# What is the Chow group of projective irreducible singular curve?

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 ?

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 $$\mathbb Z\stackrel \sim \to A_0(C):n\mapsto n\cdot[P]$$