----
> [!theorem] Theorem. ([[Noether's normalization theorem]])
> Let $k$ be a [[field]], and [[every nonzero ring homomorphism out of a field is injective|let]] $k \hookrightarrow A \neq 0$ be a [[subalgebra generated by a subset|finite-type]] $k$-[[algebra]].
>
Then there are $k$-[[algebraically independent]] $x_{1},\dots,x_{n} \in A$, $n \geq 0$, such that $A$ is [[finite algebra|finite]] over the [[subalgebra]] $A':=k[x_{1},\dots,x_{n}]$ [[subalgebra generated by a subset|generated by]] $x_{1},\dots,x_{n}$.
>
"[[subalgebra generated by a subset|Finitely generated algebras]] over a [[field]] are always [[finite algebra|finite]] [[module|modules]] over some [[polynomial 4|polynomial]] [[subalgebra]]."
^theorem
> [!basicexample] Example of the proof method.
> [[Laurent polynomial|Put]] $A=k[T, T^{-1}]=\frac{k[X , Y]}{XY-1}$. Then $k[T] \subset k[T, T ^{-1}]$ is *not* a [[finite algebra|finite extension]]. Indeed, $T^{-1}$ is not [[integral element of an algebra|integral over]] $k[T]$: if it were, then[^2]
$(T^{-1})^{n} \in \span_{k[T]}\{ (T ^{-1})^{0}, \dots, (T ^{-1})^{n-1} \} \text{ for some } n\geq1.$
But multiplying through by $T^{n}$ yields $1 \in \span_{k[T]}(T^{n},T^{n-1},\dots,T^{1})$, a contradiction.
>
However, choose a scalar $c \in k$. **Claim:** $k[T ^{-1}- cT] \subset k[T, T ^{-1} - cT]$ *is* finite, "for most $c
quot;.
>
To see this, first note that $\textcolor{Skyblue}{\{ T,T ^{-1}-cT \}}$ generates $A=k[T, T ^{-1}]$ as a $k$-[[algebra]][^3]. We want to show that $k[T ^{-1} - cT] \subset k[T, T ^{-1}]=\textcolor{Skyblue}{k[T, T ^{-1} - cT]}$ is a finite extension. By [[integral algebra|the equivalence here]] ($2 \implies 3$), it suffices to show $T$ and $T ^{-1} - cT$ are [[integral element of an algebra|integral]] over $k[T ^{-1}-cT]$. Well, $T^{-1}-cT$ definitely is. As for $T$... we have the obvious relation $T ^{-1} T-1=0$. Then 'doing nothing', we have $\big((T ^{-1} - cT)+cT\big)T-1=0$. Now expand: $cT^{2}+(T ^{-1} - cT) - 1 = 0.$
If $c \neq 0$, then we can divide by it to obtain an integrality equation witnessing that $T$ is integral over $k[T^{-1} - cT]$.
^basic-example
> [!proof]- Proof. ([[Noether's normalization theorem]])
> We'll assume $k$ is infinite; see Mumford's Red Book for Nagata's proof that works over every field.
>
> Our proof will be based on the example above, together with an induction on the minimal number of generators of $A$ as a $k$-[[algebra]].
>
> *Base case (0 generators).* Here $A=k$. Set $n=0$, $A'=A$.
>
> *Induction step.* Assume that $\{ x_{1},\dots,x_{m} \} \subset A$ generate $A$ as a $k$-algebra, and that the theorem holds when $A$ is generated as a $k$-algebra by a set with less than $m$ elements.
>
> If $x_{1},\dots,x_{m}$ are $k$-[[algebraically independent]] already, we are done (take $A=A'$). So assume they are not.
>
> *CLAIM: In this case, there exist $c_{1},\dots,c_{m-1} \in k$ such that $x_{m}$ is integral over $B=k[x_{1}-c_{1}x_{m},\dots,x_{m-1}-c_{m-1}x_{m}]$.* Once we show this we are done: by induction hypothesis, there exists $k$-algebraically independent $z_{1},\dots,z_{n} \in B$ such that $B$ is finite over $A'=k[z_{1},\dots,z_{n}]$; the extension $A'\subset B \subset B[x_{m}]=A$ is then finite[^1] by [[transitivity of finiteness and integrality and finite-typedness for algebras|transitivity of finiteness]] and we are done.
>
> *Proof of claim.*
> Using that $x_{1},\dots,x_{m}$ are not [[algebraically independent]], obtain a nonzero [[polynomial 4|polynomial]] $f \in k[T_{1},\dots,T_{m}]$ of degree $r$ such that $f(x_{1},\dots,x_{m})=0$. Write $f$ as a sum of its [[homogeneous polynomial|homogeneous]] parts; let $F$ be the part of highest degree $r=deg f$. (e.g. $f=\underbrace{ (T_{1}^{2}T_{2} + 3T_{2}^{3}) }_{ F }+ (2T_{1}T_{3}+T_{2}T_{3}) +(T_{2}+T_{3})+1$) .
>
> For $c_{1},\dots,c_{m-1} \in k$ (to be chosen later), define the "triangular automorphism" $\sigma_{c}: T_{i} \mapsto T_{i}+ c_{i}T_{m} \ (1 \leq i \leq m-1), T_{m} \mapsto T_{m}$
> of the polynomial ring. Define $g:=\sigma_{c}(f)=f(T_{1}+c_{1}T_{m},\dots, T_{m-1}+c_{m-1} T_{m}, T_{m}),$
> an element of $k[T_{1}+c_{1}T_{m}, \dots, T_{m-1}+c_{m-1}T_{m}][T_{m}]$.
> It follows from the fact that $F$ is homogeneous of degree $r$ that
>
> $g(T_{1},\dots,T_{m})=\overbrace{ F(c_{1},\dots,c_{m-1}, 1)T_{m}^{r} }^{ \in k } + \{ \text{terms of degree }<r \text{ in }T_{m} \}.$
>
>
>
[^1]: $B \subset B[x_{m}]$ is finite extension by $(2)\implies (3)$ [[integral algebra|here]]. $B[x_{m}]=A$ because once we have $x_{m}$ as a generator we can cancel it with $-c_{1}x_{m},\dots,-c_{m-1}x_{m}$ to leave us with $k[x_{1},\dots,x_{m}]=A$.
----
####
[^2]: Write out the [[integral element of an algebra|integrality equation]]: $(T ^{-1})^{n} + a_{1} (T^{-1})^{n-1} + \dots + a_{n}(T^{-1})^{0}=0$
for some $a_{1},\dots,a_{n} \in k[T]$. Rearranging demonstrates $(T ^{-1})^{n}$ to be a $k[T]$-[[linear combination]] of $(T ^{-1})^{n-1},\dots,(T ^{-1})^{0}$, hence the containment claimed.
[^3]: Indeed, $T^{-1}=cT + (T ^{-1} - cT)$.
#### wrong
Unpacking a bit, the [[subalgebra generated by a subset|finite-type]] premise gives a finite subset $\{ a_{1},\dots,a_{n} \} \subset A$ such that every $a \in A$ can be written in the form $a= \sum_{j} r_{j} a_{1}^{\alpha_{1},j} \cdots a_{n}^{\alpha_{n},j}\text{ for some } r_{j} \in R, \alpha_{k,j} \in \mathbb{N} .$
NNT says there exist $x_{1},\dots,x_{m} \in A$ such that $a$ can be written as a $k[x_{1},\dots,x_{n}]$-[[linear combination]] of $a_{1},\dots,a_{m}$: $a=p_{a_{1}}(x_{1},\dots,x_{n}) a_{1}+\dots p_{a_{m}}(x_{1},\dots,x_{n})a_{m}$
for polynomials $p_{a_{1}},\dots,p_{a_{m}} \in k[T_{1},\dots,T_{n}]$.
-----
#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```