Noether Normalization of $\mathbb{C}$-algebra $\mathbb C[x,y,z]/(xy+z^2,x^2y−xy^3+z^4−1)$.

Question

I'm trying to find a Noether normalization of the $\mathbb{C}$-algebra $$\mathbb C[x,y,z]/(xy+z^2,x^2y−xy^3+z^4−1).$$ The proof of Noether Normalization Theorem is not constructive. I choose $x = {x_1} - {y_1},$ $y = {x_1} + {y _1}$ and $z = {z_1}$ but I failed. How I can find it?

Answer 1

Let $A = \mathbb C[x,y,z]/(xy+z^2, x^2y-xy^3 +z^4-1)$.

Substituting the first relator into the second, we get $$x^2y - xy^3 + x^2y^2 - 1 = 0$$ or rearranging $$f(x,y) = xy^3 - x^2y^2 - x^2y + 1 = 0.$$

One possible way of proving Noether normalisation asks us to now find an integer $a$ such that whenever we have a term $x^iy^j$ in $f$, the values $ai + j$ are all distinct. For example, we can take $a = 3$. We then consider the expression $$g(X,Y) = (X+Y^3)Y^3 - (X+Y^3)^2Y^2 - (X+Y^3)^2Y + 1,$$ noting that $g(x - y^3, y) = f(x,y) = 0$.

If we expand $g$ out we get $$g(X, Y) = -Y^8-Y^7+Y^6-2 X Y^5-2 X Y^4+X Y^3-X^2 Y^2-X^2 Y+1$$ Therefore $y$ is integral over $\mathbb C[x']$ where $x' = x-y^3$, since $g(x', Y)$ is a monic poly satisfied by $y$. Also $x$ is integral over $\mathbb C[x']$, since $x = x' + y^3$. Therefore $\mathbb C[x'] \subseteq \mathbb C[x,y]$ is an integral extension, and $\mathbb C[x,y] \subseteq \mathbb C[x,y,z]$ is also integral because of the first relator.

We conclude that either $\mathbb C[x']$ is the subring you're looking for, or $\mathbb C$ itself, depending on whether or not $x'$ is algebraic over $\mathbb C$. I can't off the top of my head see an easy way to show this, but hopefully someone else will be able to find one!

Edit: thanks to Moos, who provided a way to finish the argument.

Krull's generalised ideal theorem says if $R$ is Noetherian with $I$ a proper ideal generated by $n$ elements, then the height of any minimal prime over $I$ is at most $n$. This means that if we instead quotient by a minimal prime over $I$, say $P$, then the catenary property gives us that $\dim \mathbb C[x,y,z]/P = \dim \mathbb C[x,y,z] - ht P \geq 1$, so in particular is not just $\mathbb C$.

Therefore $A$ is integral over the polynomial subring $\mathbb C[x']$.

Answer 2

We have $(xy+z^2,x^2y-xy^3+z^4-1)=(xy+z^2,x^2y-xy^3+x^2y^2-1)$, i.e. your algebra is isomorphic to $R[\sqrt{xy}]$ with $R=\mathbb C[x,y]/(x^2y-xy^3+x^2y^2-1)$.

Any noether normalization of $R$ is also a noether normalization of $R[\sqrt{xy}]$, so we should find one of $R$.

To that account, let $x=-X$ and $y=X+Y$. We get

$$x^2y-xy^3+x^2y^2-1=X^2(X+Y)+X(X+Y)^3+X^2(X+Y)^2-1=2X^4 + \text{lower terms in } X$$

Hence $X$ is integral over $\mathbb C[Y]$ and we are done: A noether normalization is

$$\mathbb C[y+x] \hookrightarrow R.$$