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 $k$-algebras:

$k[x_1, \ldots, x_n] \to A$

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

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, so assume that $y_{m+1}$ is depends algebraically on $y_1, \ldots, y_m$.

So 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 \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 $\mu \in \N^n$ 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 $e$ to be $0$.

We proceed by induction on $e$:

• The case $e=0$ is easy since we then have a polynomial $f \in k[X]$ and thus $\mu = (0,\ldots, 0)$ will suffice.
• Suppose it holds for $e$ and suppose $P$ depends only on $x_1,\ldots, x_e,x_{e+1}$. Then we can write $P$ as

$\displaystyle P = \sum_{i=0}^l a_i x_{e+1}^i$

where $a_i \in k[x_1,\ldots,x_e, X ]$ and $a_l \neq 0$.

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

Note 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 $i \neq e+1$.

Now observe that:

$\displaystyle 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 that:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f_\mu(P)(z_1,\ldots, z_m, z_{m+1})\)	\(=\)	\(\displaystyle f_\mu(P)(f_\mu^{-1}(y_1),\ldots, f_\mu^{-1}(y_m), f_\mu^{-1}(y_{m+1}))\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle P(y_1,\ldots, y_m, y_{m+1})\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

By the induction hypothesis, have 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.

$\blacksquare$

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

$\alpha : k[x_1,\ldots, x_n ] \to L$

If we can prove that $n = 0$, we are done.

Suppose that $n > 0$, so $x_1 \in k[x_1,\ldots, x_n ]$ 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 $m \in \N$ and $a_0,\ldots, a_{m-1} \in k[x_1,\ldots, x_n ]$ such that

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

If we multiply this by $\alpha(x_1)^m$, we find that

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

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

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

which measn that $x_1$ is invertible.

This contradiction shows that $n=0$.

$\blacksquare$

Corollary 2

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 $B$.

This induces an injective morphism:

$\displaystyle \frac{A}{f^{-1}(\mathfrak{M})} \to \frac{B}{\mathfrak{M}}$

and thus $\displaystyle \frac{A}{f^{-1}(\mathfrak{M})}$ cannot have zero-divisors, which means that $f^{-1}(\mathfrak{M}) $ is prime.

But now $\displaystyle \frac{B}{\mathfrak{M}}$ is a field extension of $k$ and finitely generated, which implies by Corollary 1 that $\displaystyle \frac{B}{\mathfrak{M}}$ is a finite field extension.

Hence the inverse of every element in $\displaystyle \frac{B}{\mathfrak{M}}$ can be written as a linear combination of the element, and thus any sub-k-algebra must also be a field.

In particular $\displaystyle \frac{A}{f^{-1}(\mathfrak{M})} $ is a field and thus $f^{-1}(\mathfrak{M})$ is a maximal ideal.

$\blacksquare$

Source of Name

This entry was named for Emmy Noether.