[Math] Noether’s normalization lemma in practice (example)

commutative-algebranoetherianpolynomialsproof-verification

I would like to know how to use the Noether's normalization lemma in practice.

Noether's normalization lemma

Let $k$ an infinite field, and $k[a_1,\dots ,a_n]$ be a finite $k$-algebra. There exist an integer $m\in\lbrace 0,\dots ,n\rbrace$ and $\lbrace b_1,\dots ,b_m\rbrace$ such that :

  1. $\lbrace b_1,\dots ,b_m\rbrace$ is algebraically independent over $k$;
  2. $k[a_1,\dots ,a_n]$ is a finite $k[b_1,\dots ,b_m]$-module.

I am working on the example
$$k[X,Y,Z]/\left<XY+YZ+XZ\right>=k[a_1,a_2 ,a_3]$$
with $a_1=\overline{X}$, $a_2=\overline{Y}$ and $a_3=\overline{Z}$. Let the morphism
$$\varphi : k[X,Y,Z]\longrightarrow k[a_1,a_2 ,a_3]$$
defined by $\varphi (X)=a_1$, $\varphi (Y)=a_2$ and $\varphi (Z)=a_3$ (as in the proof).

We have $XY+YZ+XZ\in \ker\varphi$, hence $\ker\varphi \neq \lbrace 0\rbrace $. So as in the induce process of the proof, I note

$$\varphi' : k[X,Y]\longrightarrow k[a_1,a_2]$$
define by $\varphi' (X)=a_1$ and $\varphi' (Y)=a_2$. Let $g\in\ker \varphi '$ so $\overline{g(X,Y)}=\overline{0}=\left<XY+YZ+XZ\right> =\lbrace (XY+YZ+XZ)h(X,Y) \mid h\in k[X,Y]\rbrace$.

If $g\neq 0$, for all $h\in k[X,Y]$,
$$\left\lbrace \begin{array}[ll]
&\mathrm{deg}_Z (XY+YZ+XZ)h(X,Y)=1 & \text{if } h\neq 0 \\
\deg_Z g =0 &
\end{array}\right.$$
And if $h=0$ then $(XY+YZ+XZ)h(X,Y)=0\neq g$. So $g\not\in\overline{0}$ absurd. So $g=0$ and we get that $\ker\varphi '=\lbrace 0\rbrace$.

Conclusion: I take $m=2$ and $b_1=a_1$, $b_2=a_2$ and using the Noether's normalization lemma I get that

  1. $\lbrace a_1, a_2\rbrace$ is algebraically independent over $k$;
  2. $k[a_1,a_2 ,a_3]$ is a finite $k[a_1,a_2]$-module.

Questions

Is my proof correct ? And more importantly is it this way that in practice we use the Noether's normalization lemma ?

Thank you for your help.

Best Answer

I figured out where my mistake is. (I answer my own question in case someone run into the same problem.)

$\ker \varphi \neq\lbrace 0 \rbrace$ is okay. But to continue I need that $a_3$ be an algebraic integer over $k[a_1,a_2]$, so I need a monoic polynomial $g\in k[a_1,a_2][X]$ but that isn't always exist. What is true is that if $g\in\ker \varphi$, there exist suitable $\alpha ,\beta\in k$ such that $g(X+\alpha Z,Y+\beta Z, Z)$ is monoic in $Z$. In this example $\alpha =1$ and $\beta =0$ suit. So $a_3$ is an algebraic integer over $k[a_1-a_3,a_2]$, and only now I consider $$ \varphi ' : k[X,Y] \longrightarrow k[a_1-a_3,a_2]$$ One can prove that $\ker \varphi ' =\lbrace 0 \rbrace $, hence using Noether's normalization lemma we get that $k[a_1,a_2,a_3]$ is a finite $k[a_1-a_3,a_2]$-module and $a_1-a_3,a_2$ are algebracally independent over $k$.

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