Determining the Zariski closure of some points

Question

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.

Answer 1

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 $(\alpha,\beta,\gamma)=((-r)^2,(-r)^3,(-r)^4)$.