Timeline for Counterexample to Openness of Flat Locus
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2014 at 14:50 | vote | accept | Mahdi Majidi-Zolbanin | ||
| Feb 2, 2014 at 14:16 | comment | added | Jason Starr | @user76758: Yes, of course I should have seen that. If you annihilate $1$, then you annihilate the entire ring. | |
| Feb 2, 2014 at 14:14 | comment | added | Jason Starr | @ACL: I think we probably both posted our comments about "$1$" at the same time: the element $1$ in $B$ shows that $B\otimes_A \mathbb{Q}$ cannot vanish. | |
| S Feb 2, 2014 at 9:33 | history | suggested | gaoxinge | CC BY-SA 3.0 |
more clear
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| Feb 2, 2014 at 8:57 | review | Suggested edits | |||
| S Feb 2, 2014 at 9:33 | |||||
| Feb 2, 2014 at 7:21 | comment | added | user76758 | @JasonStarr: I was looking at the distinguished element 1 in $B$ (which is what makes $B$ more tractable than a random $A$-module). | |
| Feb 2, 2014 at 6:54 | answer | added | anonymous | timeline score: 4 | |
| Feb 2, 2014 at 6:06 | comment | added | ACL | @JasonStarr: The constant sequence $(1)_p$ is obviously non-torsion. | |
| Feb 2, 2014 at 5:35 | comment | added | Jason Starr | @user76758: Here is a direct way of getting a contradiction from vanishing of $B\otimes_A \mathbb{Q}$, rather than using ultrafilters or the vanishing of $B[1/N]$ (which I still do not immediately see). The natural ring homomorphism $\mathbb{Z} \to B$ is clearly injective. Thus, by flatness of $\mathbb{Q}$ over $\mathbb{Z}$, also the induced homomorphism $\mathbb{Q} \to B\otimes_A \mathbb{Q}$ is injective. Therefore $B\otimes_A \mathbb{Q}$ is nonvanishing. | |
| Feb 2, 2014 at 5:28 | comment | added | Jason Starr | @user76758: How do you go from the assertion that $B\otimes_A \mathbb{Q}$ vanishes to the assertion that there exists an integer $N$ that annihilates $B$? Of course you can directly show that $B\otimes_A \mathbb{Q}$ is nonzero using ultrafilters, etc. But I do not see how to directly conclude that $B$ is annihilated by a single integer $N$. | |
| Feb 2, 2014 at 5:15 | answer | added | Jason Starr | timeline score: 4 | |
| Feb 2, 2014 at 4:42 | comment | added | user76758 | Let $A = \mathbf{Z}$, $B = \prod_p \mathbf{F}_p$. Assume the non-flat locus $Y$ in Spec($B$) is closed. It contains the evident clopen points Spec($\mathbf{F}_p$), so if $J$ is an ideal in $B$ cutting out $Y$ then $J$ has vanishing image in each direct factor ring $\mathbf{F}_p$, so $J=0$. Thus, $B$ would be nowhere flat over $A$, so $B \otimes_A \mathbf{Q}$ would vanish and hence $B[1/N]=0$ for some $N > 0$. But this is false. | |
| Feb 2, 2014 at 3:11 | history | edited | Mahdi Majidi-Zolbanin | CC BY-SA 3.0 |
added 64 characters in body
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| Feb 2, 2014 at 3:04 | history | asked | Mahdi Majidi-Zolbanin | CC BY-SA 3.0 |