I'm trying to prove that the curve parametrized by $(t^2, t^3, t^4)$ where $t$ lies in the reals is an affine variety. I believe the variety should be $V(z-x^2, y^2-x^3)$. I can get the curve above is a subset of $V$ but I'm struggling to show the other inclusion.
I tried both a direct proof (which I believe won't work nicely because when you try to parametrize your $(a, b, c)\in V$ you'll need to set $t=\sqrt(a)$ which won't work because $\sqrt(a)$ could take on positive or negative values). And I tried the contrapositive (I said if $c \notin C$, our curve, then $c=(t^2, t^3 +a, t^4+b)$ where not both a and b are zero. I then plugged this $c$ into the equations, found that $b=0$ but couldn't find any good reason that $a=0$ as well.
I feel like I've exhausted all options...can someone please offer some insight?
Suppose we have real solution of $z-x^2=y^2-x^3=0$. If $x=0$, then $y=z=0$. Otherwise, set $t=y/x$. Then $y^2-x^3=0$ gives $t^2 x^2=x^3$, so that (as $x\ne0$), $x=t^2$. Then we get $y=tx=t^3$ and $z=x^3=t^4$. (No cube roots!)