I do not know whether my logic is correct. I just want to have some confirmation as I do not want to talk nonsense.
I think $CP^1$ can be embedded into $CP^2$ as a curve. Let $[x,y,z]\in CP^2$. Choose $z=1$. Then we have $[\frac{x}{z},\frac{y}{z},1]$. Now choose $y$ fixed, we have a copy of $CP^1$. Now to see $CP^1$ as an algebraic curve in $CP^2$, it suffices to get a homogeneous degree $d$ polynomial. So we just choose $y=0$ as the above. So $CP^1=V(y,z-1)$. Is my reasoning correct? I think $CP^1$ is just a curve parametrized by $\frac{x}{z}$ but it is not clear to me this variety is a curve.