Timeline for About the Dimension of a complete local ring
Current License: CC BY-SA 3.0
7 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 4, 2014 at 18:13 | comment | added | YCor | Yes, that's part of the remaining case $n=2$. | |
| Oct 4, 2014 at 18:10 | comment | added | Antoine Ducros | No, ok. I made a mistake: I thought you meant that the case where dim A=1 is trivial. And I do not know whether k[[x,y]] can be embedded in k[[t]]. | |
| Oct 4, 2014 at 16:02 | comment | added | YCor | Case $n=1$: a negative answer would mean that there is an injective continuous $k$-algebra homomorphism from $k[[X]]$ into a 0-dimensional $A$, which would be of finite length over $k$. Do I miss something? | |
| Oct 4, 2014 at 10:21 | comment | added | Antoine Ducros | Is the case n=1 that trivial, YCor ? | |
| Oct 4, 2014 at 8:32 | comment | added | YCor | The example given there is essentially that if $f\in vk[[v]]$ is transcendental over $k(v)$, then $x\mapsto u$, $y\mapsto uv$, $z\mapsto uf(v)$ defines a continuous injective $k$-algebra embedding of $k[[x,y,z]]$ into $k[[u,v]]$. Note that this answers negatively the question for all $n\ge 3$. This leaves the case $n=2$ open (the case $n\le 1$ being trivial). | |
| Oct 4, 2014 at 4:14 | history | answered | Jesse Elliott | CC BY-SA 3.0 |