Noether normalization in algebraically closed field 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

*

*$y_1,...,y_m$ are algebraically independent over $k$

*$A$ 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.
 A: 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 $\pi(\alpha)=P$.
