# Rational equivalence and Hilbert flag scheme

Given a smooth surface $X \subset \mathbb{P}^3$ if we have two curves $C_1, C_2$ that are rationally equivalent is it true that both $(C_1,X)$ and $(C_2,X)$ will be in the same irreducible component of the Hilbert flag scheme $Hilb_{P,Q}$ where $C_i \in Hilb_P$ and $X \in Hilb_Q$? Intuitively, I would think this to be true since $C_1 \sim_{rat} C_2$ implies that there exists a flat family of curves parametrized by $\mathbb{P}^1$ such that $C_i$ is one of the fibers. But $\mathbb{P}^1$ is irreducible.

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It sounds to me like you are answering your own question. Certainly your argument proves that there exists an irreducible component of the flag Hilbert scheme containing both $(C_1,X)$ and $(C_2,X)$. If there is a second (or third, ...) irreducible component containing $(C_1,X)$, say, it does not necessarily follow that this irreducible component also contains $(C_2,X)$.