An exercise in Fulton's book Algebraic curves (2.8) asks to prove that the algebraic set in ${\mathbb C}^3$ defined by the ideal $I=(xz -y^2,yz-x^3,x^2y-z^2)$ (meaning $V=\{P\in {\mathbb C}^3: F(P) \mbox{ for all } F\in I\}$) is irreducible. I did that by first defining $J$ to be the radical ideal of $I$, that is $J=\{F\in {\mathbb C}[x,y,z]: F^n \in I \mbox{ for some } n\}$, so that $V$ is also defined by $J$, and then defining the polynomial map $\varphi: {\mathbb C} \longrightarrow {\mathbb C}^3$, $\varphi(a) =(a^3,a^4,a^5)$ onto $V$, from which it follows that the induced map on coordinate rings $\hat{\varphi}: \Gamma(V)={\mathbb C}[x,y,z]/J \longrightarrow \Gamma({\mathbb C}) \eqsim {\mathbb C}[x]$, $\hat{\varphi}(F)(x) = F(x^3,x^4,x^5)$ is one-to-one, and so $\Gamma(V)$ is isomorphic to a subring of ${\mathbb C}[x]$, that is a domain, from which it follows that the ideal corresponding to $V$ (which by the Hilbert Nullstellensatz is $J$) is prime, and therefore $V$ is irreducible.
But suppose I also want to prove that the ideal $I$ is prime. It seems to me that the above argument is useless because an algebraic sets does not see the difference between an ideal and its radical. Is there some other argument involving algebraic geometry to prove that $I$ is prime?