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Let $A=\mathbb C[x,y_1,y_2,\ldots]$ and $I=(y_1^2, y_2^2, y_3^2\cdots(x-1)y_1, (x-2)y_2, (x-3)y_3\cdots)$.

Let $X=\operatorname{Spec}(A/I)$. It can be shown that $X$ is homeomorphic to $\operatorname{Spec}(\mathbb C[x])$ in Zariski topology. Then Vakil claims that:

the nonreduced points of $X$ are precisely the closed points corresponding to the positive integers

I am not very convinced of this claim. "The closed points corresponding to the positive integers" are prime ideals like $(x-n,y_1,y_2,\ldots)/I$ where $n$ is a positive integer. They are clearly nonreduced rings.

However, $(x-0.5,y_1,y_2,\ldots)/I$ is also a nonreduced ring and it corresponds to "0.5". So I guess my mistake must be that I misinterpreted his meaning. So what is the correct way to interpret and verify this?

Michael Hardy
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No One
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    As I understand it, a "nonreduced point" is one whose local ring on $X$ is not reduced. Localizing at $(x-0.5, y_1,\dots)/I$ makes $x-n$ invertible for each integer $n$, so all the $y_i$'s vanish in the local ring. – Laurent Moret-Bailly Apr 15 '18 at 12:39

1 Answers1

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Let $P=(x-n, y_1, \cdots)\subset A$, it is a prime ideal containing $I$. To localize $A/I$ at $P$ is to make anything not in $P$ invertible. So $x-m, \forall m\ne n$ becomes invertible now, hence $y_m$ becomes $0$ in the localization (because $(x-m)y_m=0$ in $A/I$). The only non-zero $y$ is $y_n$, and its square vanishes. This shows that the localization has one nilpotent: the localized image of $y_n$.

If we take instead $P=(x-0.5, y_1, ...)$, then all $x-n$ become invertible in the localization and all $y_n$ become $0$: you have no nilpotent any more.

Xipan Xiao
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