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I am confused on the following example. What is $A_0(C)$ is $C$ is an irreducible projective curve ? For example, let $C=V(x^2z-y^3)$ be a cusp curve, let $P=[0:0:1]$ be the cusp. Is $P$ rational equivalent to other points of $C$ ? I have the following argument. Consider the variety $D=V(x^2-y^3, Ax+Bz)$ in $\mathbb{P}^2\times \mathbb{P}^1$ with coordinates $[x:y:z]\times [A:B]$, then this is a subvariety of $C\times \mathbb{P}^1$, where $D_0=3[1:0:0]$ is a smooth point and $D_\infty=P$. Is this right ?

I know that the zero Chow group of union of two $\mathbb{P}^1$ is $\mathbb{Z}$, but how about the zero Chow group of a nodal curve ?

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Yes, $A_0(C)=\mathbb Z\cdot[P]\stackrel \sim \to \mathbb Z$.
To see this consider the normalization $$n:\mathbb P^1=\tilde C\to C:(u:v)\mapsto (x:y:z)=(u^2v:u^3:v^3)$$ which is a regular bijection but not an isomorphism.
The point $\tilde P=(0:1)\in \tilde C$ has as image the singularity of the cusp $n(\tilde P)=P \in C$.
The other points $\tilde Q\in \tilde C$ have images $n(\tilde Q)=Q\in C$ which run through all smooth points $Q\in C$.
Since $\tilde P$ is rationally equivalent to $\tilde Q$, we deduce that $ P$ is rationally equivalent to $Q$, because rational equivalence is preserved by the proper morphism $n$.
Hence we have $A_0(C)=\mathbb Z \cdot[P]$.
Finally we remark that $[P]$ has no torsion because it has degree $1$ under the group morphism $\operatorname {deg}:A_0(C)\to \mathbb Z$. In conclusion we obtain the desired isomorphism of groups $$\mathbb Z\stackrel \sim \to A_0(C):n\mapsto n\cdot[P]$$

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