Show the ideal $I=(x^{2}-y,z-1)$ is prime in $K[x,y,z]$ 
I am tasked to show that the ideal $I=(x^{2}-y,z-1)\subset K[x,y,z]$ is it's own radical where $K$ is an algebraically closed field. 

I tried to proceed in the obvious fashion. 
Let $\varphi:K[x,y,z]\rightarrow K[x]$ be evaluation $f(x,y,z)\mapsto f(x,x^{2},1)$. It is trivial to show that $I\subset\ker\varphi$ but the other inclusion is not so trivial. This is what I've tried. Write $f(x,y,z)$ as a polynomial in $z$, $f(x,y,z)=\sum_{i=0}^{N}f_{i}(x,y)z^{i}$. Then $f(x,x^{2},1)=\sum_{i=0}^{N}f_{i}(x,x^{2})=0$. From here I am stuck because we have no idea how to show each $f_{i}(x,y)$ is divisible by $x^{2}-y$.
Another approach that I tried to take was to show that $I$ is its own radical by element argument but that failed miserably.  
I would like to show this without the use of Groebner basis or multinomial long division. The reason being this is one of the exercises in Hartshorne and he doesn't touch on those things at all. This is not homework to be graded just cultural enrichment. Thanks in advance. 
 A: Quotient out by $I$ precisely means that you can replace $z$ by $1$ and $y$ by $x^2$, so the inclusion $k[x] \subset k[x,y,z]$ induces an isomorphism
$k[x] \cong k[x,y,z]/I$.
A: The ideal is prime if and only if $K[x,y,z]/I$ is a domain. We have $$ \frac{K[x,y,z]}{(x^2-y, z-1) } \cong \frac{ K[x,y] }{(y-x^2)}.$$ You will be done if you show that $y-x^2\in K[x,y]$ is irreducible.

Here is a proof of the isomorphism I claimed above. 
Define $$\phi : K[x,y,z] \to \frac{K[x,y]}{(y-x^2)}$$ by the mapping $p(x,y,z) \mapsto p(x,y,1) + (y-x^2).$ By direct substitution, $z-1$ and $y-x^2$ are in $\ker \phi$ so $(z-1, y-x^2) \subseteq \ker \phi.$ 
Conversely, suppose $p(x,y,z) \in \ker \phi.$ Then $p(x,y,1) \in (y-x^2).$ To be explicit, $p(x,y,1) = g(x,y)(y-x^2)$ for some $g(x,y)\in K[x,y].$ 
Writing out $p(x,y,z) = \sum a_i b_j c_k x^i y^j z^k$ we have 
$$ p(x,y,z) - p(x,y,1) = \sum a_i b_j c_k x^i y^j (z^k -1)$$
and since $z^k-1 = (z-1)(1+z+z^2+\cdots z^{k-1})$ we have $$p(x,y,z) - p(x,y,1) = (z-1)h(x,y,z)$$ for some $h\in K[x,y,z].$ Therefore, $$p(x,y,z) = g(x,y)(y-x^2) + (z-1)h(x,y,z),$$ i.e. $p(x,y,z) \in (y-x^2, z-1).$ Thus, $\ker \phi = (y-x^2, z-1)$ and the First Isomorphism Theorem gives us the isomorphism we are after. 
