Timeline for On a condition on ideals viwed as a Zariski open condition on co-tangent space
Current License: CC BY-SA 4.0
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S Jun 21, 2020 at 0:05 | history | bounty ended | CommunityBot | ||
S Jun 21, 2020 at 0:05 | history | notice removed | CommunityBot | ||
Jun 20, 2020 at 2:15 | vote | accept | uno | ||
Jun 13, 2020 at 0:49 | answer | added | Hailong Dao | timeline score: 1 | |
S Jun 12, 2020 at 22:21 | history | bounty started | uno | ||
S Jun 12, 2020 at 22:21 | history | notice added | uno | Draw attention | |
Apr 29, 2020 at 17:50 | comment | added | R. van Dobben de Bruyn | Sorry, I didn't mean to suggest that this is a full answer, but it is certainly a strategy in case $R$ is a $k$-algebra. Then $R/I$ is a finite dimensional $k$-algebra, so you can view it as a ring object in affine $k$-schemes (in particular, addition and multiplication are given by polynomials; to make this explicit use the structure constants of $R/I$). For example this immediately shows that for a single $f\not\in I$, the set of $x$ with $f\in(\mathfrak mI:x)$ is open, because $x \mapsto fx \in R/\mathfrak mI$ is continuous. | |
Apr 29, 2020 at 16:31 | comment | added | uno | @R. van Dobben de Bruyn: I'm having some issues seeing as to how the multiplication $R/I \times \mathfrak m/\mathfrak mI\to R/\mathfrak m I$ being a morphism of schemes would be helpful (and what scheme structure exactly are you talking about on $ \mathfrak m/\mathfrak mI$ ?) ... a priori, I don't see any connection between the usual Zariski topology of Spec $(R/I)$ and the Zariski topology I've defined on $\mathfrak m/\mathfrak m^2$ ... | |
Apr 21, 2020 at 2:32 | comment | added | R. van Dobben de Bruyn | You always have $(\mathfrak mI \colon x) \supseteq I$, right? So the question is when is there nothing else. You would like to have access to $R/J$ for any $\mathfrak m$-primary ideal $J$ as an affine space over $k$ (or a successive fibration of affine spaces), and use some sort of structure constants to show that multiplication $m \colon R/I \times \mathfrak m/\mathfrak m I \to R/\mathfrak m I$ is a morphism of varieties. This is certainly ok if $R$ is a $k$-algebra, but probably also works in other settings. Then study the map $m$ geometrically. | |
Apr 20, 2020 at 23:43 | history | edited | uno | CC BY-SA 4.0 |
added 60 characters in body
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Apr 13, 2020 at 23:24 | history | asked | uno | CC BY-SA 4.0 |