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?

  • $\begingroup$ $k[x,y,z]/(yz, z(z-1))$ which is the same as $k[x,y]\times k[z]$ seems to be wrong: the unit group of the first ring is $k^\times$. $\endgroup$ – user26857 Sep 4 at 8:18
  • $\begingroup$ @user26857 I see. The second unit group is $k^* \times k^*$ $\endgroup$ – Lada Dudnikova Sep 4 at 8:30
  • $\begingroup$ @user26857 am I wrong in writing ideals or stating that "The ring of functions of a disjoint set is the direct sum of two function rings"? $\endgroup$ – Lada Dudnikova Sep 4 at 8:32
  • $\begingroup$ @user26857 is it ok to write (1+z)(1-z+z^2-z^3...)=1. It would be the second unit ring generator you want. $\endgroup$ – Lada Dudnikova Sep 5 at 5:26

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.

  • $\begingroup$ Do you mean that the branched cover mentioned in the question doesn't need to have the same amount of layers everywhere? Is it true that the amount of elements in Noeter normalization is equal to $dim(Spec A)$? $\endgroup$ – Lada Dudnikova Sep 4 at 5:38
  • $\begingroup$ @Lada Dudnikova yes and yes. $\endgroup$ – jgon Sep 4 at 14:05
  • $\begingroup$ I have edited the question and there may be a mistake, but looks like it's not a plane and a parallel line that have the needed function ring. $\endgroup$ – Lada Dudnikova Sep 4 at 14:30
  • $\begingroup$ @LadaDudnikova your edit confuses me. The original ring is not isomorphic to the quotient ring you've considered. $z^2-z$ isn't zero for one thing. Also $z$ isn't integral over $k[x,y]$. $\endgroup$ – jgon Sep 4 at 15:29
  • $\begingroup$ Do you know a way to construct this finitely generated algebra $k[x,y]\times k[z]$ as a factor of polynomial ring over an ideal? $\endgroup$ – Lada Dudnikova Sep 4 at 17:11

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