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(Source: Ravi Vakil FOAG Exercise 5.5.H)

Suppose $X = \operatorname{Spec} \mathbb C[x,y]/I$, and that the (only, added by myself) associated points of X are $[(y - x^2)], [(x - 1; y - 1)]$, and $[(x - 2; y - 2)]$. The question asks us to sketch $X$ as a subset of $\mathbb A^2_{\mathbb C}$ including fuzz.

By "fuzz" he means "points where the stalks are non-reduced". So the only task is to find the nonreduced locus of $X$, which, by his previous exercises is the closure of the union of non-reduced associated points. Now it suffices to find all non-reduced associated points.

It is clear from here that $[(x - 1; y - 1)]$ is the only embedded point. I guess (but not sure) $[(x - 1; y - 1)]$ is the only place where $X$ has a non-reduced stalk (fuzz) but I don't know how to prove this (or in general whether embedded points are non-reduced). If my guess is wrong, please let me know.

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  1. The locus with non-reduced stalk is given by the closure of the associated points whose stalks are non-reduced. So with only the information you're given here, you cannot deduce that $[(x-1,y-1)]$ is the only place where $X$ has non-reduced stalks. For example, the whole component on $y=x^2$ might be non-reduced, or that the point $(2,2)$ might also be non-reduced.

  2. The stalks at embedded points are non-reduced. To see this, let's suppose that we are working on an irreducible affine $\mathrm{Spec} A$. Then $Nilrad(A)$ is a prime ideal and we have that its closure is equal to $\mathrm{Spec}A$. Let $\mathfrak{p}$ be your embedded point. By assumption, there exists an element $a\in A$ such that the closure of $\mathfrak{p}$ is an irreducible component of $Supp(a)$. Since $\mathfrak{p}$ is an embedded point, this is not equal to $\mathrm{Spec}(A)$. It follows that we must have $a_{Nilrad(A)} = 0 \in A_{Nilrad(A)}$. This implies that there exists $b\in A-Nilrad(A)$ such that $ab=0$. But this implies that $a\in Nilrad(A)$. So we see that $A_{\mathfrak{p}}$ is non-reduced.

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