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Geometric meaning of Noether normalization theorem


Nov 20, 2012
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
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
and then restrict to $X$:
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
for every $i=1,\ldots,n$, or equivalently
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
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?