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Let $G$ be a finite subgroup of $\text{SL}(2,\mathbb{C})$ and let $Y=\text{GHilb}(\mathbb{C}^2)$ be the minimal resolution of $X=\mathbb{C}^2/G$ where $\text{GHilb}(\mathbb{C}^2)$ is the Nakamura $G$-Hilbert scheme. According to Bridgeland-King-Reid there is a universal closed subscheme $\mathfrak{Z} \subset Y \times \mathbb{C}^2$ which makes a certain diagram commutative. What is really known about $\mathfrak{Z},$ for example

Is $\mathfrak{Z}$ reduced?

I was thinking, if one can prove that $Z \subset \text{Hilb}^{|G|}(\mathbb{C}^2) \times \mathbb{C}^2$ where $Z$ is the universal closed subscheme associated to the Hilbert scheme of $|G|$ points on $\mathbb{C}^2,$ is reduced, because $Y \times \mathbb{C}^2 \hookrightarrow \text{Hilb}^{|G|}(\mathbb{C}^2) \times \mathbb{C}^2$ we would then know that $\mathfrak{Z}$ is reduced. So does this hold?

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$Z$ is certainly reduced. In fact, it is smooth at any point lying over a point in the Hilbert scheme corresponding to a collection of distinct points, so it is generically smooth (and generically reduced). But it is also flat over the Hilbert scheme, so there are no embedded components, and it is reduced. (In fact, I believe it is actually smooth, but this is slightly more subtle.) I don't see, however, why knowing $Z$ is reduced gives what you want; reduced schemes have lots of nonreduced subschemes. –  Jack Huizenga Mar 7 '13 at 7:02
    
Dear Jack, why can't one argue similarly for $\mathfrak{Z}?$ As for the last statement, I thought, because $G$-invariant part of a reduced scheme is reduced, am I wrong? –  Ehsan M. Kermani Mar 7 '13 at 8:35
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The $G$-invariant part of a reduced scheme is not reduced in general: take $\mathbb{Z}/2\mathbb{Z}$ acting on the curve $xy=0$ by exchanging coordinates. –  Laurent Moret-Bailly Mar 7 '13 at 8:55
    
Fair enough. @Laurent Moret-Bailly, what condition(s) implies that the $G$-invariant part is reduce, then? –  Ehsan M. Kermani Mar 9 '13 at 17:48

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