Two rational equivalent closed subschemes in $\mathbb{P}^n$ have the same Hilbert polynomial?

Question

Let $X$ and $Y$ be two closed subschemes in projective space $\mathbb{P}^n$. I wonder if $X$ and $Y$ are rationally equivalent, must it be true that the Hilbert polynomials of $X$ and $Y$ are the same? For some simple situations, it is true. For instance, if both $X$ and $Y$ are hypersurfaces, then they have the same degree since they are rationally equivalent. Therefore, they have the same Hilbert polynomials. Furthermore, if both $X$ and $Y$ are complete intersection of $r$ hypersurfaces $f_1,\cdots,f_r$ of degree $d_1\leq d_2\leq\cdots\leq d_r$, then they also have the same Hilbert polynomials.

So is this true for the general case? If it is not true, can someone give me a counterexample? Any help is appreciated.

Answer 1

No, this is not true in general. Recall that a curve of degree $d$ and genus $g$ in (any) $\Bbb P^n$ has Hilbert polynomial $P(t)=dt+1-g$ by Riemann-Roch, and that any two varieties of the same dimension and degree are rationally equivalent inside $\Bbb P^n$ since they're both rationally equivalent to $d$ times a linear space of the appropriate dimension. So all we have to do here to demonstrate a counterexample is to find two curves of the same degree but different genera inside a common $\Bbb P^n$.

For one example, take the twisted cubic $C\subset\Bbb P^3$ and an elliptic curve $E\subset\Bbb P^2\subset\Bbb P^3$, where the $\Bbb P^2$ is included as a linear subspace. Both have degree three, but one is rational and the other isn't.