Noether Normalization lemma for $k[x,y]\times k[z]$ 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?
 A: Geometrically, what Noether normalization says is that for any affine variety of finite type over a field $k$, you can find a (surjective) projection to an affine space that turns $\operatorname{Spec}A$ into a finite branched covering of the affine space.
Also, geometrically, $k[x,y]\times k[z]$ is the disjoint union of affine two space with the affine line. Thus you are correct to expect that we are looking for two elements. Really what we want though is a projection to affine space. We should probably take this projection to be the identity on $\operatorname{Spec}k[x,y]$, and take it to be the inclusion of the affine line into the plane as the $x$-axis on $\operatorname{Spec}k[z]$.
This leads us to take the elements $u_1=(x,z)$, $u_2=(y,0)$ as our candidates for the elements we need for Noether normalization.
Now if $\alpha=(p(x,y),q(z))\in k[x,y]\times k[z]$, then $(\alpha - q(u_1))(\alpha -p(u_1,u_2))=0$, so all $\alpha\in k[x,y]\times k[z]$ are integral over $k[u_1,u_2]$, as desired.
