Let $A=\Bbb C[x,y,z,t]/I$ where $I=(x^2 +xy-z,xy^2-t^2+z,x^2y+xy^2+t^2)$.
Find the Noether Normalization of A.
My solution:
We are looking for new variables $x_1,x_2,x_3$. I found:
$x\mapsto x_1$
$y\mapsto -x_1$
$t\mapsto x_2$
$z\mapsto x_3$
New variables are independent - important.
Then, our generators of ideal go to $(x_3,-x_1^3-x_2^2+x_3,x_2^2)$, so vanishing set of them is set $\{ z=0, t=0, (xy=0)\}$. So under that coordinate change, ring $A$ goes to $\Bbb C[x,y]/(xy)$, which is a direct sum of $\Bbb C[x]$ and $\Bbb C[y]$.
So, is the Noether Normalization of $A$, $\Bbb C[x,y]/(xy)$???
Thanks in advance for help.
As discussed in the comments, the material in the post is not a solution. Here's an approach which does produce a solution.
Step 1: We can rewrite $\Bbb C[x,y,z,t]/(x^2+xy-z,xy^2-t^2+z,x^2y+xy^2+t^2)$ as a ring with no $z$ by replacing $z=x^2+xy$ everywhere it appears. Specifically, $\Bbb C[x,y,t]/(xy^2+x^2+xy-t^2,x^2y+xy^2+t^2)$ is isomorphic to our original ring by the map which sends $x,y,t$ to themselves and $z\mapsto x^2+xy$.
Step 2: We can rewrite the ideal $(xy^2+x^2+xy-t^2,x^2y+xy^2+t^2)$ as $(x^2y+2xy^2+x^2+xy,x^2y+xy^2+t^2)$, which shows us that the projection from the variety of this ideal to the curve given by $V(x^2y+2xy^2+x^2+xy)\subset \Bbb A^2$ is 2-to-1 (In algebraic terms, this corresponds to the fact that $\Bbb C[x,y,t]/(x^2y+2xy^2+x^2+xy,x^2y+xy^2+t^2)$ is a finite $\Bbb C[x,y]/(x^2y+2xy^2+x^2+xy)$ module with basis $\{1,t\}$.) So all we need to do is find a Noether normalization of this and we're in business.
Step 3: the equation $x^2y+2xy^2+x^2+xy$ factors as $x(2y^2+xy+y+x)$, so the variety it cuts out is the union of the $y$-axis and a conic. We can see easily from the equation of this conic that the projection on to the $y$-axis is 2-to-1, using the same reasoning as in step 2. The projection of the $y$-axis to the $y$-axis is the identity, so the projection from the curve $V(x^2y+2xy^2+x^2+xy)$ to the $y$-axis is finite.
Now we put it all together. By composing steps 1, 2, and 3, we get a map from our original curve on to $\Bbb A^1$ which is finite: it's the composition of an isomorphism and two finite maps. Tracing what this means on coordinate algebras, we get that the map $\Bbb C[y]\to \Bbb C[x,y,z,t]/I$ by $y\mapsto y$ is a Noether normalization of this ring.