Timeline for Flat cover by a locally Noetherian scheme
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2012 at 13:58 | vote | accept | hadimath | ||
| Dec 17, 2011 at 19:41 | comment | added | Jason Starr | @Akhil: I should say a bit more. Since $B$ is $A$-flat, the induced map $B\otimes_A (m_A/m_A^2) \to m_AB/m_A^2B$ is an isomorphism. In particular, $m_AB/m_AB^2$ is a free $B/m_AB$-module of infinite rank. But that implies that $m_AB$ cannot be a finitely generated ideal in $B$, thus $B$ is not Noetherian. | |
| Dec 17, 2011 at 19:02 | comment | added | Jason Starr | @Akhil: You are right. That map may be zero. But the map from $m_A/m_A^2$ to $m_AB/m_A^2B$ is nonzero, which gives the same result. | |
| Dec 16, 2011 at 8:06 | answer | added | Laurent Moret-Bailly | timeline score: 12 | |
| Dec 16, 2011 at 1:55 | comment | added | Akhil Mathew | @Jason: If $A \to B$ is a local extension of DVR's (automatically faithfully flat), then the map $m_A/m_A^2 \to m_B/m_B^2$ may be zero (e.g. if there is ramification). | |
| Dec 15, 2011 at 19:52 | comment | added | Martin Brandenburg | @Jason: This is not a comment - it is an answer. | |
| Dec 15, 2011 at 16:27 | comment | added | Jason Starr | No, there is not necessarily such a $T$. For instance, let $S$ be $\text{Spec}(A)$ where $A=k[[x_1,x_2,...]]$. If there is such a faithfully flat $T$, then there is a point $t$ of $T$ which maps to the closed point of $S$. Let $B$ be the local ring of $T$ at $t$. Then there is a faithfully flat local homomorphism $A\to B$. In particular, the induced map $m_A/m_A^2 \to m_B/m_B^2$ is injective. Since $m_A/m_A^2$ is infinite dimensional, so is $m_B/m_B^2$, contradicting that $B$ is Noetherian. | |
| Dec 15, 2011 at 15:52 | history | asked | hadimath | CC BY-SA 3.0 |