noether normalization theorem geometric meaning

Question

Let $X\subset \mathbb{A}^n$ be an affine variety, let $I(X)=\{f\in k[X_1,\ldots,X_n]:f(P)=0,\ \forall P \in X\}$. We consider the ring $$A=k[a_1,\ldots,a_n]=\frac{k[X_1,\ldots,X_n]}{I(X)}$$ where $a_i=X_i \mod I(X)$.

Noether normalization says that there are algebraically indipendent linear forms $y_1,\ldots,y_m$ in $a_1,\ldots,a_n$ such that $A$ is a finitely generated $k[y_1,\ldots,y_m]$-module. These linear forms lift to linear forms $\tilde{y_1},\ldots,\tilde{y_m}$ in $X_1,\ldots,X_n$. Define $$\pi:=(\tilde{y_1},\ldots,\tilde{y_m}):\mathbb{A}^n\longrightarrow\mathbb{A}^m$$ and then restrict to $X$: $$\Phi:=\pi:X\longrightarrow\mathbb{A}^m$$ We want to show that $\Phi^{-1}(P)$ is finite and non-empty for every point $P\in\mathbb{A}^m$. Since $A$ is a f.g. $k[y_1,\ldots,y_m]$-module, then $A$ is integral over $k[y_1,\ldots,y_m]$, hence we have $$a_i^{N}+f^i_{N-1}(y_1,\ldots,y_m)a_i^{N-1}+\ldots+f^i_0(y_1,\ldots,y_m)=0$$ for every $i=1,\ldots,n$, or equivalently $$X_i^{N}+f^i_{N-1}(\tilde{y_1},\ldots,\tilde{y_m})X_i^{N-1}+\ldots+f^i_0(\tilde{y_1},\ldots,\tilde{y_m})=g_i(X_1,\ldots,X_n)$$ for some $g_i\in I(X)$. If $(x_1,\ldots,x_n)$ is a point of $X$, the $g_i(x_1,\ldots,x_n)=0$ thus $x_i$ is a solution of $f^i(x)=0$, where $$f^i(x)=x^{N}+f^i_{N-1}(y_1,\ldots,y_m)x^{N-1}+\ldots+f^i_0(y_1,\ldots,y_m)$$ Now, $X$ irreducible implies $I(X)$ prime and so $A$ is an integral domain. We can take field of fractions and consider $f^i(x)\in k(a_1,\ldots,a_n)[X]$. Now by fundamental theorem of algebra, we get that there are only finitely many solutions $x_i^0$ of $f^i(x)=0$ This is what i understood. Now what follows is obscure for me: for every point $y=(y_1,\ldots,y_m)\in\mathbb{A}^m$ we have only finitely many points $x=(x_1^0,\ldots,x_n^0)\in X$ such that $\Phi(x)=y$. Why this? How does the previous argument imply this conclusion?

Answer 1

Assume that $y\in\mathbb A^m$ is some point and $\Phi(x)=y$. We're clear on the part that $x_i$ is a zero of the nonzero polynomial $f^i$ under these assumptions. Let $Z_i$ be the set of zeros of $f^i$ - all of these are finite sets. But then, then there are only finitely many choices for $x$, namely the points in $Z_1\times\cdots\times Z_n$.

Edit: Think of $\pi:X\to\mathbb A^m$ in the following way: $X\subseteq\mathbb A^2$ is a plane curve, and $\pi:\mathbb A^2\to\mathbb A^1$ is the linear projection to some line, maybe the ordinate:

Back to the general case, though: The surjective map $\pi$ corresponds to an inclusion of coordinate rings $\pi^\sharp:k[\tilde y_1,\ldots,\tilde y_m]\hookrightarrow k[X_1,\ldots,X_m]$ wich really just maps $\pi^\sharp(\tilde y_i)=\tilde y_i$. It factors as a map $\Phi^\sharp:k[\tilde y_1,\ldots,\tilde y_m]\hookrightarrow A$ with $\Phi^\sharp(\tilde y_i)=y_i$, i.e. together with the canonical projection $\theta: k[X_1,\ldots,X_m]\twoheadrightarrow A$, we have $\theta\circ\Phi^\sharp = \pi^\sharp$. Now, if we have $\Phi(P)=Q=(q_1,\ldots,q_m)\in\mathbb A^m$, then this means that $y_i(P)=q_i$ and $\tilde y_i(P)=q_i$, interpreting $P$ as a point of $X$ or $\mathbb A^n$ respectively. Also, we have $P=(p_1,\ldots,p_n)\in\mathbb A^n$ and this just means $X_i(P)=p_i$. Now, the $X_i$ satisfy the algebraic relation $$X_i^N + f_{N-1}^i(\tilde y_1,\ldots,\tilde y_m)X_i^{N-1} + \cdots + f_0^i(\tilde y_1,\ldots,\tilde y_m) = g_i$$ for $g_i \in I(X)$. Now we think of $Q$ as fixed, and we evaluate this polynomial at $P$. We get $$p_i^N + f_{N-1}^i(Q)\cdot p_i^{N-1} + \cdots + f_0^i(Q) = 0$$ Since $Q$ was fixed, the expressions $f_k^i(Q)$ are just scalars, and we can interpret this as a univariate polynomial in $p_i$.