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Let $\{Z_i\}_{i=1}^\infty$ be a sequence of smooth irreducible $k$-dimensional submanifolds of $\mathbb{C}\mathbb{P}^n$ which converges to a closed subscheme $Z$ in the sense of the Hilbert scheme of $\mathbb{C}\mathbb{P}^n$. As far as I understand (please correct me if this is wrong) the dimension of $Z$ also equals $k$.

Question. Is it true that any irreducible component of $Z$ is $k$-dimensional?

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  • $\begingroup$ By the going down theorem for flat morphisms, for every irreducible component of the Hilbert scheme, if the general point of the component parameterizes a closed subscheme of pure dimension $k$, then every subscheme parameterized by that component has pure dimension $k$. $\endgroup$
    – Jason Starr
    Commented Mar 11, 2017 at 10:26
  • $\begingroup$ @JasonStarr: I am not an algebraic geometer, so let me make the terminology more precise: when one says that a subscheme has pure dimension $k$, does it mean that every irreducible component of it is $k$-dimensional? $\endgroup$
    – asv
    Commented Mar 11, 2017 at 10:41
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    $\begingroup$ Yes, that is what it means to say that a scheme has pure dimension $k$. $\endgroup$
    – Jason Starr
    Commented Mar 11, 2017 at 11:04
  • $\begingroup$ @JasonStarr: Thanks very much. Then the answer to my question seems to be 'yes' unless I missed something. Why would not you write it as a final answer? $\endgroup$
    – asv
    Commented Mar 11, 2017 at 13:28

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