Noether normalization lemma states that for every finitely-generated algebra $A$ over a field $k$ it's possible to find a set of algebraically independent elements $y_1 \cdots y_n \in A$ such that $A$ is integral over $k[y_1 \cdots y_n]$
I am asked to show it explicitly for $A =k[x,y]\times k[z]$. Which $y_1,y_2$ (I think there are two of them) we can take and why are they the ones needed?
I partly understand how to find the set of algebraically independent elements if we have the factor of a polynomial ring in some polynomials, but I don't see how this can be interpreted this way. It's not just $k[x,y,z]/(xz,yz)$ because there are two constant terms in $A$ but only one in $k[x,y,z]/(xz,yz)$.
Les us have a plane $z=0$ and a line $y=0, z=1$. The ring of functions of a disjoint set is the direct sum of two function rings. So we get an ideal $(yz, z(z-1))$
UPD: I take $k[x,y,z]/(yz, z(z-1))$ which is the same as $k[x,y]\times k[z]$
Now I use $a = y - z$ It leads to $k[x,y,z]/(a+z)z, z(z-1) $ which is integral over $k[x,a]$. So $y_1 = x, y_2=y-z$.
Is it true that we can say that $k[x,y,z]/(yz,z(z-1)$ is integral over $k[x,y]$ because $z^2-z$ is the polynomial making $z$ integral. But this means that we can take $y_1=x,y_2=y$. Is this reasoning correct?