Timeline for complete ring as union of finite type algebras
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2011 at 14:27 | vote | accept | unknown | ||
| Feb 23, 2011 at 15:08 | comment | added | Leo Alonso | Perhaps the paper you mention refers to the fact that $Spf(\hat{R})$ is the direct limit (or increasing union) of the finite type schemes $Spec(R/m^i)$, provided that $R$ is finite type over some base scheme ($m$ denotes the maximal ideal of $R$). | |
| Feb 23, 2011 at 15:03 | history | edited | Leo Alonso | CC BY-SA 2.5 |
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| Feb 23, 2011 at 12:06 | comment | added | unknown | @Leo Alonso I found this as I wrote as "well known fact" on a paper. The fact that this is an union is crucial in the proof. So I would like to understand this at least in easy cases. | |
| Feb 23, 2011 at 12:03 | comment | added | Leo Alonso | I was thinking on the description $\hat{R}=k[[x]]=k[x][[T]]/\langle T−x \rangle$. So the stages are $R_i:=k[x,T]/\langle T^i,T−x \rangle = k[x]/\langle x^i\rangle$, as usual, but this is not an increasing union... Can you provide more detail to your question? I still think that the metric is the key. | |
| Feb 23, 2011 at 11:48 | comment | added | unknown | @Leo Alonso in the case $R=k[x]$ completed along $I=(x)$ what do you get as these algebras? | |
| Feb 23, 2011 at 11:40 | history | answered | Leo Alonso | CC BY-SA 2.5 |