How to prove $ \left\{ t^2,t^3 \right\}$ equals the vanishing set of $y^2-x^3$? Exercise 3.2 in Hartshorne is about proving that morphisms of varieties may be underlain by homeomorphisms without being isomorphisms of varieties. The morphism in consideration is $\varphi:t\mapsto (t^2,t^3)$ where the image is to be thought of as the curve $y^2-x^3$.
I don't understand over which fields $\Bbbk$ the set $ \left\{ t^2,t^3 \right\}$ equals the vanishing set  of $y^2-x^3\in \Bbbk[x,y]$ - only why it's contained in the vanishing set. Indeed if $(a,b)$ satisfies $a^3=b^2$ we at least need the existence of square/cube roots, since we want $t$ such that $t^2=a,t^3=b$. Suppose for convenience the field is algebraically closed. Then we have some creature worthy of the name $\sqrt a$ that satisfies $\sqrt a^2=a$ and by assumption $\sqrt a^6=b^2$. But why should we have $\sqrt a^3=b$? (more accurately, why can we choose $\sqrt a$ to have this property?)
Maybe this is elementary field/Galois theory, but better late than never.

Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.
Otherwise $x\ne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.
Not a square or cube root in sight!
