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First of all, i'm sorry for my bad english. I'm from a foreign country. :-)

I have some questions about a paragraph appearing in page : $205$ of the following electronic textbook : https://scholar.harvard.edu/files/joeharris/files/000-final-3264.pdf

  • The paragraph says :

Let $S = K[x_0 , \dots , x_n]$ be the homogeneous coordinates ring of : $ X = \mathbb{P}^n $.

The Hilbert scheme $ \mathcal{H}_P (X) $ is constructed as a subscheme of the Grassmannian of $ P(d) $ - dimentional subspaces of $S_d$, the space of homogeneous forms of degree $ d $, for suitably large $d$.

  • Question :

I don't understand properly this definition above of $ \mathcal{H}_P (X) $, for $ X = \mathbb{P}^n $. if we believe in this definition, does it-mean that $ \mathcal{H}_P (X) $ is possibly constructed on the one hand as a subscheme of the Grassmannian of $ P(d_1 ) $ - dimentional subspaces of $ S_{d_{1}} $, the space of homogeneous forms of degree $ d_1 $, for suitably large $d_1$, and in the same time, as a subscheme of the Grassmannian of $ P(d_{2}) $ - dimentional subspaces of $ S_{d_{2}} $, the space of homogeneous forms of degree $ d_{2} $, for suitably large $d_{2}$, such that : $ d_1 \neq d_2 $ ?

Thanks in advance for your help.

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  • $\begingroup$ What is the problem? The same scheme can be constructed as a subscheme of two different schemes. $\endgroup$ – danneks Apr 21 '18 at 10:27
  • $\begingroup$ If we assume that $ \mathcal{H}_P (\mathbb{P}^n ) \subset G(P(d_1 ) , S_{d_{1}} ) $ and $ \mathcal{H}_P (\mathbb{P}^n ) \subset G(P(d_2 ) , S_{d_{2}} ) $, and since : $ G(P(d_1 ) , S_{d_{1}} ) \cap G(P(d_2 ) , S_{d_{2}} ) = \emptyset $, so $ \mathcal{H}_P ( \mathbb{P}^n ) $ doesn't exist in this case, no ? $\endgroup$ – YoYo Apr 21 '18 at 11:33
  • $\begingroup$ The Hilbert scheme is defined uniquely only up to isomorphism, as is usual with universal properties. It can be embeddded in different schemes. And it does not make sense to talk about intersection of two schemes, if they are not considered as subschemes of a third scheme. $\endgroup$ – danneks Apr 23 '18 at 5:53

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