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Let $Z =C \cup F$ be a disjoint scheme union of closed subschemes of Pn.

Let $p_C$ be the Hilbert polynomial; assume

(1) $F$ finite, reduced,

(2) all irreducible components of $C$ are positive dimensional

Q. does it follow that $C$ is the sole subscheme of $Z$ with same Hilbert polynomial $p_C$?

OK if $C$ is integral: indeed, if $C'$ is a subscheme of $Z$ with $p_{C'} = p_C$, we get

$C'_{red} = C \cup F'$, disjoint union, with $F'$ finite, hence

$p_{C'_red }=p_C+deg F'\leq p_{C'}=p_C$,

thus $F'$ is empty and $C'=C$.

I wish it were true say for $C$ a local complete intersection.

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Equivalently, does $C$ have a subscheme $C'$ such that $p_C - p_{C'}$ is finite?

$C$ satisfies Serre's condition S1 iff $C$ is the closure of the union of the generic points of its geometric components ("no embedded components"). If $C'$ misses one of those generic points, $p_C - p_{C'}$ will grow like the Hilbert polynomial of the missing component, so by your positive-dimensionality condition the difference wouldn't be finite. One place to read about S1 is the book [Eisenbud], which includes the exercise "Show $X$ reduced $\iff$ it is R0 (generically reduced) + S1."

Local complete intersection $\implies$ Cohen-Macaulay $\implies$ S1, so you're good.

For a non-example, let $C = \{[x,y,z] : x^2 = xy = 0\}$, $C' = \{[x,y,z] : x = 0\}$.

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  • $\begingroup$ $C'$ is assumed to be a subscheme of $Z$, not necessarily of $C$ proper. Of course if this were so, the equality of hilbpols would imply $C=C'$ at once. The difficulty is how to control possible embedded points of $C'$. $\endgroup$
    – ivainsencher
    Commented Aug 29, 2013 at 15:39
  • $\begingroup$ Sorry, you didn't use $C'$ in your question, and I didn't notice you were using it in your "OK if $C$ is integral" section. My $C'$ is not your $C'$. Since you assume $Z$ and $F$ disjoint, your question is equivalent to "Does $C$ have a subscheme only finitely much smaller?" to which one then adds points from $F$. $\endgroup$
    – Allen Knutson
    Commented Sep 1, 2013 at 17:19
  • $\begingroup$ I think that mod a typo you've clarified it all. $\endgroup$
    – ivainsencher
    Commented Sep 2, 2013 at 20:34

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