Let X be a smooth projective variety. We consider the Hilbert scheme X_[n] of points on X. Denote Z the uiversal subscheme in X×X_[n]. We know that Z|(X×ζ)=ζ, where ζ belongs to X[n]. But does the ideal sheaf I_Z have the same universal property, i.e. I_Z|_(X×ζ)=I_ζ?
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Yes. The point is that the structure sheaf of $Z$ is flat over $X[n]$, so the formation of its sheaf of ideals commutes with base change on $X[n]$.
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$\begingroup$ Sorry, I can't understand. Could you give the details? Why 0→I_Z→O_(X×X_[n])→O_Z→0 restrict to ζ is still exact? Why we don't use subsheaves to construct Quot scheme? $\endgroup$– MessiCommented Sep 1, 2010 at 13:35
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$\begingroup$ Restricting to $X \times \zeta$ means tensoring with the structure sheaf of $X \times \zeta$; because of the symmetry of the Tor_1, the exact sequence above stays exact. We don't use subsheaves precisely because they don't pull back to subsheaves, so they don't define a functor. $\endgroup$– AngeloCommented Sep 1, 2010 at 16:36