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.


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.

  • $\begingroup$ Your "proof" that $\ker\phi'=0$ seems not to make sense. ALso, what you wrote seems not to include a proof of the scond point. $\endgroup$ – Mariano Suárez-Álvarez Apr 18 '16 at 13:11
  • $\begingroup$ I didn't provide a proof of 2. because it is automatically true since we have $\ker\varphi' =\lbrace 0\rbrace$ (using the exact same arguments than in lemma's proof). $\endgroup$ – Zanzi Apr 18 '16 at 13:24
  • $\begingroup$ Well, since you did not tell us what proof of the lemma you have in mind, it is impossible to know. $\endgroup$ – Mariano Suárez-Álvarez Apr 18 '16 at 13:25
  • $\begingroup$ If $\ker\varphi' =\lbrace 0\rbrace$ , we can show that $a_3$ is an algebraic integer over $k[a_1,a_2]$, 2. follows. $\endgroup$ – Zanzi Apr 18 '16 at 13:29
  • $\begingroup$ Are you sure about 2, the module finiteness? I did not see an argument. $\endgroup$ – Mohan Apr 18 '16 at 13:41

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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