Why is $(z^2-x^2y, y^2-xz, x^3-yz)$ a prime ideal? How can I prove that the ideal $I = (z^2-x^2y, y^2-xz, x^3-yz)$ is a prime ideal of $ K[x,y,z]$. I want to construct a morphism $\phi:K[x,y,z] \rightarrow K[t,s]$ whose kernel is equal to $I$ but from the picture I just don't see a parametrization or how to do it.
 A: I think an explanation of where lhf got those numbers would be helpful. I am going to create a matrix in which each row corresponds to one of the generating binomials. Each column will be associated to $x,y,$ and $z$ respectively. An exponent before the minus sign will result in a positive entry and an exponent after the minus sign will correspond to a negative entry.
For instance: $z^2 - x^2y$ will correspond to the row $\begin{pmatrix}-2 & -1 & 2 \end{pmatrix}$.
So the matrix we get from this ideal is:
$$
A = \begin{pmatrix}
-2 & -1 & 2 \\
-1 & 2 & -1 \\
3 & -1 & -1
\end{pmatrix}_.$$
Now if we define a morhpism $\varphi : K[x,y,z] \to K[t]$ by $\varphi(x) = t^a$, $\varphi(y) = t^b$, and $\varphi(z) = t^c$; then $I$ will be the kernel precisely when $$\begin{pmatrix} a \\ b \\ c \end{pmatrix}$$ is in the kernel of $A$.
It happens that $\ker(A)$ is spanned by $$\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}_.$$
Disclaimer: When I said "precisely when", that statement takes a bit of proof. As lhf pointed out, one inclusion is obvious. The other might take a bit of work to convince yourself.
A: Try $\phi: K[x,y,z] \rightarrow K[t]$ with $\phi(x)=t^3$, $\phi(y)=t^4$, $\phi(z)=t^5$.
It is immediate that $I \subseteq \ker \phi$. I haven't checked the reverse inclusion.
