6
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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 $\pi(\alpha)=P$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.