Noether Normalization Lemma

Theorem
Let $k$ be a field.

Let $A$ be a non-trivial finitely generated $k$-algebra.

Then there exists $n \in \N$ and a finite injective morphism of $k$-algebra:


 * $k \left[{x_1, \dotsc, x_n}\right] \to A$

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

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 elements $y_1, \dotsc, y_m, y_{m+1}$.

If $y_1, \dotsc, y_m, y_{m+1}$ are algebraically independent, then $A$ is isomorphic to $k \left[{x_1, \dotsc, x_m, x_{m+1} }\right]$.

In this case the theorem is obvious, so assume that $y_{m+1}$ depends algebraically on $y_1, \ldots, y_m$.

So there exists a polynomial $P \in k \left[{x_1, \dotsc, x_m, X}\right]$ such that $P \left({y_1, \dotsc, y_n, y_{n+1} }\right) = 0$ in $A$.

Let $\mu \in \N^n$ and define:


 * $f_\mu: k \left[{x_1, \dotsc, x_n, X}\right] \to k \left[{x_1, \dotsc, x_n, X}\right]: x_i \mapsto x_i + X^{\mu_i}, X \mapsto X$

This is easily seen to be an isomorphism.

Next, we want to show that there exists $\mu \in \N^n$ such that $f_\mu \left({P}\right)$ 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 $e$ to be $0$.

We proceed by induction on $e$.

The case $e = 0$ is easy since we then have a polynomial $f \in k \left[{X}\right]$.

Thus $\mu = \left({0, \dotsc, 0}\right)$ will suffice.

Suppose it holds for $e$ and suppose $P$ depends only on $x_1, \dotsc, x_e, x_{e+1}$.

Then we can write $P$ as:
 * $\displaystyle P = \sum_{i \mathop = 0}^l a_i x_{e + 1}^i$

where $a_i \in k \left[{x_1, \dotsc, x_e, X}\right]$ and $a_l \ne 0$.

Hence by the induction hypothesis, there exists a $\mu' \in \N^n $ such that $f_{\mu'} \left({a_l}\right) $ has a leading coefficient in $k$.

Note that we can choose that $\mu'_l$ for $l > e$ since $f_\mu \left({a_i}\right)$ is independent of these components.

Hence we define $\mu$ as $\mu_i = \mu'_i$ for every $i \ne e+1$.

Now observe that:


 * $\displaystyle f_\mu \left({P}\right) = \sum_{i \mathop = 0}^l f_\mu \left({a_i}\right) f_\mu \left({x_{e+1} }\right)^i =  \sum_{i \mathop = 0}^l f_{\mu'} \left({a_i}\right) \left({x_{e+1} + X^{\mu_{e+1} } }\right)^i$

We have that $f_{\mu'} \left({a_l}\right)$ has a leading coefficient in $k$.

Thus by choosing $\mu_{e+1}$ big enough, $f_\mu \left({P}\right)$ has a leading coefficient in $k$.

Consider now thus the elements $z_i = f_\mu^{-1} \left({y_i}\right) \in A$.

These elements still generate $A$ since $f_\mu$ is an isomorphism.

On the other hand, we find that:

By the induction hypothesis, we have the existence of:


 * $\alpha : k \left[{x_1, \dotsc, x_n}\right] \to A'$

where $A'$ is generated by $z_1, \dotsc, z_m$.

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 $f_\mu \left({P}\right)$.

Hence it is the wanted finite injective morphism.

Corollary 1
Let $k$ be a field and let $L / k$ be an 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:


 * $\alpha: k \left[{x_1, \ldots, x_n}\right] \to L$

If we can prove that $n = 0$, the proof is complete.

Let $n > 0$.

Then:
 * $x_1 \in k \left[{x_1, \dotsc, x_n}\right]$

and:
 * $\alpha \left({x_1}\right) \ne 0$

We have that $\alpha \left({x_1}\right)^{-1}$ is integral over $k \left[{x_1, \dotsc, x_n}\right]$.

Thus there exists a $m \in \N$ and $a_0, \dotsc, a_{m-1} \in k \left[{x_1, \dotsc, x_n}\right]$ such that:


 * $\displaystyle \alpha \left({x_1}\right)^{-m} + \sum_{i \mathop = 0}^{m-1} \alpha \left({a_i}\right) \alpha \left({x_1}\right)^{-i} = 0$

If we multiply this by $\alpha \left({x_1}\right)^m$, we find that:

and thus, since $\alpha$ is injective, we find that:


 * $\displaystyle 1 = x_1 \left({- \sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }\right)$

which means that $x_1$ is invertible.

This contradiction shows that $n = 0$.

Corollary 2
Let $k$ be a field.

Let $f: A \to B$ be a morphism of finitely generated $k$-algebras.

Then $f^{-1} \left({\mathfrak M}\right)$ is a maximal ideal in $A$ for every maximal ideal $\mathfrak M$ in $B$.

Proof
Let $\mathfrak M$ be a maximal ideal of $B$.

This induces an injective morphism:


 * $\dfrac A {f^{-1} \left({\mathfrak M}\right)} \to \dfrac B {\mathfrak M}$

and thus $\dfrac A {f^{-1} \left({\mathfrak M}\right)}$ cannot have zero-divisors.

Thus $f^{-1} \left({\mathfrak M}\right)$ is prime.

We have that $\dfrac B {\mathfrak M}$ is a field extension of $k$ which is finitely generated

Thus, by Corollary 1, $\dfrac B {\mathfrak M}$ is a finite field extension.

Hence the inverse of every element in $\dfrac B {\mathfrak M}$ can be written as a linear combination of the element.

Thus any sub-$k$-algebra must also be a field.

In particular $\dfrac A {f^{-1} \left({\mathfrak M}\right)}$ is a field.

Thus $f^{-1} \left({\mathfrak M}\right)$ is a maximal ideal.