Noether Normalization Lemma

Theorem
Let $$k$$ be a field, and $$A$$ a finitely generated $k$-algebra. Assume that $$A$$ is not the trivial ring. There exist an element $$n \in \N$$ and a finite injective morphism of $$k$$-algebras
 * $$k[x_1, \ldots, x_n] \to A$$.

Proof
Since $$A$$ is finitely generated, we prove this by induction on the number of generators as a $$k$$-algebra. We denote this number by $$m$$.

Basis for the induction
If $$m=0$$, then $$A = k$$ and thus there is nothing to prove.

Induction step
Let $$A$$ be a $$k$$-algebra generated by the element $$y_1,\ldots, y_m, y_{m+1}$$. If $$y_1,\ldots, y_m, y_{m+1}$$ are algebraically independent, then $$A$$ is isomorphic to $$k[x_1, \ldots, x_m,x_{m+1}]$$. In this case the theorem is obvious. Otherwise we can assume that $$y_{m+1}$$ is algebraically dependent on $$y_1, \ldots, y_m$$.

Hence we have found that there exists a polynomial $$P \in k[x_1,\ldots, x_m, X] $$ such that $$P(y_1,\ldots, y_n, y_{n+1}) = 0$$ in $$A$$. Let $$ \mu \in \mathbb{N}^n$$ and define
 * $$f_\mu : k[x_1,\ldots, x_n, X] \to k[x_1,\ldots, x_n, X] : x_i \mapsto x_i + X^{\mu_i}, X \mapsto X$$.

This is easily seen to be an isomorphism. Next, we want to show that there exists $$\mu \in \mathbb{N}$$ such that $$f_\mu(P)$$ is a polynomial in $$X$$ with leading coefficient in $$k$$. Let $$e$$ be the biggest natural number such that $$P$$ involves $$x_e$$ non-trivially. If there is no such number, define it as $$0$$. We prove this by induction on $$e$$:
 * The case $$e=0$$ is easy since we then have a polynomial $$f \in k[X]$$ and thus $$\mu = (0,\ldots, 0)$$ will suffice.
 * Suppose it holds for $$e$$ and suppose $$P$$ depends only on $$x_1,\ldots, x_e,x_{e+1}$$. Hence we can write $$P$$ as
 * $$P = \sum_{i=0}^l a_i x_{e+1}^i$$

where $$a_i \in k[x_1,\ldots,x_e, X ]$$ and $$a_l \neq 0$$. Hence there exists a $$\mu' \in \mathbb{N}^n $$ such that $$f_{\mu'}(a_l) $$ has a leading coefficient in $$k$$. Remark that we can choose that $$\mu'_l $$ for $$l > e$$ since $$f_\mu(a_i)$$ is independent of these components. Hence we define $$\mu $$ as $$\mu_i = \mu'_i$$ for every $$i \neq e+1$$. Remark now that
 * $$ f_\mu(P) = \sum_{i=0}^l f_\mu(a_i) f_\mu(x_{e+1})^i  =  \sum_{i=0}^l f_{\mu'}(a_i)  (x_{e+1} + X^{\mu_{e+1}})^i  $$

and thus since $$f_{\mu'}(a_l)$$ has a leading coefficient in $$k$$, we find that by choosing $$\mu_{e+1}$$ big enough, $$f_\mu(P)$$ has a leading coefficient in $$k$$.

Consider now thus the elements $$z_i = f_\mu^{-1}(y_i) \in A$$. These elements still generate $$A$$ since $$f_\mu$$ is an isomorphism. On the other hand, we find
 * $$f_\mu(P)(z_1,\ldots, z_m, z_{m+1}) =  f_\mu(P)(f_\mu^{-1}(y_1),\ldots, f_\mu^{-1}(y_m),  f_\mu^{-1}(y_{m+1})) = P(y_1,\ldots, y_m, y_{m+1}) = 0 $$

By the induction hypothesis, we find the existence of
 * $$\alpha : k[x_1,\ldots, x_n] \to A'$$

where $$A'$$ is generated by $$z_1,\ldots, z_m$$. By extending this $$alpha$$ to $$A$$ by the natural inclusion, we find that $$\alpha$$ is finite, since $$z_i$$ is integral by construction and $$z_{m+1}$$ is integral over the others since it satisfies $$f_\mu(P)$$. Hence it is the wanted finite injective morphism.

Statement
Let $$k$$ be a field and let $$L$$ be a field extension of $$k$$. If $$L$$ is finitely generated as a $$k$$-algebra, then $$L$$ is a finite extension

Proof
By Noether normalization, we find a finite and injective morphism $$\alpha : k[x_1,\ldots, x_n ] \to L$$. If we prove that $$n = 0$$, we have found the wanted. Suppose now $$n > 0$$. Hence $$x_1 \in k[x_1,\ldots, x_n ]$$ and $$\alpha(x_1) \neq 0$$. We have that $$\alpha(x_1)^{-1}$$ is integral over $$k[x_1,\ldots, x_n ]$$, which means that there exists a $$m \in \mathbb{N}$$ and $$a_0,\ldots, a_{m-1} \in k[x_1,\ldots, x_n ]$$ such that
 * $$\alpha(x_1)^{-m} + \sum_{i=0}^{m-1} \alpha(a_i)\alpha(x_1)^{-i} = 0$$.

If we multiply this by $$\alpha(x_1)^m$$, we find that
 * $$ 0 = 1 + \sum_{i=0}^{m-1} \alpha(a_i)\alpha(x_1)^{m-i} = \alpha\left( 1 + x_1 \left( \sum_{i=0}^{m-1} a_i x_1^{m-i-1} \right)\right)$$

and thus, since $$\alpha$$ is injective, we find that $$1 =  x_1 \left( - \sum_{i=0}^{m-1} a_i x_1^{m-i-1}\right)$$, which means that $$x_1$$ would be invertible. This cannot be and thus $$n=0$$.

Statement
Let $$k$$ be a field. If $$f: A \to B$$ is a morphism of finitely generated $$k$$-algebras, then $$f^{-1}(\mathfrak{M}) $$ is a maximal ideal in $$A$$ for every maximal ideal $$\mathfrak{M}$$ in $$B$$.

Proof
Let $$\mathfrak{M}$$ be a maximal ideal of $$B$$. Hence this induces an injective morphisms
 * $$\frac{A}{f^{-1}(\mathfrak{M})} \to \frac{B}{\mathfrak{M}}$$

and thus $$\frac{A}{f^{-1}(\mathfrak{M})}$$ cannot have zero-divisors, which means that $$f^{-1}(\mathfrak{M}) $$ is prime. But now $$\frac{B}{\mathfrak{M}}$$ is a field extension of $$k$$ and finitely generated, which implies by Corollary 1 that $$\frac{B}{\mathfrak{M}}$$ is a finite field extension. Hence the inverse of every element in $$\frac{B}{\mathfrak{M}}$$ can be written as a linear combination of the element, and thus any sub-k-algebra must also be a field. In particular $$\frac{A}{f^{-1}(\mathfrak{M})} $$ is a field and thus $$f^{-1}(\mathfrak{M})$$ is a maximal ideal.