# Determining the Zariski closure of some points

Question: Determine the Zariski closure in $$\mathbb{C}^3$$ of the points on the curve $$\{(a^2, a^3, a^4) : a \in \mathbb{C}\}$$.

My work: If I denote the set of these points by $$A$$, then I know that the Zariski closure of $$A$$ is the smallest algebraic set containing $$A$$. I also know that the Zariski closure of $$A$$ is $$\mathcal{Z}(\mathcal{I}(A))$$ where $$\mathcal{Z}(\mathcal{I}(A))=\{(a^2, a^3, a^4)\in A:f(a^2, a^3, a^4)=0 \text{ and }f\in\mathbb{C}[x_1,x_2,x_3]\}.$$ If I take polynomials $$f(x_1,x_2,x_3)-x_i$$ where $$i=1,2,3$$ then $$\mathcal{I}(A)=\langle f(x_1,x_2,x_3)-x_1, f(x_1,x_2,x_3)-x_2, f(x_1,x_2,x_3)-x_3 \rangle.$$

The common solutions of the polynomials in $$\mathcal{I}(A)$$ are the points $$(a^2, a^3, a^4)$$ where $$f((a^2, a^3, a^4))-a^i=0$$ for all $$i$$. Can i take now a specific function that $$f(x_1,x_2,x_3)=0$$, and to solve the corresponding system?

Is my approach on the right track.

I claim $$\{(a^2,a^3,a^4):a\in\mathbb C\}=Z(x^2-z,y^2-x^3),$$ so in particular $$\{(a^2,a^3,a^4):a\in\mathbb C\}\subset\mathbb C^3$$ is Zariski closed. Indeed, $$\subseteq$$ is clear.
Conversely, suppose $$(\alpha,\beta,\gamma)\in\mathbb C^3$$ is such that $$\gamma=\alpha^2$$ and $$\alpha^3=\beta^2$$. Let $$\alpha=r^2$$ for some $$r\in\mathbb C$$. Thus $$\beta^2=r^6$$, i.e., $$\beta=\pm r^3$$. If $$\beta=r^3$$ then $$(\alpha,\beta,\gamma)=(r^2,r^3,r^4)$$. Similarly, if $$\beta=-r^3$$ then $$(\alpha,\beta,\gamma)=((-r)^2,(-r)^3,(-r)^4)$$.