | added | Laurent Moret-Bailly | If you exclude 0 as you say, the result is obviously false: take $X=\{x\}$ (a closed point of $Y$), $f$ the inclusion, and $y_0\neq x$. | |
| Mar 24, 2020 at 21:33 | comment | added | Johnny T. | @PraphullaKoushik Right, so I was wondering because he writes a $0$-divisor in $R^{*}$, is he excluding the zero element? | |
| Mar 24, 2020 at 20:20 | comment | added | Praphulla Koushik | $R^*$ usually denotes non zero elements in a ring $R$.. | |
| Mar 24, 2020 at 19:45 | history | edited | YCor | CC BY-SA 4.0 |
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| Mar 24, 2020 at 19:40 | comment | added | Johnny T. | @LaurentMoret-Bailly That is what I first thought as well, but then he does put the $*$, as in a $0$-divisor in $O_{X,x}^{*}$... Do you think that changes the meaning? | |
| Mar 24, 2020 at 19:36 | history | edited | Johnny T. | CC BY-SA 4.0 |
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| Mar 24, 2020 at 19:28 | comment | added | Laurent Moret-Bailly | To me, a "zero divisor" in a ring $R$ is an element $a$ such that multiplication by $a$ in $R$ is not injective. This includes $0$ if $R$ is not the zero ring. | |
| Mar 24, 2020 at 19:10 | history | asked | Johnny T. | CC BY-SA 4.0 |