Noether Normalization Lemma

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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:

$k \sqbrk {x_1, \dotsc, x_n} \to A$


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

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

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 $P \in k \sqbrk {x_1, \dotsc, x_m, X}$ such that $\map P {y_1, \dotsc, y_n, y_{n + 1} } = 0$ in $A$.


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

$f_\mu: k \sqbrk {x_1, \dotsc, x_n, X} \to k \sqbrk {x_1, \dotsc, x_n, X}: x_i \mapsto x_i + X^{\mu_i}, X \mapsto X$

This is easily seen to be an isomorphism.


Next, we want to show that there exists $\mu \in \N^n$ such that $\map {f_\mu} P$ 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 $f \in k \sqbrk X$.

Thus $\mu = \tuple {0, \dotsc, 0}$ 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 $a_i \in k \sqbrk {x_1, \dotsc, x_e, X}$ and $a_l \ne 0$.

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


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

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

Now observe that:

$\displaystyle \map {f_\mu} P = \sum_{i \mathop = 0}^l \map {f_\mu} {a_i} \, \map {f_\mu} {x_{e + 1} }^i = \sum_{i \mathop = 0}^l \map {f_{\mu'} } {a_i} \paren {x_{e + 1} + X^{\mu_{e + 1} } }^i$

We have that $\map {f_{\mu'} } {a_l}$ has a leading coefficient in $k$.

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


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

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

On the other hand, we find that:

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

By the induction hypothesis, we have the existence of:

$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to A'$

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 $\map {f_\mu} P$.



Hence it is the wanted finite injective morphism.

$\blacksquare$


Source of Name

This entry was named for Emmy Noether.


Also see