The Noether normalization lemma states that if $k$ is a field, and $A$ a finitely generated $k$-algebra, then there exist elements $y_1,...,y_m\in A$ such that

  1. $y_1,...,y_m$ are algebraically independent over $k$
  2. $A$ is finite over $B=k[y_1,...,y_m]$.

The following problem is Exercise 3.16 from Undergraduate Algebraic Geometry by Reid:

Let $I=\ker \{k[x_1,\dots,x_n]\rightarrow k[a_1,\dots,a_n]=A\}$, and consider $V=V(I)$ in $k^n$.

Let $Y_1,\dots,Y_m$ be general linear forms in $X_1,\dots,X_m$, and write $\pi: k^n\rightarrow k^m$ for the linear projection defined by the $Y$'s. Set $p = \pi |V : V → k^m$. Prove that for every $P \in k^m$, $p^{−1}(P)$ is a finite set, and nonempty if $k$ is algebraically closed.

I showed the first part by using that for each $X_i$, there is a monic equation in $I$ in terms of $X_i$. So the solution set has to be finite. I am stuck on showing the nonemptiness.

I am confused about the concept finite extension. I saw an example of $A=\mathbb C[x_1,x_2]/(x_1x_2-1)$. Apparently the assumption is $x_1$ is transcendental over $\mathbb C$. In this case, we see that $A$ is not a finite extension of $\mathbb C[x_1]$. However, if we make a change of variables, let $x_1=y_1+y_2, x_2=y_1-y_2$, then $A$ becomes $\mathbb C[y_1,y_2]/(y_1^2-y_2^2-1)$, in which case $A$ is finite over $\mathbb C[y_1]$. So in this case, $m=1$.

Applying the result of this question to the example, does it mean for any $y_1$, there exists value of $y_2$ in $V(y_1^2-y_2^2-1)$? And how to show the general case?

Sorry if this is a stupid question.

If someone can explain intuitively what Noether normalization implies, that will be great.

Any help would be greatly appreciated.


About nonemptyness: following the hint given by the book we start with a point $P=(b_1,\dots,b_m)\in K^m$, and consider the ideal $J_P=I+(Y_1-b_1,\dots,Y_m-b_m)$, where $Y_i$ are linear forms in $K[X_1,\dots,X_n]$ such that $y_i=Y_i(a_1,\dots,a_n)$ are algebraically independent over $K$ and $K[y_1,\dots,y_m]\subset A$ is finite.
We want to show that $J_P\ne(1)$. Suppose the contrary, and write $$1=f+(Y_1-b_1)g_1+\cdots+(Y_m-b_m)g_m$$ in $K[X_1,\dots,X_n]$. (Notice that $f\in I$, that is, $f(a_1,\dots,a_n)=0$.) It follows that $1=(y_1-b_1)g_1(a)+\cdots+(y_m-b_m)g_m(a)$, where $a= (a_1,\dots,a_n)$. In particular, $(y_1-b_1,\dots,y_m-b_m)=(1)$ in $A$. But this is not possible since $(y_1-b_1,\dots,y_m-b_m)$ is a maximal ideal in $K[y_1,\dots,y_m]$, and the extension $K[y_1,\dots,y_m]\subset A$ is integral. (See exercise 3.15 from the book.)
Then $J_P\ne(1)$, and whenever $K$ is algebraically closed we have $V(J_P)\ne\emptyset$. Then there is $\alpha\in K^n$ such that $g(\alpha)=0$ for all $g\in J_P$. In particular, $f(\alpha)=0$ for all $f\in I$ hence $\alpha\in V(I)$, and $Y_i(\alpha)=b_i$ for all $i$ hence $\pi(\alpha)=P$.

  • $\begingroup$ Thank you so much for the answer! Just one question, what are the relations of $y_i$ and $Y_i$? Why did you say that "$y_i=Y_i(a_1,…,a_n)$ are algebraically independent over $K$"? I basically didn't understand what $Y_i$ stands for in terms of $y_i$, hence the question about the example $\mathbb{C}[x_1,x_2]/(x_1x_2-1)$. $\endgroup$ – KittyL Feb 1 '16 at 1:06
  • $\begingroup$ @KittyL Noether normalization provides some linear forms $y_i$ in $a_1,\dots,a_n$, say $y_i=\sum_{j=1}^na_{ij}a_j$ with $a_{ij}\in K$. Then $Y_i:=\sum_{j=1}^na_{ij}X_j$, and this is why $y_i=Y_i(a_1,\dots,a_n)$. (In your example $a_i=x_i\bmod(x_1x_2-1)$, and $y_i=\frac{a_1\pm a_2}{2}$.) $\endgroup$ – user26857 Feb 1 '16 at 7:40
  • $\begingroup$ ...so $Y_1=\frac{X_1+ X_2}{2}$, and $Y_2=\frac{X_1-X_2}{2}$. $\endgroup$ – user26857 Feb 1 '16 at 7:46
  • $\begingroup$ Oh, I see! Thank you! $\endgroup$ – KittyL Feb 1 '16 at 9:50

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