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Let $X,Y$ be two projective variety over an algebraically closed field, then we know $Hom(X,Y)$ has a scheme structure by considering the embedding as a open subscheme of $Hilb(X\times Y)$ using the graph.

We know for each fixed polynomial $P$, the Hilbert scheme with Hilbert polynomial $P$ is projective hence finite type. And it seems that the graph of any $f:X\longrightarrow Y$ is isomorphic to $X$ hence has the same Hilbert polynomial viewed as a closed subscheme of $X \times Y$. However, Hom scheme can have infinitely many component (for example consider $End(E)$ where $E$ is an elliptic curve). So where does my intuition goes wrong? If $X,Y$ are both smooth curves, can we decide all of the connected components?

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    $\begingroup$ Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $\mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $\mathbb{P}^2$. However, you could have also embedded $X$ into $\mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $\mathbb{P}^2$. So, subvarieties of $\mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree. $\endgroup$
    – Ariyan Javanpeykar
    Jan 20, 2019 at 15:05

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To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $X \times Y$ is $L_X \boxtimes L_Y$, where $L_X$ and $L_Y$ are line bundles on $X$ and $Y$. Then if $f \colon X \to Y$ is a morphism, the restriction of $L_X \boxtimes L_Y$ to the graph $\Gamma(f)$ of $f$ corresponds to the line bundle $L_X \otimes f^*L_Y$ on $X$, hence the Hilbert polynomial of $\Gamma(f)$ equals the Hilbert polynomial of $X$ with respect to $L_X \otimes f^*L_Y$. In particular, it does depend on $f$.

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  • $\begingroup$ I forget the same line bundle can be different after restriction, thank you very much! $\endgroup$
    – user395911
    Jan 20, 2019 at 15:35

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