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We say that two subschemes are rationally equivalent if their associated cycles are rationally equivalent. Two cycles are rationally equivalent if their difference is in $Rat(X)$. If $X$ is a scheme, we define $Z(X)$ the group generated by the reduced irreducible subschemes of $X$. I denote with $Rat(X) \subset Z(X)$ the subgroup generated by differences of the form $$ \langle \Phi \cap (\{t_0\} \times X) \rangle - \langle \Phi \cap (\{t_1\} \times X) \rangle ,$$ where $t_0,t_1 \in \mathbb{P}^1$ and $\Phi$ is a subvariety of $\mathbb{P}^1 \times X$ not contained in any fiber ${t} \times X$ (I took this definition from Intersection theory in algebraic geometry of Harris). Now I have to prove rigorously that in $\mathbb{P}^2$ an hyperbola and the union of two lines are rationally equivalent. How can I do it? Thanks in advance.

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    $\begingroup$ This is the same question as that one. Reposting a question under another username is bad étiquette. $\endgroup$ – Georges Elencwajg Dec 2 '13 at 12:49
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Hint: Take the family of hyperbolae (I checked the plural of hyperbola and apparently hyperbolas works as well) $$p(t)=x^2-y^2-tz^2=0$$ for $t\in\mathbb{A}^1$. For $t\neq 0$, $p(t)$ is an actual hyperbola, but for $t=0$ we get the lines $x=y$ and $x=-y$.

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