Dimension of the affine variety associated to $\langle zw-y^2, xy-z^3 \rangle $ 
Find the dimension of the affine variety $V(I)$, where $I=\left\langle zw-y^2,xy-z^3\right\rangle \subseteq k[x,y,z,w]$, with $k$  algebraicaly closed field.

I tried to solve the system $zw-y^2=0$, $xy-z^3=0$ and find the maximal dimension of coordinate subspaces, but I did not succeed. Could anyone help me? Thanks!
 A: Let's prove that $k[x,y,z,w]/I$ is a domain and find its dimension by considering the trascendence degree of its fraction field.
First of all I introduce two easy facts on dimension of $k$-algebras:
Theorem Let $R$ be domain and $S \subset R$ a multiplicative subset. Then $K(R)=K(S^{-1}R)$, where $K( - )$ are the respective fraction fields
Corollary Let $R$ be a finitely generated $k$-algebra and a domain. Let $S \subset R$ a multiplicative subset. Then $trdeg_k \, K(R)=trdeg_k \,K(S^{-1}R)$
We are ready to show that $\dim V(I)=2$:


*

*If we call $A=\frac{k[x,y,z]}{(xy-z^3)}$ we obtain $$k[x,y,z,w]/I \simeq \frac{A[w]}{(zw-y^2)}$$

*$xy-z^3$ is irreducible as element of $k[x,y,z]$ so $A$ is a domain. Moreover $z$ is integral over $A$ and so $\dim A= \dim k[x,y] =2$

*The ring $\frac{K(A)[w]}{(zw-y^2)}$ is a domain because is isomorphic to $K(A)$ and $\frac{A[w]}{(zw-y^2)}$ is one of its subrings, so it is a domain.

*Let's consider now the ring $A_z$ (the ring $A$ in wich we have inverted all the powers of $z$), this is a domain (localization of a domain is also a domain) and $\dim A_z = 2$ (by properties of localization). 
We obtain easy $$A_z\simeq \frac{A_z[w]}{(w-y^2z^{-1})} \simeq \left(\frac{A[w]}{(zw-y^2)}\right)_z$$


Now let's conclude considering $$trdeg_k \,K\left(\frac{A[w]_z}{(zw-y^2)_z}\right)=trdeg_k \,K\left(\frac{A[w]}{(zw-y^2)}\right)= trdeg_k \, K(A_z) = \dim A_z = 2$$
A: A computational approach. 
Set $R=k[x,y,z,w]$ and $I=\langle zw-y^2,xy-z^3\rangle$.


*

*Show that $zw-y^2,xy-z^3$ is a Grobner basis for $I$ with respect to the grevlex order. Thus the initial ideal $\langle{\rm LT}(I)\rangle$ of $I$ is generated by $y^2,z^3$. 

*As it is well known (see e.g. Cox et al., Ideals, Varieties, and Algorithms, Chapter 9), if $k$ is an algebraically closed field $$\dim V(I)=\deg{}^aHP_{R/I},$$ where ${}^aHP_{R/I}$ is the affine Hilbert polynomial of $R/I$. 
But ${}^aHP_{R/I}={}^aHP_{R/\langle{\rm LT}(I)\rangle}={}^aHP_{R/\sqrt{\langle{\rm LT}(I)\rangle}}$. In our case $\sqrt{\langle{\rm LT}(I)\rangle}=\langle y,z\rangle$ and thus we get $\dim V(I)=\deg{}^aHP_{R/I}=2$.
