Noether Normalization Lemma
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Theorem
Let $k$ be a field.
Let $A$ be a non-trivial finitely generated $k$-algebra.
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Then there exists $n \in \N$ and a finite injective morphism of $k$-algebra:
- $k \sqbrk {x_1, \dotsc, x_n} \to A$
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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.
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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:
- $\ds P = \sum_{i \mathop = 0}^l a_i x_{e + 1}^i$
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:
- $\ds \map {f_\mu} P = \sum_{i \mathop = 0}^l \map {f_\mu} {a_i} \, \map {f_\mu} {x_{e + 1} }^i = \sum_{i \mathop = 0}^l \map {f_{\mu'} } {a_i} \paren {x_{e + 1} + X^{\mu_{e + 1} } }^i$
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:
| \(\ds \map {f_\mu} P \tuple {z_1, \dotsc, z_m, z_{m + 1} }\) | \(=\) | \(\ds \map {f_\mu} P \tuple {\map {f_\mu^{-1} } {y_1}, \dotsc, \map {f_\mu^{-1} } {y_m}, \map {f_\mu^{-1} } {y_{m + 1} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map P {y_1, \dotsc, y_m, y_{m + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
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 $\map {f_\mu} P$.
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Hence it is the wanted finite injective morphism.
$\blacksquare$
Source of Name
This entry was named for Emmy Noether.