Suppose that $k$ is an algebraically closed field and $A$ is the ring $k[a,b,c,d]/(ac-b^2,bd-c^2,ad-bc)$. Let $X$ be Spec$A$, and $m$ be the maximal ideal of $A$ generated by the quotient images of $a,b,c,d$ in $A$.Let $x_0$ be the closed point of $X$ corresponding to the maximal ideal $m$ of $A$.Then is $x_0$ a local complete intersection closed subscheme of $X$?How do I justify whether it is or not? Thank you very much in advance for your kind help.
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$\begingroup$ 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 $d=0$. $\endgroup$– Jason StarrCommented Aug 30, 2022 at 16:35
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$\begingroup$ 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. $\endgroup$– BorisCommented Aug 30, 2022 at 17:25
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$\begingroup$ 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. $\endgroup$– BorisCommented Aug 30, 2022 at 17:34
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$\begingroup$ 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. $\endgroup$– Jason StarrCommented Aug 30, 2022 at 18:51
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$\begingroup$ Thank you very much for your kind guidance and help. I will try my best to digest your comments. $\endgroup$– BorisCommented Aug 30, 2022 at 19:04
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