Timeline for Check whether a closed point of a Noetherian affine scheme is a local complete intersection
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 30, 2022 at 19:04 | comment | added | Boris | Thank you very much for your kind guidance and help. I will try my best to digest your comments. | |
| Aug 30, 2022 at 18:51 | comment | added | Jason Starr | Yes, for the second claim, I am just saying that the unique prime component of the ideal $\langle ac-b^2, bd-c^2, ad-bc, a,d \rangle$ is the maximal ideal $\langle a,b,c,d\rangle$. For the first claim, if you adjoin the elements $a$ and $d$ to the ideal, the quotient ring is zero-dimensional and Cohen-Macaulay. From this it follows that the original ring is Cohen-Macaulay and that pair of elements is a regular sequence. Since it is a regular sequence, the original ideal has LCI quotient if and only if the zero-dimensional, Cohen-Macaulay quotient is LCI. It is not even Gorenstein. | |
| Aug 30, 2022 at 17:34 | comment | added | Boris | For your comment that the point $x_0$ of $X$ is set-theoretically "cut out" by the 2 equations $a=0$ and $d=0$, could I justify this by saying that in the ring $A$, if the quotient images of $a,d$ in $A$ are both set to be zero, then the three relations defining $A$ imply that the quotient images of $b,c$ in $A$ also vanish?Thanks. | |
| Aug 30, 2022 at 17:27 | history | edited | Boris | CC BY-SA 4.0 |
clarified question
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| Aug 30, 2022 at 17:25 | comment | added | Boris | Thank you very much for your kind guidance. I was asking whether $x_0$ is a local complete intersection closed subscheme of $X$. Could you explain why every Zariski open neighborhood of the point $x_0$ in $X$ is not a local complete intersection scheme?I have trouble justifying this. And could you explain why the point $x_0$ is set-theoretically cut out by the two equations $a=0$ and $d=0$?Thank you very much. | |
| Aug 30, 2022 at 16:35 | comment | added | Jason Starr | Could you please clarify your question? Certainly every Zariski open neighborhood of that point in $X$ is not a local complete intersection scheme. However, the point you mention is set-theoretically "cut out" by the two equations $a=0$ and $d=0$. | |
| Aug 30, 2022 at 13:35 | history | asked | Boris | CC BY-SA 4.0 |